Mahesh

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Functions Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1A Functions Important Questions

I.
Question 1.
If f: R – {0} → R is defined by f(x) = x + \(\frac{1}{x}\), then prove that (f(x))² = f(x²) + f(1).
Solution:
f : R – {0} → R and
f(x) = x + \(\frac{1}{x}\)
Now f(x²) + f(1) = (x² + \(\frac{1}{x^{2}}\)) + (1 + \(\frac{1}{1}\))
= x² + 2 + \(\frac{1}{x^{2}}\) = (x + \(\frac{1}{x}\))² = (f(x))²
∴ (f(x))² = f(x²) + f(1)

Question 2.
If the function f is defined by

then find the values, if exist, of f(4), f(2.5), f(-2), f(-4), f(0), f(-7).  (Mar. ’14)
Solution:
Domain of f is(-∞, -3) ∪ [-2, 2] ∪ (3, ∞)
i) Since f(x) = 3x – 2 for x > 3
f(4) = 3(4) – 2 = 10

ii) 2.5 does not belong to domain off, hence
f(2.5) is not desired.

iii) ∵ f(x) = x² – 2 for -2 ≤ x ≤ 2
f(-2) = (-2)² – 2 = 4 – 2 = 2

iv) ∵ f(x) = 2x + 1 for x < -3
f(-4) = 2(-4) + 1 = – 8 + 1 = -7

v) ∵ f(x) = x² – 2, for -2 ≤ x ≤ 2
f(0)= (0)² – 2 = 0 – 2 = -2

vi) ∵ f(x) = 2x + 1 for x < -3
f(-7) = 2(-7) + 1 = -14 + 1 = -13

Question 3.
If A = {0, \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\), \(\frac{\pi}{2}\)} and f : A → B is a surjection defined by f(x) = cos x, then find B.  (A.P. Mar. ’16, ’11; May ’11)
Solution:
∵ f : A → B is a sujection defined by
f(x) = cos x
B = Range of f = f(A)
= {f(0), f(\(\frac{\pi}{6}\)), f(\(\frac{\pi}{4}\)), f(\(\frac{\pi}{3}\)), f(\(\frac{\pi}{2}\))}
= {cos 0, cos \(\frac{\pi}{6}\), cos \(\frac{\pi}{4}\), cos \(\frac{\pi}{3}\), cos \(\frac{\pi}{2}\)}
= {1, \(\frac{\sqrt{3}}{2}\), \(\frac{1}{\sqrt{2}}\), \(\frac{1}{2}\), 0}

Question 4.
Determine whether the function f: R → R defined by f(x) = \(\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}\) is an injection or a surjection or a bijection.
Solution:
f(x) = \(\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}\)
f(0) = \(\frac{e^{0}-e^{-0}}{e^{0}+e^{-0}}\) = \(\frac{1-1}{1+1}\) = 0
f(-1) = \(\frac{e^{1}-e^{1}}{e^{-1}+e^{1}}\) = 0
∵ f(0) = f(-1)
⇒ f is not an injection.
Let y = f(x) = \(\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}\)
When y = 1, there is no x ∈ R such that f(x) = 1
⇒ f is not a surjection
if there is such x ∈ R then
\(\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}\) = 1
⇒ e^|x| – e^-x = e^x + e^-x, clearly x ≠ 0
for x > 0, this equation gives
e^x – e^-x = e^x + e^-x ⇒ -e^-x = e^-x which is not possible
for x < 0, this equation gives
e^-x – e^-x = e^x + e^-x
⇒ e^-x = e^x which is also not possible.

Question 5.
Determine whether the function f: R → R defined by

is an injection or a surjection or a bijection
Solution:
∵ f(x’) = x for x > 2
⇒ f(3) = 3
∵ f(x) = 5x – 2, for x ≤ 2
⇒ f(1) = 5(1) – 2 = 3
∵ 1 and 3 have same f-image.
Hence f is not an injection.
Let y ∈ R then y > 2 (or) y ≤ 2.
If y > 2 take x = y ∈ R so that
f(x) = x = y
If y ≤ 2 take x = \(\frac{y+2}{5}\) ∈ R and
x = \(\frac{y+2}{5}\) < 1
∴ f(x) = 5x – 2 = 5\(\left[\frac{y+2}{5}\right]\) – 2 = y
∴ f is a surjection
∵ f is not an injection ⇒ It is not a bijection.

Question 6.
Find the domain of definition of the function y(x), given by the equation 2^x + 2^y = 2.
Solution:
2^x = 2 – 2^y < 2 (∵ 2^y > 0)
⇒ log₂ 2^x < log₂2
⇒ x < 1
∴ Domain = (-∞, 1)

Question 7.
It f: R → R is defined as f(x + y) = f(x)+ f(y) ∀ x, y ∈R and f(1) = 7, then find \(\sum_{r=1}^{n} f(r)\).
Solution:
Consider f(2) = f(1 + 1) = f (1) + f (1) = 2f(1)
f(3) =f(2 + 1) = f(2) + f(1) = 3f(1)
Similarly f(r) = rf(1)
∴ \(\sum_{r=1}^{n} f(r)\) = f(1) + f(2) + ….. + f(n)
= f(1) + 2f(1)+ ….. + nf(1)
= f(1) (1 + 2 + ….. + n)
= \(\frac{7 n(n+1)}{2}\)

Question 8.
If f(x) = \(\frac{\cos ^{2} x+\sin ^{4} x}{\sin ^{2} x+\cos ^{4} x}\) ∀ x ∈ R then show that f(2012) = 1.)
Solution:
f(x) = \(\frac{\cos ^{2} x+\sin ^{4} x}{\sin ^{2} x+\cos ^{4} x}\)
= \(\frac{1-\sin ^{2} x+\sin ^{4} x}{1-\cos ^{2} x+\sin ^{4} x}\)

Question 9.
If f : R → R, g : R → R are defined by f(x) = 4x – 1 and g(x) = x² + 2 then find
i) (gof)(x)
ii) (gof)\(\left(\frac{a+1}{4}\right)\)
iii) (fof)(x)
iv) go(fof)(0)  (Mar. ’05)
Solution:
Given f : R → R and g : R → R and
f(x) = 4x – 1 and g(x) = x² + 2

(i) (gof) (x) = g (f(x))
= g (4x – 1), ∵ f(x) = 4x – 1
= (4x – 1)² + 2, ∵ g(x) = x² + 2
= 16x² – 8x + 1 + 2
= 16x² – 8x + 3

(ii) (gof)\(\left(\frac{a+1}{4}\right)\) = \(g\left(f\left(\frac{a+1}{4}\right)\right)\)
= \(g\left(4\left(\frac{a+1}{4}\right)-1\right)\)
= g(a)
= a² + 2

(iii) (fof)(x) = f(f(x))
= f(4x – 1), ∵ f(x) = 4x – 1
= 4(4x – 1) – 1
= 16x – 4 – 1 = 16x – 5

(iv) (fof)(0) = f(f(0))
= f(4 × 0 – 1)
= f(-1)
= 4(-1) – 1 = -5
Now go(fof)(0)
= g(-5) = (-5)² + 2 = 27

Question 10.
If f: [0, 3] → [0, 3] is defined by

then show that f[0,3] ⊆ [0, 3] and find fof.
Solution:
0 ≤ x ≤ 2 ⇒ 1 ≤ 1 + x ≤ 3 ——— (1)
2 < x ≤ 3 ⇒ -3 ≤ -x ≤ -2
⇒ 3 – 3 ≤ 3 – x ≤ 3 – 2
⇒ 0 ≤ 3 – x < 1 ——- (2)
from (1) and (2).
f[0, 3] ⊆ [0, 3]
When 0 ≤ x ≤ 1, we have
(fof) (x) = f(f(x))
= f(1 + x) = 1 + (1 + x) = 2 + x
[∵ 1 ≤ 1 + x ≤ 2]
When 1 < x ≤ 2, we have
(fof) (x) = f(f(x))
= f(1 + x)
= 3 – (1 + x)
= 2 – x, [∵ 2 < 1 + x ≤ 3]
When 2 < x ≤ 3, we have
(fof) (x) = f(f(x))
= f(3 – x)
= 1 + (3 – x)
= 4 – x, [∵ 0 ≤ 3 – x < 1]

Question 11.
If f, g : R → R are defined

and
then find (fog)(π) + (gof)(e).
Solution:
(fog)(π) = f(g(π)) = f(0) = 0
(gof)(e) = g(f(e)) = g (1) = -1
∴ (fog)(π) + (gof) (e) = -1.

Question 12.
Let A = {1, 2, 3), B = {a, b, c}, C = (p, q, r}
If f: A → B, g: B → C are defined by
f = {(1, a), (2, c), (3, b)},
g = {(a, q), (b, r), (c, p)} then
show that f^-1og^-1 = (gof)^-1
Solution:
Given that
f = {(1, a), (2, c), (3, b)} and
g = {(a, q), (b, r), (c, p)}
then g f = {(1, q), (2, p), (3, r)}
⇒ (gof)^-1 = {(q, 1), (p, 2), (r, 3)}
⇒ g^-1 = {(q, a), (r, b), (p, c)} and
f^-1 = {(a, 1), (c, 2), (b, 3)}
f^-1 g^-1 = {(q, 1), (r, 3), (p, 2)}
⇒ (gof)^-1 = f^-1og^-1

Question 13.
If f: Q → Q, is defined by f(x) = 5x + 4 for all x ∈ Q, show that f is a bijection and find f^-1.(A.P Mar. ’16, May ’12, ’05)
Solution:
Let x₁, x₂ ∈ Q
Now f(x₁) = f(x₂)
⇒ 5x₁ + 4 = 5x₂ + 4
⇒ 5x₁ = 5x₂
⇒ x₁ = x₂
∴ f is an injection.
Let y ∈ Q then x = \(\frac{y-4}{5}\) ∈ Q and
f(x) = f\(\left(\frac{y-4}{5}\right)\) = 5\(\left(\frac{y-4}{5}\right)\) + 4 = y
∴ f is a surjection.
Hence f is a bijection
∴ f^-1 : Q → Q is also bijection
We have (fof^-1)(x) = I(x)
⇒ f(f^-1(x)) = x, ∵ f(x) = 5x + 4
⇒ 5f^-1(x) + 4 = x
⇒ f^-1(x) =\(\frac{x-4}{5}\) for ∀ x ∈ Q

Question 14.
Find the domains of the following real valued functions.

i) f(x) = \(\frac{1}{6 x-x^{2}-5}\)
Solution:
f(x) = \(\frac{1}{6 x-x^{2}-5}\) = \(\frac{1}{(x-1)(5-x)}\) ∈ R
⇔ (x – 1) (5 – x) ≠ 0
⇔ x ≠ 1, 5
∴ Domain of f is R – {1, 5}

ii) f(x) = \(\frac{1}{\sqrt{x^{2}-a^{2}}}\), (a > 0)  (A.P.) (Mar. ’15)
Solution:
f(x) = \(\frac{1}{\sqrt{x^{2}-a^{2}}}\) ∈ R
⇔ x² – a² > 0
⇔ (x + a)(x – a) > 0
⇔ x ∈ (-∞, -a) ∪ (a, ∞)
∴ Domain of f is (-∞, -a) ∪ (a, ∞) = R – [-a, a]

iii) f(x) = \(\sqrt{(x+2)(x-3)}\)
Solution:
f(x) = \(\sqrt{(x+2)(x-3)}\) ∈ R
⇔ (x + 2)(x – 3) ≥ 0
⇔ x ∈ (-∞, -2) ∪ [3, ∞)
∴ Domain of f is
(-∞, -2] ∪ [3, ∞) = R – (-2, 3)

iv) f(x) = \(\sqrt{(x-\alpha)(\beta-x)}\) (0 < α < β)
Solution:
f(x) = \(\sqrt{(x-\alpha)(\beta-x)}\) ∈R
⇔ (x – α) (β – α) ≥ 0
⇔ α ≤ x ≤ β ; (∵ α < β)
⇔ x ∈ [α, β]
∴ Domain of f is [α, β]

v) f(x) = \(\sqrt{2-x}\) + \(\sqrt{1+x}\)
Solution:
f(x) = \(\sqrt{2-x}\) + \(\sqrt{1+x}\) + x ∈ R
⇔ 2 – x ≥ 0 and 1 + x ≥ 0
⇔ 2 ≥ x and x ≥ -1
⇔ -1 ≤ x ≤ 2
⇔ x ∈ [-1, 2]
∴ Domain of f is [-1, 2].

vi) f(x) = \(\sqrt{x^{2}-1}\) + \(\frac{1}{\sqrt{x^{2}-3 x+2}}\)
Solution:
f(x) = \(\sqrt{x^{2}-1}\) + \(\frac{1}{\sqrt{x^{2}-3 x+2}}\) ∈ R
⇔ x² – 1 ≥ 0 and x² – 3x + 2 > 0
⇔ (x + 1)(x – 1) ≥ 0 and (x – 1)(x – 2) > 0
⇔ x ∈ (-∞, -1] ∪ [1, ∞) and x ∈ (-∞, 1)u(2, ∞)
⇔ x ∈ (R – (-1, 1)) ∩ (R – [1, 2])
⇔ x ∈ R – {(-1, 1) ∪ [1,2]}
⇔ x ∈ R – (-1, 2]
⇔ x ∈ (-∞, -1] ∪ (2, ∞)
∴ Domain of f is (-∞, -1) ∪ (2, ∞) = R – (-1, 2]

vii) f(x) = \(\frac{1}{\sqrt{|x|-x}}\)
Solution:
f(x) = \(\frac{1}{\sqrt{|x|-x}}\) ∈ R
⇔ |x| – x > 0
⇔ |x| > x
⇔ x ∈ (-∞, 0)
∴ Domain of f is (-∞, 0)

viii) f(x) = \(\sqrt{|x|-x}\)
Solution:
\(\sqrt{|x|-x}\) ∈ R
⇔ |x| – x ≥ 0
⇔ |x| ≥ x
⇔ x ∈ R
∴ Domain of f is R

Question 15.
If f = ((4, 5), (5, 6), (6, -4)} and g = ((4, -4), (6, 5), (8, 5)} then find
(i) f + g
(ii) f – g
(iii) 2f + 4g
(iv) f + 4
(v) fg
(vi) \(\frac{\mathbf{f}}{\mathbf{g}}\)
(vii) |f|
(viii) \(\sqrt{f}\)
(ix) f²
(x) f³
Solution:
Given that f = ((4, 5), (5, 6), (6, -4)}
g {(4, —4), (6, 5), (8, 5)}
Domain of f = {4, 5, 6} = A
Domain of g = (4, 6, 8} = B
Domain of f ± g = A ∩ B = {4, 6}
i) f + g = {4, 5 + (-4), (6, -4 + 5)}
= {(4, 1), (6, 1)}

ii) f – g = {(4, 5 – (-4)), (6, -4, -5)}
= {(4, 9), (6, -9)}

iii) Domain of 2f = A = {4, 5, 6}
Domain of 4g = B = {4, 6, 8}
Domain of 2f + 4g = A ∩ B = {4, 6}
∴ 2f = {(4, 10), (5, 12), (6, -8)}
4g = {(4, —16), (6, 20), (8, 20)}
∴ 2f + 4g = {(4, 10 + (-16), 6, -8 + 20)}
= {(4, -6), (6, 12)}

iv) Domain of f + 4 = A= {4, 5, 6}
f + 4 ={4, 5 + 4), (5, 6 +4), (6, -4 + 4)}
= ((4, 9), (5, 10), (6, 0)}

v) Domain of fg = A ∩ B = {4, 6}
fg = {(4, (5) (-4), (6, (-4) (5))}
= {(4, -20), (6, -20)}

vi) Domain of \(\frac{f}{g}\) = {4, 6}
∴ \(\frac{f}{g}\) = {(4, \(\frac{-5}{4}\)), (6, \(\frac{-4}{5}\))}

vii) Domain of |f| = {4, 5, 6}
∴ |f| = {(4, |5|), (5, |6|), (6, |-4|)}
= {(4, 5), (5, 6), (6, 4))

viii) Domain of \(\sqrt{f}\) = {4, 5}
∴ \(\sqrt{f}\) = {(4, \(\sqrt{5}\)), (5, \(\sqrt{6}\))}

ix) Domain of f² = {4, 5, 6} = A
∴ f² = ((4, (5)²) (5, (6)², (6, (-4)²)}
f² = {(4, 25), (5, 36), (6, 16)}

x) Domain of f³ = A = {4, 5, 6}
∴ f³ = {(4, 5³), (5, 6³), (6, (-4)³}
= {(4, 125) (5, 216) (6, -64)}

Question 16.
Find the domains and ranges of the following real valued functions.
i) f(x) = \(\frac{2+x}{2-x}\)
ii) f(x) = \(\frac{x}{1+x^{2}}\)
iii) f(x) = \(\sqrt{9-x^{2}}\)  (A.P.)(Mar. ’15)
Solution:
i) f(x) = \(\frac{2+x}{2-x}\) ∈ R
⇔ 2 – x ≠ 0 x ⇔ x ∈ R -{2}
∴ Domain of f is R – {2}
Let f(x) = \(\frac{y}{1}\) = \(\frac{2+x}{2-x}\).
Apply componendo and dividendo rule
⇒ \(\frac{y+1}{y-1}\) = \(\frac{(2+x)+(2-x)}{(2+x)-(2-x)}\)
⇒ \(\frac{y+1}{y-1}\) = \(\frac{4}{2 x}\)
⇒ x = \(\frac{2(y-1)}{y+1}\)
Clearly x is not defined for y + 1 = 0
(i.e.,) y = -1
∴ Range of f is R – {-1}.

ii) f(x) = \(\frac{x}{1+x^{2}}\)
Solution:
f(x) = \(\frac{x}{1+x^{2}}\) ∈ R
∵ ∀ x ∈ R, x² + 1 ≠ 0
Domain of f is R
Let f(x) = y = \(\frac{x}{1+x^{2}}\)
⇒ x²y – x + y = 0
⇒ x = \(\frac{-(-1) \pm \sqrt{1-4 y^{2}}}{y}\) is a real number.
iff 1 – 4y² ≥ 0; y ≠ 0
⇔ (1 – 2y)(1 + 2y) ≥ 0 and y ≠ 0
⇔ y ∈ \(\left[-\frac{1}{2} ; \frac{1}{2}\right]\) – {0}
Also x = 0 ⇒ y = 0
∴ Range of f = \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)

iii) f(x) = \(\sqrt{9-x^{2}}\)
Solution:
f(x) = \(\sqrt{9-x^{2}}\) ∈ R
⇔ 9 – y² ≥ 0
⇔ x ∈ [-3, 3]
∴ Domain of f is [-3, 3]
Let f(x) = y = \(\sqrt{9-x^{2}}\)
⇒ x = \(\sqrt{9-y^{2}}\) ∈ R
⇔ 9 – y² ≥ 0 ⇔ (3 + y)(3 – y) ≥ 0
∴ -3 ≤ y ≤ 3
But f(x) attains only non negative values
∴ Range of f = [0, 3].

Question 17.
If f(x) = x² and g(x) = |x|, find the following functions.
i) f + g
ii) f – g
iii) fg
iv) 2f
v) f²
vi) f + 3
Solution:
Given f(x) = x²

Domain of f = Domain of g = R
Hence domain of all the functions (i) to (vi) is R

iv) 2f(x) = 2 f(x) = 2x²

v) f²(x) = (f(x))² = (x²)² = x⁴

vi) (f + 3)(x) = f(x) + 3 = x² + 3.

Question 18.
Determine whether the following functions are even or odd.
i) f(x) = a^x – a^-x + sin x
Solution:
Given f(x) = a^x – a^-x + sin x
∴ f(- x) = a^-x – a^x + sin (-x)
= a^-x – a^x – sin x
= – (a^x – a^x – sin x) = – f(x)
∴ f(x) is an odd function.

ii) f(x) = x\(\left(\frac{e^{x}-1}{e^{x}+1}\right)\)
Solution:
f(x) = x\(\left(\frac{e^{x}-1}{e^{x}+1}\right)\)

∴ f is an even function.

iii) f(x) = log (x + \(\sqrt{x^{2}+1}\))
Solution:
Given f(x) = log (x + \(\sqrt{x^{2}+1}\))
Then f(-x) = log (-x + \(\sqrt{(-x)^{2}+1}\))
= log (\(\sqrt{x^{2}+1}\) – x)

∴ f is an odd function.

Question 19.
Find the domains of the following real valued functions.
i) f(x) = \(\frac{1}{\sqrt{[x]^{2}-[x]-2}}\)
Solution:
f(x) = \(\frac{1}{\sqrt{[x]^{2}-[x]-2}}\) ∈ R
⇔ [x]² – [x] – 2 > 0
⇔ ([x] + 1) ([x] – 2) > 0
⇔ [x] < -1, (or) [x] > 2
But [x] < -1 ⇒ [x] = -2, -3, -4, ……..
⇒ x < -1 [x] > 2 ⇒ [x] = 3, 4, 5, ……
⇒ x ≥ 3
∴ Domain of f = (-∞, -1) ∪ [3, ∞]
= R – [-1, 3)

ii) f(x) = log (x – [x])
f(x) = log (x – [x]) ∈ R
⇔ x – [x] > 0
⇔ x > [x]
⇔ x is a non integer
∴ Domain of f is R – Z

iii) f(x) = \(\sqrt{\log _{10}\left(\frac{3-x}{x}\right)}\)
Solution:
f(x) = \(\sqrt{\log _{10}\left(\frac{3-x}{x}\right)}\) ∈ R
⇔ log₁₀\(\left(\frac{3-x}{x}\right)\) ≥ 0 and \(\frac{3-x}{x}\) > 0
⇔ \(\frac{3-x}{x}\) ≥ 10° = 1 and 3 – x > 0, x > 0
⇔ 3 – x ≥ x and 0 < x < 3
⇔ x ≤ \(\frac{3}{2}\) and 0 < x < 3
⇔ x ∈ (-∞, \(\frac{3}{2}\)] ∩ (0, 3) = (0, \(\frac{3}{2}\)]
∴ Domain of f is (0, \(\frac{3}{2}\)]

iv) f(x) = \(\frac{\sqrt{3+x}+\sqrt{3-x}}{x}\)
Solution:
f(x) = \(\frac{\sqrt{3+x}+\sqrt{3-x}}{x}\) ∈ R
⇔ 3 + x ≥ 0, 3 – x ≥ 0 and x ≠ 0
⇔ x ≥ -3, x ≤ 3 and x ≠ 0
⇔ -3 ≤ x ≤ 3, and x ≠ 0
⇔ x ∈ [-3, 3] and x ≠ 0
⇔ x ∈ [-3, 3] – {0}
∴ Domain of f is [-3, 3] – {0}

Question 20.
If f : A → B and g : B → C are two injective functions then the mapping gof : A → C is an injection.
Solution:
f : A → B and g: B → C are one—one.
∴ gof : A → C
To prove that g o f is one — one function
Let a₁, a₂ ∈ A ∴ f(a₁), f(a₂) ∈ B and g(f(a₁)), g(f(a₂)) ∈ C i.e., (gof) (a₁), gof(a₂) ∈ C
Now (gof) (a₁) = gof (a₂)
⇒ g(f(a₁)) = g(f(a₂))
⇒ f(a₁) = f(a₂) (∵ g is one -one)
⇒ a₁ = a₂ (∵ f is one-one)
Hence gof: A → C is a one-one function.
Note : The converse of the above theorem is not true.
If f : A → B, g: B → C and go f is one-one then both f and g need not be one-one.
For, consider
A = {1, 2}, B = {p, q, r), C = {s, t}
Let f = {(1, p), (2, q)} and
g = {(p, s), (q, t), (r, t)}
Now gof = {(1, s), (2, t)} ⇒ gof is one—one
from A to C
Observe that g: B → C is not one—one.

Question 21.
If f : A → B and g : B → C are two onto (surjective) functions then the mapping gof : A → C is a surjection. (May. ’08)
Solution:
f : A → B and g : B → C are onto.
∴ gof : A → C
To prove that gof is onto. Let c be any element of C.
Since g : B → C is an onto function there exists an element b ∈ B such that g(b) = c
Since f : A → B is an onto function, there exists an element a ∈ A such that f(a) = b
Now g(b) = c ⇒ g [f(a)] c = (gof) (a) = c
Thus for any element c ∈ C there is an element a ∈ A such that (gof) (a) = c
∴ gof : A → C is a surjection

Question 22.
If f : A → B and g : B → C are two bijective functions then the mapping gof : A → C is a bijection. (Mar. ’16, May ’12)
Solution:
f and g are injections gof: A → C is an injection.
f and g are surjections= gof : A → C is a surjection.
Hence it follows that if f and g are bijections,
gof is also a bijection.
Note : The converse of the above theorem is not true.

Question 23.
If f : A → B and g : B → C are such that gof is a surjection, then g is necessarily a surjection.
Solution:
Let c ∈ C. Since gof is a surjection, from A to
C there exists an element a ∈ A such that (gof) (a) = c, i.e., g(f(a)) = c
Since g: B → C and f(a) ∈ B, ∀ c ∈ C there exists an element belonging to B.
Hence g is a surjection.

Question 24.
If f : A → B and g : B → C and h : C → D are functions then ho(gof) = (hog)of (Mar. ’12, ’08)
Solution:
f : A → B and g : B → C gof : A → C
Now gof : A → C and h : C → D
⇒ ho(gof) : A → D
Similarly (hog) of : A → D
Thus ho (gof) and (hog) of both exist and have the same domain A and co-domain D.
Let a be an element of A.
Now [ho(gof)](a) = h[(gof)(a)] = h[g(f(a))]
= (hog) [f(a)] = [(hog)of] (a)
∴ ho(gof) = (hog)of
Note : Thus composition of mappings is associative.

Question 25.
Let f : A → B, I_A and I_B are identity functions on A and B respectively. Then for I_A = f = I_B of Mar.’12, ’08
Solution:
∵ I_A : A → A and f : A → B are functions foI_A is a function from A to B. Hence foI_A and f are definded on same domain A.
i) Let a ∈ A then (foI_A)(a) = f(I_A(a))
= f(a)
[∵ I_A(a) = a, ∀ a ∈ A]
∴ foI_A = f ————- (1)

ii) ∵ f : A → B, I_B : B → B are functions, I_B of is a function from A to B
∴ The functions I_B of and f are defined on the same domain A.
Let a ∈ A, then (I_B of)(a)
= I_B(f (a)) = f(a)
∵ f : A → B, we have f(a) ∈ B
∴ I_B of = f ———– (2)
From(1) & (2) foI_A = I_Bof = f

Question 26.
Let A and B be two non-empty sets. If f : A → B is a bijection, then f^-1 : B → A is also a bijection.
Solution:
f : A → B is a bijection.
∴ f^-1 : B → A is a unique function.
(i) To prove that f^-1 is one—one:
Let b₁ and b₂ be any two different elements of B. i.e., b₁ ≠ b₂.
Then, we have to prove that
f^-1(b₁) ≠ F^-1(b₂)
Let f^-1(b₁) = a₁ and f^-1(b₂) = a₂
such that a₁, a₂ ∈ A
Then b₁ = f(a₁) and b₂ = f(a₂)
Now b₁ ≠ b₂ ⇒ f(a₁) ≠ f(a₂)
⇒ a₁ ≠ a₂ (∵ f is a bijection)
⇒ f^-1(b₁) ≠ f^-1(b₂)
∴ f^-1 is one—one.

(ii) To prove that f^-1 is onto:
Let ‘a’ be an element of A.
Then there exists an element b ∈ B such that f(a) = b (or) f^-1(b) = a
(or) a = f^-1(b)
Thus ‘a’ is the f^-1 – image of the element b ∈ B
Hence f^-1 is onto.
∴ f : B → A is a bijection.

Question 27.
If f : A → B is a bijection, then f^-1of = I_A and fof^-1 = I_B
(A.P) (Mar. 15,12, ’07; May ’07, ‘06)
Solution:
f : A → B is a bijection ⇒ f^-1 : B → A is also a bijection.
By definition, fof^-1 : B → B and f^-1of : A → A are bijection
Also I_A : A → A and I_B : B → B
To prove that f^-1of = I_A
Let a ∈ A
Since f : A → B there exists a unique element
b ∈ B such that f(a) = b
a = f^-1 (b) (∵ f is a bijection)
∴ (f^-1of) (a) = f^-1[f(a)]
= f^-1(b) = a = I_A (a)
f^-1 of = I_A
Similarly it can be shown that fof^-1 = I_B

Question 28.
If f : A → B and g : B → A are two functions such that gof = I_A and fog = I_B then g = f^-1.
Solution:
(i) To prove that f is one—one:
Let a₁, a₂ ∈ A.
Since f: A → B, f(a₁), f(a₂) ∈ B
Now f(a₁) = f(a₂)
⇒ g[f(a₁)] = g[f(a₂)]
⇒ (gof)(a₁) = gof(a₂)
⇒ I_A (a₁) = I_A (a₂)
∴ a₁ = a₂ = ∴ f is one — one

(ii) To prove that f is onto :
Let b be an element of B
∴ b = I_B fog(b)
⇒ b = f{(g(b)} ⇒ f{g(b)} = b
i.e., there exists a pre-image g(b) ∈ A for b, under the mapping f. ∴ f is onto
Thus f is one-one, onto and hence
f^-1 : B → A exists and is also one — one and onto.

(iii) To prove g = f^-1:
Now g : B → A and f^-1 : B → A
Let a ∈ A and b be the f – image of a, where b ∈ B
∴ f(a) = b ⇒ a =f^-1(b)
Now g(b) = g[f(a)] (gof) (a)
= I_A(a) = a = f^-1(b)
∴ g = f^-1

Question 29.
If f : A → B and g : B → C are bijective functions, then (gof)^-1 = f^-1og^-1(AP) (Mar. ‘16’14, ‘11; May ‘11)
Solution:
f : A → B, g : B → C are bijections
⇒ gof : A → C is a bijection
Also g^-1 : C → B and f^-1 : B → A are bijections
⇒ f^-1og^-1 : C → A is a bijection.
Let c be any element of C.
Then ∃ an element b ∈ B such that g(b) = c
⇒ b = g^-1(c)
Also ∃ an element a A such that f(a) = b
⇒ a = f^-1(b)
Now (gof) (a) = g(f(a) = g(b) = c
⇒ a = f^-1(b)
Now (gof) (a) = g(f(a) = g(b) = c
⇒ a = (gof)^-1 (c) ⇒ (gof) (c) = a
Also (f^-1og^-1) (c) ⇒ f^-1 (g^-1 (c)) = f^-1 (b) = a
∴ From (1) and (2);
(gof)^-1 (c) = (f^-1og^-1(c))
⇒ (gof)^-1 = f^-1og^-1

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 10 Properties of Triangles to solve questions creatively.

Intermediate 1st Year Maths 1A Properties of Triangles Formulas

→ Sine Rule :
In ΔABC \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\) = 2R where R is the circumradius of ΔABC.

→ Cosine Rule :
a² = b² + c² – 2bc. cos A ;
b² = c² + a² – 2ca.cos B;
c² = a² + b² – 2ab. cos C.

→ cos A = \(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\),
cos B = \(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\),
cos C = \(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\)

→ a = b cos C + c cos B,b = c cos A + a cos C and c = a cos B + b cos A (Projection rule)

→ tan \(\frac{B-C}{2}=\frac{b-c}{b+c}\) cot\(\frac{A}{2}\) (Napier’s analogy or tangent rule)

• sin\(\frac{A}{2}\) = \(\sqrt{\frac{(s-b)(s-c)}{b c}}\)
• cos\(\frac{A}{2}\) = \(\sqrt{\frac{s(s-a)}{b c}}\)
• tan\(\frac{A}{2}\) = \(\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\frac{\Delta}{s(s-a)}\)

→ Δ = area of ΔABC = \(\frac{1}{2}\) bc sin A = \(\frac{1}{2}\) ca sin B = \(\frac{1}{2}\) ab sin C
= \(\sqrt{s(s-a)(s-b)(s-c)}=\frac{a b c}{4 R}\)
= 2R² sin A sin B sin C

• r = \(\frac{\Delta}{s}\)
• r₁ = \(\frac{\Delta}{s-a}\)
• r₂ = \(\frac{\Delta}{s-b}\)
• r₃ = \(\frac{\Delta}{s-c}\)

→ r = 4 R sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) sin \(\frac{C}{2}\); r₁ = 4Rsin \(\frac{A}{2}\) cos \(\frac{B}{2}\) cos \(\frac{C}{2}\)

→ r = (s – a) tan \(\frac{A}{2}\);
r₁ = s tan \(\frac{A}{2}\) = (s – c) cot \(\frac{B}{2}\) = (s – b) cot \(\frac{C}{2}\)

Mollweide rule:
In ΔABC \(\frac{a+b}{c}=\frac{\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}}\)
\(\frac{b+c}{a}=\frac{\cos \left(\frac{B-C}{2}\right)}{\sin \frac{A}{2}}\)
\(\frac{c+a}{b}=\frac{\cos \left(\frac{C-A}{2}\right)}{\sin \frac{B}{2}}\)

Sine rule :
In ΔABC, \(\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}\) = 2R Where R is the circum – radius.
⇒ a = 2R sin A, b = 2R sin B, c = 2R sin C
a : b : c = sin A : sin B : sin C.

Cosine rule :
In ΔABC,
a² = b² + c² – 2bc cos A
b² = c² + a² – 2ac cos B
c² = a² + b² – 2ab cos C
or
cos A = \(\frac{\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{a}^{2}}{2 \mathrm{bc}}\)
cos B = \(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\)
cos C = \(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\) ⇒ cos A : cos B : cos C
= a(b² + c² – a²) : b(c² + a² – b²) : c(a² + b² – c²)

Projection rule :
In ΔABC

• a = b cos C + c cos B,
• b = c cos A + a cos C,
• c = s cos B + b cos A

Mollwiede’s rule :
In ΔABC

• \(\frac{a-b}{c}=\frac{\sin \frac{A-B}{2}}{\cos \frac{C}{2}}\)
• \(\frac{a+b}{c}=\frac{\cos \frac{A-B}{2}}{\sin \frac{C}{2}}\)

Similarly the other two can be written by symmetry.

Tangent rule (or) Napier’s analogy :
In ΔABC

Half angle formulae :

cot A = \(\frac{\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{a}^{2}}{4 \Delta}\)
cot B = \(\frac{c^{2}+a^{2}-b^{2}}{4 \Delta}\)
cot C = \(\frac{a^{2}+b^{2}-c^{2}}{4 \Delta}\)

→ Area of ΔABC 1s given by

• Δ = \(\frac{1}{2}\)ab sin C = \(\frac{1}{2}\)bc sin A = \(\frac{1}{2}\) ca sin B
• Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\).
• Δ = \(\frac{a b c}{4 R}\)
• Δ = 2R² sin A sin B sin C
• Δ = rs
• Δ = \(\sqrt{\mathrm{rr}_{1} \mathrm{r}_{2} \mathrm{r}_{3}}\)

→ If ‘r’ is radius of in circle and r₁, r₂, r₃ are the radii of ex-circles opposite to the vertices A, B, C of ΔABC respectively then
i. r = \(\frac{\Delta}{\mathrm{s}}\), r₁ = \(\frac{\Delta}{\mathrm{s-a}}\), r₂ = \(\frac{\Delta}{\mathrm{s-b}}\), r₃ = \(\frac{\Delta}{\mathrm{s-c}}\)

→ r = 4R sin\(\frac{\mathrm{A}}{2}\)sin\(\frac{\mathrm{B}}{2}\) sin\(\frac{\mathrm{C}}{2}\)

• r₁ = 4R sin\(\frac{\mathrm{A}}{2}\)cos\(\frac{\mathrm{B}}{2}\)cos\(\frac{\mathrm{C}}{2}\)
• r₂ = 4R cos\(\frac{\mathrm{A}}{2}\)sin\(\frac{\mathrm{B}}{2}\) cos\(\frac{\mathrm{C}}{2}\)
• r₃ = 4R cos\(\frac{\mathrm{A}}{2}\)cos\(\frac{\mathrm{B}}{2}\) sin\(\frac{\mathrm{C}}{2}\)

→ r = (s – a)tan\(\frac{A}{2}\) = (s – b)tan\(\frac{B}{2}\) = (s – c)tan\(\frac{C}{2}\)

• r₁ = s tan\(\frac{A}{2}\) = (s – b)cot\(\frac{C}{2}\) = (s – c)cot\(\frac{B}{2}\)
• r₂ = s tan\(\frac{B}{2}\) = (s – c)cot\(\frac{A}{2}\) = (s – a)cot\(\frac{c}{2}\)
• r₁ = s tan\(\frac{A}{2}\) = (s – a)cot\(\frac{B}{2}\) = (s – b)cot\(\frac{A}{2}\)

→ \(\frac{1}{r}=\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}\)

→ rr₁r₂r₃ = Δ²

→ r₁r₂ + r₂r₃ + r₃r₁ = s²

→ r(r₁ + r₂ + r₃) = ab + bc + ca – s²

→ (r₁ – r)(r₂ + r₃) = a²

→ (r₂ – r)(r₃ + r₁) = b²

→ (r₃ – r)(r₁ + r₂) = c²

→ a = (r₂ + r₃)\(\sqrt{\frac{r r_{1}}{r_{2} r_{3}}}\)

→ b = (r₃ + r₁)\(\sqrt{\frac{r r_{2}}{r_{3} r_{1}}}\)

→ c = (r₁ + r₂)\(\sqrt{\frac{r r_{3}}{r_{1} r_{2}}}\)

→ r₁ – r = 4Rsin²\(\frac{A}{2}\)

→ r₂ – r = 4Rsin²\(\frac{B}{2}\)

→ r₃ – r = 4Rsin²\(\frac{C}{2}\)

→ r₁ + r₂ = 4R cos²\(\frac{C}{2}\)

→ r₂ + r₃ = 4R cos²\(\frac{A}{2}\)

→ r₃ + r₁ = 4R cos²\(\frac{B}{2}\)

→ r₁ + r₂ + r₃ = 4R

→ r + r₂ + r₃ – r₁ = 4R cos A

→ r + r₁ + r₃ – r₂ = 4R cos B

→ r + r₁ + r₂ – r₃ = 4R cos C

→ In an equilateral triangle of side ‘a’
area = \(\frac{\sqrt{3} a^{2}}{4}\)
R = a/√3
r = R/2
r₁ = r₂ + r₃ = 3R/2
r : R : r₁ = 1 : 2 : 3

In circle:
The circle that touches the three sides of a triangle ABC internally is called the “in circle” or inscribed” of its triangle. The centre of the circle is called Incentre denoted by I the radius of the circle is denoted by inradius denoted by Y

→ In a triangle ABC

Excircle:
The circle that touches the side BC (opposite to angle A) internally and the other two sides AB and AC externally is called Excircle. The centre of this circle is called excentre opposite to ‘A’. denoted by I₁. The radius of this circle is called ex-radius, denoted by r₁

|||^ly exradius opposite to angle B is denoted by r₂. The centre of this excircle is denoted by I₂ exradius opposite to angle C is denoted by r₃. The centre of this ex-circle is denoted by I₃

In a triangle ABC

• r₁ = \(\frac{\Delta}{s-a}\)
• r₂ = \(\frac{\Delta}{s-b}\)
• r₃ = \(\frac{\Delta}{s-c}\)

→ r₁ = s tan A/2

→ r₂ = s tan B/2

→ r₂ = s tan C/2

→ r₁ = (s – c)cot\(\frac{B}{2}\)
= (s- b)cot\(\frac{C}{2}\)

→ r₁ = (s – a)cot\(\frac{C}{2}\)
= (s – c)cot\(\frac{A}{2}\)

→ r₁ = (s – a)cot\(\frac{B}{2}\)
= (s – c)cot\(\frac{A}{2}\)

→ r₁ = \(\frac{a}{\tan \frac{B}{2}+\tan \frac{C}{2}}\)

→ r₂ = \(\frac{b}{\tan \frac{C}{2}+\tan \frac{A}{2}}\)

→ r₃ = \(\frac{c}{\tan \frac{A}{2}+\tan \frac{B}{2}}\)

→ r₁ = 4Rsin\(\frac{A}{2}\)cos\(\frac{B}{2}\)cos\(\frac{C}{2}\)

→ r₂ = 4Rcos\(\frac{A}{2}\)sin\(\frac{B}{2}\)cos\(\frac{C}{2}\)

→ r₃ = 4Rcos\(\frac{A}{2}\)cos\(\frac{B}{2}\)sin\(\frac{C}{2}\)

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 9 Hyperbolic Functions to solve questions creatively.

Intermediate 1st Year Maths 1A Hyperbolic Functions Formulas

→ e^x = 1 + \(\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots+\frac{x^{n}}{n !}\)+ … ∞

→ sinhx = \(\frac{e^{x}-e^{-x}}{2}\)
and coshx = \(\frac{e^{x}+e^{-x}}{2}\)

→ tanh x = \(\frac{\sinh x}{\cosh x}\),

→ coth x = \(\frac{1}{\tanh x}\),

→ sech x = \(\frac{1}{\cosh x}\)

→ cosech x = \(\frac{1}{\sinh x}\), if x ≠ 0

→ cosh²x – sinh²x – 1, sech²x – 1 – tanh²x, cosech²x = coth²x – 1

→ sinh^-1x = loge [x + \(\sqrt{x^{2}+1}\) ] for all x ∈ R

→ cosh^-1x = loge [x + \(\sqrt{x^{2}-1}\)] for all x ∈ (1, ∞)

→ tanh^-1x = \(\frac{1}{2}\)log_e\(\left(\frac{1+x}{1-x}\right)\), |x| < 1 (i.e) for all x ∈ (-1, -1)

→ coth^-1x = \(\frac{1}{2}\)log_e\(\left(\frac{1+x}{1-x}\right)\), |x| > 1 (i.e) for all x ∈ (-∞, -1) (1, ∞)

→ sech^-1x = log_e\(\left(\frac{1+\sqrt{1+x^{2}}}{x}\right)\) for all x ∈ (0, 1)

→ cosech^-1x = log_e\(\left(\frac{1-\sqrt{1+x^{2}}}{x}\right)\), if x < 0 (i.e,) x ∈ (-∞, 0) and
= log_e\(\left(\frac{1+\sqrt{1+x^{2}}}{x}\right)\), if x > 0 (i.e,) x ∈ (0, ∞)

→ sinh (x + y) = sinh x cosh y + cosh x sinh y

→ cosh (x + y)= cosh x cosh y + sinh x sinh y

→ sinh (x – y) = sinh x cosh y – cosh x sinh y

→ cosh (x – y) = cosh x cosh y – sinh x sinh y

→ tanh (x + y) = \(\frac{\tanh x+\tanh y}{1+\tanh x \tanh y}\)

→ tanh (x – y) = \(\frac{\tanh x-\tanh y}{1-\tanh x \tanh y}\)

→ sinh 2x = 2 sinh x cosh x = \(\frac{2 \tanh x}{1-\tanh ^{2} x}\)

→ cosh 2x = cosh²x + sinh²x
= 2 cosh²x – 1 = 1 + 2 sinh² x = \(\frac{1+\tanh ^{2} x}{1-\tanh ^{2} x}\)

→ tanh 2x = \(\frac{2 \tanh x}{1+\tanh ^{2} x}\)

→ sinh 3x = 3sinh x + 4sinh³ x

→ cosh 3x = 4cosh³ x – 3cosh x

→ tanh 3x = \frac{3 \tanh x+\tanh ^{3} x}{1+3 \tanh ^{2} x}

→ Inverse hyperbolic functions:

→ sin hx = \(\frac{e^{x}-e^{-x}}{2}\)

→ cos hx = \(\frac{e^{x}+e^{-x}}{2}\)

→ tan hx = \(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\)

→ cosec hx = \(\frac{2}{e^{x}-e^{-x}}\)

→ sec hx = \(\frac{2}{e^{x}+e^{-x}}\)

→ cot hx = \(\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\)

→ cos h²x – sinh²x = 1

→ 1 – tanh²x = sech²x

→ cot h²x – 1 = cosech²x

→ Prove that sinh^-1x = log{x + \(\sqrt{x^{2}+1}\)}
Proof:
Let sinh^-1x = y ⇒ x = sinh y

→ Prove that cosh^-1x = log_e(x – \(\sqrt{x^{2}-1}\)
proof:
Let cosh^-1x = y ⇒ x = cosh y

→ Prove that Tan^-1x = \(\frac{1}{2}\)log_e\(\left(\frac{4 x}{1-x}\right)\)
proof:
Let tanh^-1x = y ⇒ x = tanh y

→ sech^-1x = log\(\left\{\frac{1+\sqrt{1-x^{2}}}{x}\right\}\)

→ cosech^-1x = log\(\left(\frac{1-\sqrt{1+x^{2}}}{x}\right)\) x < 0 = log\(\left\{\frac{1-\sqrt{1+x^{2}}}{x}\right\}\) x > 0

→ sin h(x + y) = sin hx cos hy + cos hx sin hy

→ sinh(x – y) = sin hx cos hy + cos hx sin hy

→ cosh(x + y) = cos hx cos hy + sin hx sin hv

→ cosh(x – y) = cos hx cos hy – sin hx sin hy

→ sin h2x = 2 sin hx cos hx = \(\)

→ cos h2x = cosh²x + sinh²x = 2cosh²x – 1 = 1 + 2sinh²x = \(\frac{1+{Tanh}^{2} x}{1-{Tanh}^{2} x}\)

→ Tanh(x + y) = \(\frac{\text { Tanhx+Tanhy }}{1-\text { TanhxTanhy }}\)

→ Tanh(x – y) = \(\frac{\text { Tanhx-Tanhy }}{1+\text { TanhxTanhy }}\)

→ coth(x + y) = \(\frac{\cot h x \cot h y+1}{\cot h y+\cot h x}\)

→ coth(x – y) = \(\frac{\cot h x \cot h y-1}{\cot h y-\cot h x}\)

→ Tanh2x = \(\frac{2 \tan h x}{1+\tanh ^{2} x}\)

→ cot h2x = \(\frac{\operatorname{coth}^{2} x+1}{2 \cot h x}\)

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 7 Trigonometric Equations Functions to solve questions creatively.

Intermediate 1st Year Maths 1A Trigonometric Equations Formulas

→ If K ∈ [-1, 1], then the principal solution θ of sin θ = K lies in \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)and the general solution is given by nπ + (-1)ⁿ θ, n ∈ Z

→ If K ∈ [-1, 1], then the principal solution θ of cos θ = K lies in [0, π] and the general solution is given by 2nπ ± 0, n ∈ Z

→ If K ∈ R, then the principal solution θ of tan x = K lies in \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\), and the general solution is given by n\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) ± θ, n ∈ Z

→ sin θ = 0 ⇒ θ = nπ, n is an integer and its principal solution is x = 0

→ cos θ = 0 ⇒ θ = (2n + 1)\(\frac{\pi}{2}\), n is an integer and its principal solution is x = \(\frac{\pi}{2}\)

→ tan θ = 0 ⇒ θ = nπ, n is an integer and its principal solution is x = 0

→ cot θ = 0 ⇒ θ = (2n + 1)\(\frac{\pi}{2}\) n is an integer

→ cosec θ = cosec α ⇒ θ = nπ + (-1)ⁿα, n is an integer

→ sec θ = sec α ⇒ θ = 2nπ ± α, n is an integer

→ cot θ = cot α ⇒ θ = nπ ± α and the principal solution α ∈ (0, π)

→ sin²θ = sin² α ⇒ θ = nπ ± α, wherenever a solution exists.

→ cos²θ = cos² α ⇒ θ = nπ ± α, wherenever a solution exists.

→ tan²θ = tan² α ⇒ θ = nπ ± α, wherenever a solution exists.

→ a cos θ + b sin θ = c, (a, b, c) ∈ R and a² + b² ≠ 0 has a solution, iff c² ≤ a² + b².

Trigonometric Equations:
An equation consisting of the trigonometric functions of a variable angle θ ∈ R is called a trigonometric equation.

→ sin θ = 0 ⇒ θ = nπ, where n ∈ Z.

→ cos θ = 0 ⇒ θ = (2n + 1)\(\), where n ∈ Z.

→ tan θ = 0 ⇒ θ = nπ where π is any integer.

→ The solution of sin θ = k (|k|<1) lying between – \(\frac{\pi}{2}\) and \(\frac{\pi}{2}\) is called the principle solution of the equation.

→ The solution of cos θ = k (|k|<1) lying between 0 and π is called the principal solution of the equation.

→ That solution of tan θ = k, lying between – \(\frac{\pi}{2}\) and \(\frac{\pi}{2}\) is called the principal solution of the equation.

→ The general value of θ satisfying cos θ = k (|k| < 1) is given by θ = 2nπ ± α where n ∈Z.

→ The general value of θ satisfying sin θ = k (|k|< 1) is given by θ = nπ + (-1)ⁿα where n ∈ Z.

→ The general value of θ satisfying tan θ = k is given by θ = nπ + α where n ∈ Z. (in each of the above cases, α is the principal solution).

→ If sin0 = k, tan0 = k are given equations, then the general value of θ is given by θ = 2nπ + α where α is that solution lying between 0 and 2π.

→ The equation a cos θ + b sin θ = c will have no solution or will be inconsistent if |c| > \(\sqrt{a^{2}+b^{2}}\)

→ cos nπ = (-1)ⁿ , sin nπ = 0.

→ If sin²θ = sin²α or cos²θ = cos²α or tan²θ = tan²a, then θ = nπ ± α; n ∈ Z.

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 6 Trigonometric Ratios up to Transformations to solve questions creatively.

Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Formulas

→ sin θ, cos θ, tan θ, cosec θ, sec θ and cot θ are called circular functions.

→ cosec θ, sec θ and cot θ are reciprocals to sin θ, cos θ and tan θ respectively.

→ sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ

→ sec θ + tan θ and sec θ – tan θ are mutual reciprocals.
Similarly cosec θ + cot θ and cosec θ – cot θ are also mutual reciprocals.

→ |sin θ| ≤ 1, |cosec θ| ≥ 1, |cos θ| ≤ 1 and |sec θ| ≥ 1

• sin 0° = o = cos 90°
• sin 15° = \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\) = cos 75°
• sin 18° = \(\frac{\sqrt{5}-1}{4}\) = cos 72°
• sin 30° = \(\frac{1}{2}\) = cos 60°
• sin 36° = \(\frac{\sqrt{10-2 \sqrt{5}}}{4}\) = cos 54°
• sin 45° = \(\frac{1}{\sqrt{2}}\) = cos 45°
• sin 54° = \(\frac{\sqrt{5}+1}{4}\) = cos 36°
• sin 60° = \(\frac{\sqrt{3}}{2}\) = cos 30°
• sin 72° = \(\frac{\sqrt{10+2 \sqrt{5}}}{4}\) = cos 18°
• sin 75° = \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\) = cos 15°
• sin 90° = 1 = cos 0°

Also, tan 15° = 2 – √3, tan 75° = 1 + √3
Sign to the determined by “ALL SILVER CUPS” rules

• sin (90° – θ) = cos θ; sin (90° + θ) = cos θ
• cos (90° – θ) – sin θ; cos (90° + θ) = -sin θ
• sin (180° – θ) = sin θ; sin (1800 + θ)= -sin θ
• cos (180° – θ) = -cos θ; cos (180° + θ) = -cos θ
• sin (270° – θ) = -cos θ; sin (270° + θ) = -cos θ
• cos (270° – θ) = -sin θ; cos (270° + θ) = sin θ
  sin (360° – θ) = -sin θ; sin (360° + θ) = sin θ
• cos (360° – θ) = cos θ; cos (360° + θ) = cos θ When ‘n’ is a +ve integer,
• sin (n. 360° – θ) = -sin θ; sin (n. 360° + θ) = +sin θ
• cos (n. 360° – θ) = cos θ; cos (n. 360° + θ) = cos θ
• sin (-θ) = – sin θ, cos (-θ) = cos 0; tan (-θ) = -tan θ.

→ For 0°, 180°, 360° ……………….. (multplies of π), there is no change in the ratios.

→ For 90°, 270°, 450°…………. (odd multiplies of \(\frac{\pi}{2}\) we get change in the ratio.

• For sin we get cos
• For tan we get cot
• For sec we get cosec
• For cos we get sin
• For cot we get tan
• For cot we get sin

→ Any non-constant function f : R → R is said to be “Periodic”, if there exists a real number p(≠ 0) such that f(x + p) = f(x) for each x ∈ R. The least positive value of ‘p’ with this property is called the “Period” of ‘f’.

→ If ‘f’ is a periodic function with period ‘P’ then

• – f is also a periodic function with period ‘P.
• f(x – p) = f(x), f(x + pn) = f(x) ∀ n ∈ Z, x ∈ R.

→ If f(x) is a periodic function with period ‘P, then f(ax + b) is also a periodic function with period \(\frac{P}{|a|}\)

→ If y = f(x), y = g(x) are periodic functions with l, m as the periods respectively, then a, b ∈ R the function h(x) = af(x) + bg(x) is a period function and L.C.M. of {l, m}
(if exist) is a period of h.

→ \(\frac{2 \pi}{k}\) is the period of sin x, cosec x, cos x and sec x.

→ \(\frac{\pi}{k}\) is the period of tan x and cot x.

→ Range of a sin x + b cos x is \(\left[-\sqrt{a^{2}+b^{2}}, \sqrt{a^{2}+b^{2}}\right]\)

→ Range of a sin x + b cos x + c is [c – \(\sqrt{a^{2}+b^{2}}\), c + \(\sqrt{a^{2}+b^{2}}\)]

→ sin x and cos x are continuous on ‘R’

→ tan x is. discontinuous at x = (2n + 1).\(\frac{\pi}{2}\),n ∈ Z.

→ cot x is discontinuous at x = nπ, n ∈ Z.

→ sec x is discontinuous at x = (2n + 1).\(\frac{\pi}{2}\), n ∈ Z.

→ cosecx is discontinuous at x = nπ, n ∈ Z.

→ sin (A ± B) = sin A cos B ± cos A sin B

→ cos (A ± B) = cos A cos B + sin A sin B

→ tan (A ± B) = \(\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\)

→ cot (A ± B) = \(\frac{\cot A \cot B \mp 1}{\cot B \pm \cot A}\)

→ sin (A + B). sin (A-B) = sin²A – sin²B = cos²B – cos²A

→ cos (A + B). cos (A – B) = cos²A – sin²B = cos²B – sin²A

→ sin (A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C

→ cos (A + B + C) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sin B cos C

→ tan (A + B + C) = \(\frac{\sum \tan A-\pi \tan A}{1-\sum \tan A \tan B}\)

→ sin 2A = 2 sin A cos A, sin A = 2 sin \(\frac{A}{2}\) cos \(\frac{A}{2}\)

→ cos 2A = cos²A – sin²A = 1 – 2 sin² A = 2 cos²A – 1

→ cos A = cos²\(\frac{A}{2}\) sin²\(\frac{A}{2}\) = 1 – 2 sin²\(\frac{A}{2}\) = 2 cos² \(\frac{A}{2}\) – 1

→ tan 2A = \(\frac{2 \tan A}{1-\tan ^{2} A}\), tan A = \(\frac{2 \tan \frac{A}{2}}{1-\tan ^{2} \frac{A}{2}}\)(\(\frac{A}{2}\), A are not odd multiples of \(\frac{\pi}{2}\))

→ cot 2A = \(\frac{\cot ^{2} A-1}{2 \cot A}\), cot A = \(\frac{\cot ^{2} \frac{A}{2}-1}{2 \cot \frac{A}{2}}\); (A is not an integral multiple of π)

→ sin 2A = \(\frac{2 \tan A}{1+\tan ^{2} A}\), sin A = \(\frac{2 \tan \frac{A}{2}}{1+\tan ^{2} \frac{A}{2}}\); (\(\frac{A}{2}\), is not an odd multiple of \(\frac{\pi}{2}\))

→ cos 2A = \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\), cos A = \(\frac{1-\tan ^{2} \frac{A}{2}}{1+\tan ^{2} \frac{A}{2}}\)

→ sin 3A = 3sin A – 4 sin³A

→ cos 3A = 4cos³A – 3 cos A

→ tan 3A = \(\frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A}\)

→ cot 3A = \(\frac{3 \cot A-\cot ^{3} A}{1-3 \cot ^{2} A}\)

→ sin(A + B) + sin (A – B) = 2sin A cos B

→ sin(A + B) – sin(A – B) = 2cos A sin B

→ cos (A + B) + cos (A – B) = 2cos A cos B

→ cos(A + B) – cos(A – B) = -2sin A sin B

→ sin C + sin D = 2sin \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)

→ sin C – sin D = 2cos \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)

→ cos C + cos D = 2 cos \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)

→ cos C – cos D = -2 sin \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)

→ For any A ∈ R
(a) sin A = ±\(\sqrt{\frac{1-\cos 2 A}{2}}\)
(b) cos A = ±\(\sqrt{\frac{1+\cos 2 A}{2}}\)
(c) IfA ¡s not an odd multiple of \(\frac{\pi}{2}\), then tan A = ±\(\sqrt{\frac{1-\cos 2 A}{1+\cos 2 A}}\)

→ sin \(\frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{2}}\)

→ cos \(\frac{A}{2}=\pm \sqrt{\frac{1+\cos A}{2}}\)

→ If A is not an odd multiple of n, then tan \(\frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{1+\cos A}}\)

→ If a ray \(\overrightarrow{\mathrm{OP}}\) makes an angle θ with the positive direction of X-axis then

• Sin θ = \(\frac{\mathrm{y}}{\mathrm{r}}\)
• cos θ = \(\frac{\mathrm{x}}{\mathrm{r}}\)
• tan θ = \(\frac{\mathrm{y}}{\mathrm{x}}\) (x ≠ 0)
• cot θ = \(\frac{\mathrm{x}}{\mathrm{y}}\) (y ≠ 0)
• sec θ = \(\frac{\mathrm{r}}{\mathrm{x}}\)(x ≠ 0)
• cosec θ = \(\frac{\mathrm{r}}{\mathrm{y}}\)(y ≠ 0)

Relations :

• sin θ cosec θ = 1
• cos θ sec θ = 1
• tan θ cot θ = 1
• sin² θ + cos² θ = 1
• 1 + tan²θ = sec² θ → (sec θ + tan θ)(sec θ – tan θ) = 1
  → sec θ + tan θ = \(\frac{1}{\sec \theta-\tan \theta}\) = 1
• 1 + cot²θ = cosec²θ → (cosec θ + cot θ) (cosec θ – cot θ) = 1
• sec²θ + cosec²θ = sec²θ. cosec²θ
• tan²θ – sin²θ = tan²θ . sin²θ;
  cot²θ – cos²θ = cot²θ. cos²θ
• sin²θ + cos⁴θ = 1 – sin²θ cos²θ
  = sin⁴θ + cos²θ
• sin⁴θ + cos⁴θ = 1 – 2sin²θ cos²θ
• sin⁶θ + cos⁶ θ = 1 – 3sin²θ cos²θ
• sin²x + cosec²x ≥ 2
• cos²x + sec² x ≥ 2
• tan² x + cot² x ≥ 2.

Values of trigonometric ratios of certain angles

Signs of Trigonometric ratios :
If lies in I, II, III, IV quadrants then the signs of trigonometric ratios are as follows.

Note :

• 0°, 90°, 180°, 270°. 360°, 450°, etc. are called quadrant angles.
• With “ALL SILVER TEA CUPS” symbol we can remember the signs of trigonometric ratios.

Coterminal angles :
If two angles differ by an integral multiples of 360o then two angles are called coterminal angles.
Thus 30°, 390°, 750°, 330° etc., are coterminal angles.

Complementary Angles :
Two Angles A, B are said to complementary ⇒ A + B = 90°

Supplementary angles :
Two angles A, B are said to be supplementary ⇒ A + B = 180°.

→ sin C + sin D = 2sin\(\frac{\mathrm{C}+\mathrm{D}}{2}\). cos \(\frac{\mathrm{C}-\mathrm{D}}{2}\)

→ sin C – sin D = 2cos\( . sin [latex]\frac{\mathrm{C}-\mathrm{D}}{2}\)

→ cos C + cos D = 2cos \(\frac{\mathrm{C}+\mathrm{D}}{2}\). cos \(\frac{\mathrm{C}-\mathrm{D}}{2}\)

→ cos C – cos D = 2sin\(\frac{\mathrm{C}+\mathrm{D}}{2}\). sin \(\frac{\mathrm{D}-\mathrm{C}}{2}\)

→ 2sin A cos B = sin(A + B) + sin(A – B)

→ 2cos A sin B = sin(A + B) – sin(A – B)

→ 2cos A cos B = cos(A + B) + cos(A – B)

→ 2sin A sin B = cos(A – B) – cos(A + B)
(or)
cos(A – B) – cos(A + B) = 2 sin A sin B.

→ \(\frac{\sin A+\sin B}{\sin A-\sin B}\) = tan\(\left(\frac{A+B}{2}\right)\).

→ If sin A + sin B = x, and cos A + cos B = y. Then

• tan\(\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)=\frac{\mathrm{x}}{\mathrm{y}}\)
• sin(A + B) = \(\frac{2 x y}{y^{2}+x^{2}}\)
• cos (A + B) = \(\frac{y^{2}-x^{2}}{y^{2}+x^{2}}\)
• tan(A + B) = \(\frac{2 x y}{y^{2}-x^{2}}\)

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 5 Products of Vectors to solve questions creatively.

Intermediate 1st Year Maths 1A Products of Vectors Formulas

Scalar or Dot Product of Two Vectors:
The scalar or dot product of two non – zero vectors \(\bar{a}\) and \(\bar{b}\), denoted by \(\bar{a} \cdot \bar{b}\) is defined as \(\bar{a} \cdot \bar{b}=|\bar{a}||\bar{b}|\) cos \((\bar{a}, \bar{b})\). This is a scalar, either \(\bar{a}\) = 0 (or) \(\bar{b}\) = 0, then we define \(\bar{a} \cdot \bar{b}\) = 0. If we write \((\bar{a}, \bar{b})\) = 0, then \(\bar{a} \cdot \bar{b}=|\bar{a}||\bar{b}|\) cos θ, if a ≠ 0, b ≠ 0, since 0 ≤ (a, b) = θ ≤ 7 80°, we get

• 0 ≤ θ < 90° ⇒ \(\bar{a}\). b > 0.
• θ = 90° ⇒ \(\bar{a} \cdot \bar{b}\) = 0 and the vectors \(\bar{a}\) and \(\bar{b}\) are perpendicular.
• 90° < θ ≤ 180° ⇒ \(\bar{a} \cdot \bar{b}\) < 0
• \(\bar{a} \cdot \bar{b}=\bar{b} \cdot \bar{a}\)
• a̅ (b̅ + c̅) = a̅ .b̅ + a̅ .c̅
• If a̅, b̅ are parallel, a̅.b̅ = ± |a̅ | |b̅ |.
• If l, m ∈ R, (la̅).(mb̅) = lm(a̅. b̅)
• Projection of b̅ on a̅ (or) length of the projection a̅ = \(\frac{|\bar{a} \cdot \bar{b}|}{|\bar{a}|}\)
• Orthogonal projection of b̅ on a̅ = \(\frac{(\bar{a} \cdot \bar{b})_{\bar{a}}}{|\bar{a}|^{2}}\); a̅ ≠ 0
  or
  The projection vector b̅ on a̅ = \(\left(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}|^{2}}\right)\) a̅ and it is magnitude = \(\frac{|\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}|}{|\overline{\mathrm{a}}|}\)
• The component vector of b̅ along a̅ (or) parallel to a̅ is \(\left(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}|^{2}}\right)\)a̅

Component vector of a̅ along b̅ = \(\frac{(\bar{a} \cdot \bar{b}) \bar{b}}{|\bar{b}|^{2}}\), component vector of a̅ perpendicular to b̅ = a̅ – \(\frac{(\bar{a} \cdot \bar{b}) \bar{b}}{|\bar{b}|^{2}}\)

Orthogonal unit vectors :
Ifi, j, k are orthogonal unit vector triad in a right handed system, then

• i̅ .j̅ = j̅.k̅ = k̅.i̅ = 0
• i̅ .i̅ = j̅.j̅ = k̅.k̅ = 1
• If r is any vector, r̅ = (r̅.i̅)i̅ +(r̅.j̅)j̅ ≠ (r̅.k̅)k̅

Some identities :
If a̅, b̅, c̅ are three vectors, then

• (a̅ + b̅)² = |a̅|² + |b̅|² + 2(a̅ . b̅)
• (a̅ – b̅)² = |a̅|² + |b̅|² – 2(a̅.b̅)
• (a̅ + b̅)² + (a̅ – b̅)² = 2(|a̅|² + |b̅|²)
• (a̅ + b̅)² – (a̅ – b̅)² = 4(a̅. b̅)
• (a̅ + b̅). (a̅ – b̅) = |a̅|² – |b̅|²
• (a̅ + b̅ + c̅)² = |a̅|² + |b̅|² + |c̅|² + 2(a̅ . b̅) + 2(b̅ . c̅) + 2(c̅ .a̅)

→ If a̅ = a₁ i̅ +a₂j̅ + a₃k̅ and b̅ = b₁i̅ + b₂j̅ + b₃k̅, then
a̅.b̅ = a₁b₁ + a₂b₂ + a₃b₃
a̅ is perpendicular to b̅
⇔ a₁b₁ + a₂b₂ + a₃b₃ = 0

→ |a̅| = \(\), |b̅| = \(\)

→ If (a̅, b̅) = then cos θ = \(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|}=\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{\sum a_{1}^{2}} \sqrt{\sum b_{1}^{2}}}\) and sin θ = \(\sqrt{\frac{\sum\left(a_{2} b_{3}-a_{3} b_{2}\right)^{2}}{\left(\sum a_{1}^{2}\right)\left(\sum b_{1}^{2}\right)}}\)

→ a̅ is parallel to b̅ ⇔ a₁: b₁ = a₂ : b₂ = a₃: b₃

→ a̅.a̅ >0; |a̅.b̅| < |a̅| |b̅|
|a̅ + b̅| ≤ |a̅| + |b̅|; |a̅ – b̅| ≤ |a̅| + |b̅| ;
|a̅ – b̅| ≥ |a̅| – |b̅|

Vector equations of a plane :

• The equation of the plane, whose perpendicular distance from the origin is p and whose unit normal drawn from the origin towards the plane is h is n̂ is r̅.n̂ = p.
• Equation of a plane passing through the origin and perpendicular to the unit vector n̅, is r̅.n̅ = 0
• Vector equation of a plane passing through a point A with position vector a and perpendicular to a vector n̅ is (r – a̅). n̅ = 0.

Perpendicular distance from the origin to the plane (r̅ – a̅).h = 0 is a̅. n̅ . where ‘a̅’ is the position vector of A in the plane and ‘n̅’ is a unit vector perpendicular to the plane.

Angle between two planes :
If π₁ and π₂2 be two planes and \(\bar{M}_{1}, \bar{M}_{2}\) are normals drawn to them, we define the angle between M₁ and M₂ as the angle between π₁ and π₂. If the angle between \(\) and \(\) is θ, the angle between the given planes θ = cos^-1\(\left[\frac{\bar{M}_{1} \cdot \bar{M}_{2}}{\left|\bar{M}_{1}\right|\left|\bar{M}_{2}\right|}\right]\)

Work done by a constant force F:

• If a constant force F̅ acting on a particle displaces it from a position ‘A’ to the position B, then the work done ‘W by this constant force T is the dot product of the vectors
  representing the force F̅ and displacement \(\overline{A B}\), i.e., W = F̅.\(\overline{A B}\).
• If F is the resultant of the forces F̅₁, F̅₂, ……………….F̅_n, then workdone in displacing the particle from A to B is
  \(\bar{W}=\bar{F}_{1} \cdot \overline{A B}+\bar{F}_{2} \cdot \overline{A B}+\ldots \ldots+F_{n} \cdot \overline{A B}\)

Cross Product or Vector Product of two vectors :
The vector product or cross product of two non-parallel non – zero vectors ‘a̅’ and ‘b̅’ is defined as a̅ × b̅ = |a̅||b̅| sin θ n̂, where ‘ n̂’ is a unit vector perpendicular to the plane containing ‘a̅’ and ‘b̅’ such that a̅, b̅ and ‘n̂’ form a vector triad in the right handed system and (a̅, b̅) = θ, this is a vector. If either of a̅, b̅ is a zero vector or ‘a̅’ is parallel to ‘b̅’, we define a̅ × b̅ = 0.

Some important results on vector product:

• |a̅ × b̅| = |a̅||b̅|sinθ ≤ |a̅||b̅| ;
• |a̅ × b̅| = |b̅ × a̅|
• a̅ × b̅ = -(b̅ × a̅):
• -a̅ × -b̅ = a̅ × b̅
• (-a̅) × b̅ = a̅ × (-b̅) – (a̅ × b̅)
• la̅ × mb̅ = lm(a̅ × b̅) ;
• a̅ × (b̅ + c̅) = a̅ × b̅ + a̅ × c̅
• a̅ ≠ 0, b̅ ≠ 0 and a̅ × b̅ = 0 ⇔ ‘a̅’ and ‘b̅’ are parallel vectors.
• a̅, b̅ , c̅ are non-zero vectors and a̅ × c̅ = b̅ × c̅ ⇒ either a̅ = b̅ or a̅ – b̅ is parallel to c̅.

Vector product among i. i and k:
If i̅, j̅ and k̅ are orthogonal unit vectors triad in the right handed system then

• i̅ × j̅ = j̅ × j = k̅ × k̅ = 0
• i̅ × j̅ = k̅ =-j̅ × i̅ ; j̅ × k̅ = k̅ × j̅ = i̅ ; k̅ × i̅ = -i̅ × k̅ = j̅
• If a̅ = a₁ i̅ + a₂ j + a₃k ; b̅ = b₁i̅ + b₂ j̅ + b₃k̅, then
  a̅ × b̅ = a₂b₃ – a₃b₂)i̅ + (a₃b₁ – a₁b₃)j̅ + (a₁b₂ – a₂b₁)k̅

This may be represented in the form of a determinants as a̅ × b̅ = \(\left|\begin{array}{ccc}
\bar{i} & \bar{j} & \bar{k} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{array}\right|\)

• Unit vectors perpendicular to both ‘a̅’and ‘b̅’ are ± \(\frac{\bar{a} \times \bar{b}}{|\bar{a} \times \bar{b}|}\)
• If a̅ = a₁i̅ + a₂j̅ + a₃k̅ ; b̅ = b₁i̅ + b₂ j̅ + b₃k̅ and (a̅, b̅) = θ, then
  sin θ = \(\frac{\sqrt{\sum\left(a_{2} b_{3}-a_{3} b_{2}\right)^{2}}}{\sqrt{\sum a_{1}^{2}} \sqrt{\sum b_{1}^{2}}}\) cos θ = \(\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{\sum a_{1}^{2}} \sqrt{\sum b_{1}^{2}}}\)

Vector areas:

• If \(\overline{A B}=\bar{c}\) and \(\overline{A C}=\bar{b}\) are two adjacent sides of a triangle ABC, then vector area of ΔABC = \(\frac{1}{2}\)(c̅ × b̅) and the area of the ΔABC = \(\frac{1}{2}\)|c̅ × b̅| $q. units.
• If a̅, b̅, c̅ are the position vectors of A, B, C respectively then the vector area of
  ΔABC = \(\frac{1}{2}\)[(b̅ × c̅) + (c̅ × a̅) + (a̅ × b̅)]
  Area of ΔABC = \(\frac{1}{2}\)|(b̅ × c̅) + (c̅ × a̅) + (a̅ × b̅)|sq. units.
• If \(\) and \(\) are the diagonals of a parallelogram ABCD, then the vector area of the parallelogram = \(\frac{1}{2}\)|a̅ × b̅| and area = \(\frac{1}{2}\)|a̅ × b̅|sq. units.
• If AB = a̅ and AD = b̅ are two adjacent sides of a parallelogram ABCD, then its vector area = a̅ × b̅ and area = |a̅ × b̅| sq. units.
• Vector area of the quadrilateral ABCD = \(\frac{1}{2}\)\((A C \times B D)\) and area of the quadrilateral ABCD = \(\frac{1}{2}\)\(|\overline{A C} \times \overline{B D}|\)sq. units.

Some useful formulas :

• If a̅, b̅ are two non-zero and non-parallel vectors then
  (a̅ × b̅)² = a²b² – (a̅.b̅)² = \(\left|\begin{array}{cc}
  a \cdot \bar{a} & \bar{a} \cdot \bar{b} \\
  \bar{a} \cdot b & b \cdot \bar{b}
  \end{array}\right|\)
• For any vector a̅,(a̅ × i̅)² + (a̅ × j̅)² + (a̅ × k̅)² = 2|a|²
• If a̅, b̅, c̅ are the position vectors of the points A, B, C respectively, then the perpendicular distance from c to the line AB is \(\frac{|\overline{A C} \times \overline{A B}|}{|\overline{A B}|}=\frac{|(\bar{b} \times \bar{c})+(\bar{c} \times \bar{a})+(\bar{a} \times \bar{b})|}{|b-\bar{a}|}\)

Moment of a force :
Let 0 be the point of reference (origin) and \(\overline{o p}=\bar{r}\) be the position vector of a point p on the line of action of a force F̅. Then the moment of the force F about 0 is given by r̅ × F̅.

Scalar triple product:
Let a̅, b̅, c̅ he three vectors. We call (a̅ × b̅). c̅ the scalar product of a̅, b and c. This is a scalar (real number). It is written as [a̅ b̅ c̅]

• If (a̅ × b̅). c̅ = 0, then one or more of the vectors a̅, b̅ and c̅ should be zero vectors. If a ≠ 0, b ≠ 0, c ≠ 0, then c is perpendicular to a̅ × b̅. Hence the vector c̅ lies on the plane determined by a̅ and b̅. Hence a̅, b̅ and c̅ are coplanar.
• If in a scalar triple product, any two vectors are parallel (equal), then the scalar triple product is zero i.e., [a̅ a̅ b̅] = [a̅ b̅ b̅] = [c̅ b̅ c̅] = 0.
• In a scalar triple product remains unaltered if the vectors are permutted cyclically i.e., [a̅ b̅ c̅] = [b̅ c̅ a̅] = [c̅ a̅ b̅].
  However [a̅ b̅ c̅] = -[b̅ a̅ c̅] = -[c̅ b̅ a̅] = -[a̅ c̅ b̅].
• In a scalar triple product, the dot and cross are interchangeable i.e., a̅.b̅ × c̅ = a̅ × b̅.c̅

→ If i̅ , j̅ , k̅ are orthogonal unit vector triad in the right handed system, then

• [i̅ j̅ k̅ ] = [j̅ k̅ i̅ ] = [k̅ i̅ j̅ ] = 1
• [i̅ k̅ j̅ ] = [j̅ i̅ k̅ ] = [k̅ j̅ i̅ ] = -1
• If a̅ = a₁ i̅ + a₂j̅ + a₃ k̅ ; b̅ = b₁i̅ + b₂ j̅ + b₃k̅ and c̅ = c₁i̅ + c₂ j̅ + c₃k̅ then [a̅ b̅ c̅] = \(\left|\begin{array}{lll}
  a_{1} & a_{2} & a_{3} \\
  b_{1} & b_{2} & b_{3} \\
  c_{1} & c_{2} & c_{3}
  \end{array}\right|\)

→ A necessary and sufficient condition that three non-parallel (non-collinear) and non-zero vectors a, b and c to be coplanar is [a̅ b̅ c̅] = 0. If [a̅ b̅ c̅] ≠ 0, then the three vectors are non-coplanar.

→ If a̅, b̅, c̅ are three non-zero, non-coplanar vectors and V is the volume of the parallelopiped with co-terminus edges a̅, b̅ and c̅, then v = |(a × b). c|. = |[a b c]|

→ The volume of the parallelopiped formed with A, B, C, D as vertices is \(|[A B A C A D]|\) cubic units.

→ If a̅, b̅, c̅ represent the co-terminus edges of a tetrahedron, then its volume = \(\frac{1}{6}\)[a̅, b̅, c̅] cubic units.

→ If A(x₁ y₁ z₁], B(x₂, y₂, z₂] C(x₃ y₃ z₃) and D(x₄, y₄ z₄] are the vertices of a tetrahedron = \(\frac{1}{6}\)\(|[A B A C A D]|\)

• Vector equation of a plane containing three non-collinear points a̅, b̅, c̅ is r̅ .[(b̅ × c̅) + (c̅ × a̅) + (a̅ × b̅)] = [a̅ b̅ c̅]
• A unit vector perpendicular to the plane containing three non-collinear points a̅, b̅, c̅ is \(\frac{(\bar{a} \times b)+(b \times \bar{c})+(\bar{c} \times \bar{a})}{|(\bar{a} \times \bar{b})+(b \times \bar{c})+(\bar{c} \times a)|}\)
• Length of the perpendicular from the origin to the plane containing three non-collinear points a̅, b̅, c̅ is \(\frac{|[\bar{a} b c]|}{|(\bar{a} \times \bar{b})+(\bar{b} \times \bar{c})+(\bar{c} \times \bar{a})|}\)

→ Vector equation of the plane passing through three non-collinear points a̅, b̅ and c̅ is [r̅ – a̅ b̅ – a̅ c̅ – a̅] = 0

→ Vector equation of the plane passing through a given point a̅ and parallel to the vectors b̅ and c̅ is [r b̅ c̅] = [a̅ b̅ c̅]

→ Vector equation of the line passing through the point a̅ and parallel to the vector b̅ is (r – a̅) × b̅ = 0

→ Distance of the point p(c) from a line joining the points A(a) and B(b) = |(c̅ – a̅) × b̅|
(25) i) Equation of the plane passing through the point p(x₁, y₁, z₁) and perpendicular to the vector ai̅ + bj̅ + ck̅ is a (x – x₁) + b(y – y₁) + c(z – z₁) = 0.

→ The equation of the plane passing through the points (x₁ y₁ z₁), (x₂, y₂, z₂) and
(x₃, y₃, z₃) is \(\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}
\end{array}\right|\) = 0

Skew lines:
Two lines l and m are called skew lines if there is no plane passing through these lines.

Shortest distance between the skew lines:
Shortest distance between the skew lines r̅ = a̅ + tb̅ and r̅ = c̅ + sd̅ is

Vector triple product:
If a̅, b̅ and c̅ are three vectors, products of the type (a̅ × b̅) × c̅, a̅ × (b̅ × c̅) from the vector triple products. From this definition.

• If any one of a̅, b̅ and c̅ is a zero vector, a̅ × (b̅ × c̅) or (a̅ × b̅) c̅ = 0
• If a̅ ≠ 0, b̅ ≠ 0, c̅ ≠ 0 and a̅ is parallel to b, then (a̅ × b̅) × c̅ = 0
• If a̅ ≠ 0, b̅ ≠ 0, c̅ ≠ 0 and c̅ is perpendicular to the plane of a and b, then (a̅ × b̅) × c̅ = 0.

→ If a̅ ≠ 0, b̅ ≠ 0, c̅ ≠ 0, a̅ and b̅ are non-parallel vectors and c̅ is not perpendicular to the plane passing through a̅ and b, then
(a̅ × b̅) × c̅ = (a̅.c̅)b̅ – (b̅.c)a̅
a̅ × (b̅ × c̅) = (a̅.c̅)b̅ – (a.b̅)c̅

• In general, vector triple product of three vectors need not satisfy the associative law. i.e., (a̅ × b) × c ≠ a̅ × (b × c)
  For any three vectors a̅, b̅ and c
• a̅ × (b̅ × c̅) + b̅ × (c̅ × a̅) + c̅ × (a̅ × b̅) = 0
• [a̅ × b̅ b̅ × c̅ c̅ × a̅] = [a̅ b̅ c̅]²

Scalar product of four vectors :
Scalar product of a̅, b̅, c̅ and d̅ is () = (a̅. c̅) (b̅.d̅) – (a̅. d̅) (b̅ .c̅) = \(\left|\begin{array}{ll}
\bar{a} \cdot \bar{c} & \bar{a} \cdot \bar{d} \\
\bar{b} \cdot \bar{c} & \bar{b} \cdot \bar{d}
\end{array}\right|\)

Vector product of four vectors:
If a̅, b̅, c̅ and d̅ are four vectors,
(a̅ × b̅) × (c̅ × d̅) = [a̅ c̅ d̅] b̅ – [b̅ c̅ d̅]a̅ – [a b̅ d̅]c̅ – [a̅ b̅ c̅]d̅

Some important results:

• [a̅ + b̅ b̅ + c̅ c̅ + a̅] = 2[a̅ b̅ c̅]
• i̅ × (j̅ × k̅) + i̅ × (k̅ × i̅) + k̅ × (i̅ × j̅) = 0
• [a̅ × b̅ b̅ × c̅ c̅ × a̅] = [a̅ b̅ c̅]²
  [a̅ b̅ c̅] [l̅ m̅ n̅] = \(\)
• If a̅, b̅, c̅ be such that a is perpendicular to (b̅ + c̅), bis perpendicular to (c̅ + a̅), c̅ is perpendicular to (a̅ + b̅), then |a̅ + b̅ + c̅| = \(\sqrt{a^{2}+b^{2}+c^{2}}\)
• If a line makes angles α, β, γ and δ with the diagonals of a cube, then
  cos²α + cos²β + cos²γ + cos²δ = \(\frac{4}{3}\)
• i̅ × (a̅ × i̅) + j̅ × (a̅ × j̅) + k̅ × (a̅ × k̅) – 2a̅

→ Equation of the sphere with centre at c and radius ‘a’ is r² – 2r̅. c̅ + c² = a²

Scalar Product
Def: Let \(\vec{a}, \vec{b}\) be two vectors dot product (or) scalar product (or) direct product (or) inner product denoted by \(\vec{a}, \vec{b}\). Which is defined as \(|\vec{a}||\vec{b}|\)cos θ where cos θ = \((\vec{a}, \vec{b})\)
* The product \(\vec{a}, \vec{b}\) is zero when \(|\vec{a}|\) = 0 (or) \(|\vec{b}|\) = 0 (or) θ = 90°.

Sign of the scalar product :
Let \(\vec{a}, \vec{b}\) are two non-zero vectors

• If θ is acute then \(\vec{a}.\vec{b}\)> 0 (i.e 0 < θ < 90°).
• If θ is obtuse then \(\vec{a}.\vec{b}\) < 0 (i.e 90° < θ < 180°).
• If θ = 90° then\(\vec{a} \cdot \vec{b}\) = o.
• If θ = 0° then \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|\)
• If θ = 180° then \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|\)

Note:

• The dot product of two vectors is always scalar.
• \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\) i.e dot product of two vectors is commutative.
• If \(\vec{a} \cdot \vec{b}\) are two vectors then \(\vec{a} \cdot(-\vec{b})=(-\vec{a}) \cdot \vec{b}=-(\vec{a} \cdot \vec{b})\)
• \((-\vec{a}) \cdot(-\bar{b})=\vec{a} \cdot \vec{b}\)
• If l,m are two scalars and \(\vec{a} \cdot \vec{b}\) are two vectors then \((l \bar{a}) \cdot(m \bar{b})={lm}(\vec{a} \cdot \vec{b})\)
• If \(\vec{a}\) and \(\vec{b}\) are two vectors then \(\vec{a} \cdot \vec{b}=\pm|\vec{a}||\vec{b}|\)
• If \(\vec{a}\) is a vector then \(\vec{a} \cdot \vec{a}=|\vec{a}|^{2}\)
• If \(\vec{a}\) is a vector \(\vec{a}\).\(\vec{a}\) is denoted by \(\overline{(a)^{2}}\) hence \(\overline{(a)^{2}}=|\vec{a}|^{2}\)

Components and orthogonal projection:
Def: Let \(\vec{a}=\overline{O A} \quad \vec{b}=\overline{O B}\) be two non zero vectors let the plane passing through B and perpendicular to a intersect \(\overline{O A}\) ln M.

• If \((\vec{a}, \vec{b})\) is acute then OM is called component of \(\vec{b}\) on \(\vec{a}\).
• If \((\vec{a}, \vec{b})\) is obtuse then -(OM) is called the component of \(\vec{b}\) on \(\vec{a}\).
• The vector \(\overline{O M}\) is called component vector of \(\vec{b}\) on \(\vec{a}\).

Def: Let \(\vec{a}=\overline{O A}\); \(\vec{b}=\overrightarrow{P Q}\) be two vectors let the planes passing through P, Q and perpendicular to a intersect \([latex]\)[/latex] in L, M respectively then \(\overline{L M}\) is called orthogonal projection of \(\vec{b}\) on \(\vec{a}\)

Note:

• The orthogonal projection of a vector b on a is equal tb component vector of b on a .
• Component of a vector \(\vec{b}\) on \(\vec{a}\) is also called projection of \(\vec{b}\) on \(\vec{a}\)
• If A< B, C, D are four points in the space then the component of \(\overline{A B}\) on \(\overline{C D}\) is same as the projection of \(\overline{A B}\) on the ray \(\overline{C D}\).

→ If \(\vec{a}, \vec{b}\) be two vectors (\(\vec{a} \neq \vec{o}\)) then

• The component of \(\vec{b}\) on \(\vec{a}\) is \(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}\)
• The orthogonal projection of \(\vec{b}\) on \(\vec{a}\) is \(\frac{(\vec{b} \cdot \vec{a}) \vec{a}}{|\vec{a}|^{2}}\)

→ If \(\vec{i}, \vec{j}, \vec{k}\) form a right handed system of Ortho normal triad then

• \(\vec{i} \cdot \vec{j}=\vec{j} \cdot \vec{j}=\vec{k} \cdot \vec{k}\) = 1
• \(\vec{i} \cdot \vec{j}=\vec{j} \cdot \vec{i}=0 ; \vec{j} \cdot \vec{k}=\vec{k} \cdot \vec{j}=0 ; \vec{k} \cdot \vec{i}=\vec{i} \cdot \vec{k}=0\)

→ If \(\vec{a}=a_{1} \vec{i}+a_{2} \vec{j}+a_{3} \vec{k}\) \(\vec{b}=b_{1} \vec{i}+b_{2} \vec{j}+b_{3} \vec{k}\) then \(\vec{a} \cdot \vec{b}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\)

→ If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors then

→ If \(\vec{r}\) is vector then \(\vec{r}=(\vec{r} \cdot \vec{i}) i+(\vec{r}+\vec{j}) \vec{j}+(\vec{r} \cdot \vec{k}) \vec{k}\)

Angle between the planes:
The angle between the planes is defined as the angle between the normals to the planes drawn from any point in the space.

Sphere:
The vector equation of a sphere with centre C having position vector \(\vec{c}\) and radius a is \((\vec{r}-\vec{c})^{2}\) = a² i.e. \(\vec{r}^{2}-2 \vec{r} \cdot \vec{c}+c^{2}\) = a²

The vector equation of a sphere with A(a) and B(b) as the end points of a diameter is \((\vec{r}-\vec{a}) \cdot(\vec{r}-\vec{b})\) = 0 (or) \((\vec{r})^{2}-\vec{r} \cdot(\vec{a}+\vec{b})+\vec{a} \cdot \vec{b}\) = 0

Work done by a force :
If a force \(\vec{F}\) acting on a particle displaces it from a position A to the position B then work done W by this force \(\vec{F}\) is \(\vec{F} \cdot \overline{A B}\)

• The vector equation of the plane which is at a distance of p from the origin along the unit vector \(\vec{n}\) is \(\vec{r} \cdot \vec{n}\) = p.
• The vector equation of the plane passing through the origin and perpendicular to the vector m is r.m =0
• The Cartesian equation of the plane which is at a distance of p from the origin along the unit vector n = li + mj + nk of the plane is n = lx + my + nz
• The vector equation of the plane passing through the point a having position vector \(\vec{a}\) and perpendicular to the vector \(\vec{m}\) is \((\vec{r}-\vec{a}) \cdot \vec{m}\) = 0.
• The vector equation of the plane passing through the point a having position vector \(\vec{a}\) and parallel to the plane r.m=q is \((\vec{r}-\vec{a}) \cdot \vec{m}\) = 0.

Cross( Vector) Product of Vectors:
Let \(\vec{a}, \vec{b}\) be two vectors. The cross product or vector product or skew product of vectors \(\vec{a}, \vec{b}\) is denoted by \(\vec{a} \times \vec{b}\) and is defined as follows

• If \(\vec{a}\) = 0 or \(\vec{b}\) = 0 or \(\vec{a}, \vec{b}\) are parallel then \(\vec{a} \times \vec{b}\) = 0
• If \(\vec{a}\) ≠ 0, \(\vec{b}\) ≠ 0, \(\vec{a}, \vec{b}\) are not parallel then \(\vec{a} \times \vec{b}=|\vec{a}||\vec{b}|(\sin \theta) \vec{n}\) where \(\vec{n}\) is a unit vector perpendicular to a and b so that a, b, n form a right handed system.

Note:

• \(\vec{a} \times \vec{b}\) is a vector
• If \(\vec{a}, \vec{b}\) are not parallel then \(\vec{a} \times \vec{b}\) is perpendicular to both a and b
• If \(\vec{a}, \vec{b}\) are not parallel then \(\vec{a}, \vec{b}\) , \(\vec{a} \times \vec{b}\) form a right handed system .
• For any vector \(\vec{a}\) \(\vec{a} \times \vec{b}\) = o

2. If \(\vec{a}, \vec{b}\) are two vectors \(\vec{a} \times \vec{b}=-\vec{b} \times \vec{a}\) this is called “anti commutative law”

3. If \(\vec{a}, \vec{b}\) are two vectors then \(\vec{a} \times(-\vec{b})=(-\vec{a}) \times \vec{b}=-(\vec{a} \times \vec{b})\)

4. If \(\vec{a}, \vec{b}\) are two vectors then \((-\vec{a}) \times(-\vec{b})=\vec{a} \times \vec{b}\)

5. If \(\vec{a}, \vec{b}\) are two vectors l,m are two scalars then (la) x (mb) = lm(a x b)

6. If \(\vec{a}, \vec{b}, \vec{b}\) are three vectos, then

• \(\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\)
• \((\vec{b}+\vec{c}) \times \vec{a}=\vec{b} \times \vec{a}+\vec{c} \times \vec{a}\)

7. If \(\vec{l}, \vec{l}, \vec{k}\) from a right handed system of orthonormal triad then

• \(\vec{l} \times \vec{l}=\vec{j} \times \vec{j}=\vec{k} \times \vec{k}=\vec{o}\)
• \(\vec{i} \times \vec{j}=\vec{k}=-\vec{j} \times \vec{l} ; \vec{j} \times \vec{k}=\vec{l}=-\vec{k} \times \vec{j} ; \vec{k} \times \vec{l}=\vec{j}=-\vec{l} \times \vec{k}\)

→ If \(\vec{a}=a_{1} \vec{l}+a_{2} \vec{j}+a_{3} \vec{k}, \vec{b}=b_{1} \vec{l}+b_{2} \vec{j}+b_{3} \vec{k}\) then \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\vec{l} & \vec{j} & \vec{k} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{array}\right|\)

→ If \(\vec{a}=a_{1} \vec{l}+a_{2} \vec{m}+a_{3} \vec{n}, \quad \vec{b}=b_{1} \vec{l}+b_{2} \vec{m}+b_{3} \vec{n}\) where \(\vec{l}, \vec{m}, \vec{n}\) form a right system of non coplanar vectors then \(\)

→ If \(\vec{a}, \vec{b}\) are two vectors then \((\vec{a} \times \vec{b})^{2}+(\vec{a} \cdot \vec{b})^{2}\) = a²b².

Vector Area:
If A is the area of the region bounded by a plane curve and \(\vec{n}\) is the unit vector perpendicular to the plane of the curve such that the direction of curve drawn can be considered anti clock wise then A\(\vec{n}\) is called vector area of the plane region bounded by the curve.

• The vector area of triangle ABC is \(\frac{1}{2} \overline{A B} \times \overrightarrow{A C}=\frac{1}{2} \overrightarrow{B C} \times \overrightarrow{B A}=\frac{1}{2} \overline{C A} \times \overrightarrow{C B}\)
• If \(\vec{a}, \vec{b}, \vec{c}\)are the position vectors of the vertices of a triangle then the vector area of the triangle is \(\frac{1}{2}(\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a})\)
• If ABCD is a parallelogram and \(\overrightarrow{A B}=\vec{a}, \quad \overrightarrow{B C}=\vec{b}\) then the vector area of ABCD is \(\vec{a} \times \vec{b}\).
• If ABCD is a parallelogram and \(\overrightarrow{A C}=\vec{a}, \overrightarrow{B C}=\vec{b}\) then vector area of parallelogram ABCD is \(\frac{1}{2}(\vec{a} \times \vec{b})\)
• The vector equation of a line passing through the point A with position vector a and perpendicular to the vectors \(\vec{b} \times \vec{c}\) is \(\vec{r}=\vec{a}+t(\vec{b} \times \vec{c})\).

Scalar Triple Product:

• If \(\vec{a}, \vec{b}, \vec{c}\) are the three vectors, then the real numbers \((\vec{a} \times \vec{b}) \cdot \vec{c}\) is called scalar triple product denoted by \([\vec{a} \vec{b} \vec{c}]\). This is read as ‘box’ \(\vec{a}, \vec{b}, \vec{c}\)
• If V is the volume of the parallelepiped with coterminous edges a, b, c then V = |\([\vec{a} \vec{b} \vec{c}]\)|
• If \(\vec{a}, \vec{b}, \vec{c}\) form the right handed system of vectors then V = \([\vec{a} \vec{b} \vec{c}]\)
• If \(\vec{a}, \vec{b}, \vec{c}\) form left handed system of vectors then -V = \([\vec{a} \vec{b} \vec{c}]\)

Note:

• The scalar triple product is independent of the position of dot and cross. i.e. \(\vec{a} \times \vec{b} \cdot \vec{c}=\vec{a} \cdot \vec{b} \times \vec{c}\)
• The value of the scalar triple product is unaltered so long as the cyclic order remains unchanged
  \([\vec{a} \vec{b} \vec{c}]=[\vec{b} \vec{c} \vec{a}]=[\vec{c} \vec{a} \vec{b}]\)
• The value of a scalar triple product is zero if two of its vectors are equal
  \([\vec{a} \vec{a} \vec{b}]\)= 0 \([\vec{b} \vec{b} \vec{c}]\) = 0
• If a, b, c are coplanar then \([\vec{a} \vec{b} \vec{c}]\) = 0
• If a,b,c form right handed system then \([\vec{a} \vec{b} \vec{c}]\) > 0
• If a,b,c form left handed system then \([\vec{a} \vec{b} \vec{c}]\) < 0
• The value of the triple product changes its sign when two vectors are interchanged
  \([\vec{a} \vec{b} \vec{c}]\) = –\([\vec{a} \vec{c} \vec{b}]\)
• If l,m, n are three scalars \(\vec{a}, \vec{b}, \vec{c}\) are three vectors then \(\)

→ Three non zero non parallel vectors abc nare coplanar iff \([l \vec{a} m \vec{b} \quad n \vec{c}]={lmn}\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)= 0

→ If \(\vec{a}=a_{1} \vec{l}+a_{2} \vec{m}+a_{3} \vec{n}, \vec{b}=b_{1} \vec{l}+b_{2} \vec{m}+b_{3} \vec{n}, \vec{c}=c_{1} \vec{l}+c_{2} \vec{m}+c_{3} \vec{n}\) where \(\vec{l}, \vec{m}, \vec{n}\) form a right handed system of non coplanar vectors, then \([\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}
\vec{m} \times \vec{n} & \vec{n} \times \vec{l} & \vec{l} \times \vec{m} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\)

→ The vectors equation of plane passing through the points A, B with position vectors \(\vec{a}, \vec{b}\) and parallel to the vector \(\vec{c}\) is \([\vec{r}-\vec{a} \vec{b}-\vec{a} \vec{c}]\) (or) \([\vec{r} \vec{b} \vec{c}]+[\vec{r} \vec{c} \vec{a}]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)

→ The vector equation of the plane passing through the point A with position vector \(\vec{a}\) and parallel to \(\vec{b}, \vec{c}\) is \([\vec{r}-\vec{a} \vec{b} \vec{c}]\) = 0 i.e. \(\left[\begin{array}{lll}
\vec{r} & \vec{b} & \vec{c}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)

Skew lines:
Two lines are said to be skew lines if there exist no plane passing through them i.e. the lines lie on two difference planes
Def:- l₁ and l₂ are two skew lines. If P is a point on l₁ and Q is a point on l₂ such that \(\overleftarrow{P Q}\) ⊥ l₁ and PQ ⊥ l₂ then \(\overleftarrow{P Q}\) is called shortest distance and \(\overleftarrow{P Q}\) is called shortest distance line between the lines l₁ and l₂.

The shortest distance between the skew lines \(\vec{r}=\vec{a}+t \vec{b}\) and \(\vec{r}=\vec{c}+t \vec{d}\) is \(\frac{|[\vec{a}-\vec{c} \vec{b} \vec{d}]|}{|\vec{b} \times \vec{d}|}\)

Vector Triple Product:
Cross Product of Three vectors : For any three vectors \(\bar{a}, \bar{b}\) or \(\bar{c}\) then cross product or vector product of these vectors are given as \(\bar{a} \times(\bar{b} \times \bar{c}),(\bar{a} \times \bar{b}) \times \bar{c}\) or \((\bar{b} \times \bar{c}) \times \bar{a}\) etc.

vi. If \(\bar{a}, \bar{b}\) or \(\bar{c}\) are non zero vectors and \(\) then b and c are parallel (or collinear) vectors.

vii. If \(\bar{a}, \bar{b}\) or \(\bar{c}\) are non zero and non parallel vectors then \(\bar{a} \times(\bar{b} \times \bar{c}), \quad \bar{b} \times(\bar{c} \times\bar{a})\) and \(\bar{c} \times(\bar{a} \times \bar{b})\) are non collinear vectors.

viii. If \(\bar{a}, \bar{b}\) or \(\bar{c}\) are any three vectors then \(\bar{a}(\bar{b} \times \bar{c})+\bar{b} \times(\bar{c} \times \bar{a})+\bar{c} \times(\bar{a} \times \bar{b})=\overline{\mathrm{O}}\)

ix. If \(\bar{a}, \bar{b}\) or \(\bar{c}\) are any three vectors then \(\bar{a}(\bar{b} \times \bar{c})+\bar{b} \times(\bar{c} \times \bar{a})+\bar{c} \times(\bar{a} \times \bar{b})\) are coplanar. [since sum of these vectors is zero]

x. \(\bar{a}(\bar{b} \times \bar{c})\) is vector lies in the plane of \(\bar{b}\) and \(\bar{c}\) or parallel to the plane of \(\bar{b}\) and \(\bar{c}\).

Product of Four Vectors:
Dot product of four vectors : The dot product of four vectors a̅, b̅, c̅ and d̅ is given as \((\bar{a} \times \bar{b}) \cdot(\bar{c} \times \bar{d})=(\bar{a} \cdot \bar{c})(\bar{b} \cdot \bar{d})-(\bar{a} \cdot \bar{d})(\bar{b} \cdot \bar{c})=\left|\begin{array}{ll}
\bar{a} \cdot \bar{c} & \bar{a} \cdot \bar{d} \\
\bar{b} \cdot \bar{c} & \bar{b} \cdot \bar{d}
\end{array}\right|\)

→ Cross product of four vectors : If a̅, b̅, c̅ and d̅ are any four vectors then

→ The vectorial equation of the plane passing through the point a and parallel to the vectors b̅, c̅is \([\overline{\mathrm{r}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]\)

→ The vectorial equation of the plane passing through the points a̅, b̅ and parallel to the vector c̅ is \([\overline{\mathrm{rb}} \overline{\mathrm{c}}]+[\overline{\mathrm{r}} \overline{\mathrm{c}} \overline{\mathrm{a}}]=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]\)

→ The vectorial equation of the plane passing through the points a̅, b̅, c̅ is \([\overline{\mathrm{r}} \overline{\mathrm{b}} \overline{\mathrm{c}}]+[\overline{\mathrm{r}} \overline{\mathrm{c}} \overline{\mathrm{a}}]+[\overline{\mathrm{ra}} \overline{\mathrm{b}}]=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]\)

→ If the points with the position vectorsa̅, b̅, c̅, d̅ are coplanar, then the condition is \([\overline{\mathrm{a}} \overline{\mathrm{bd}}]+[\overline{\mathrm{b}} \overline{\mathrm{c}} \overline{\mathrm{d}}]+[\overline{\mathrm{c}} \overline{\mathrm{a}} \overline{\mathrm{d}}]=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]\)

→ Length of the perpendicular from the origin to the plane passing through the points a̅, b̅, c̅ is \(\frac{|[\overline{\mathrm{a} b} \overline{\mathrm{c}}]|}{|\overline{\mathrm{b}} \times \overline{\mathrm{c}}+\overline{\mathrm{c}} \times \overline{\mathrm{a}}+\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}\)

→ Length of the perpendicular from the point c̅ on to the line joining the points a̅, b̅ is \(\frac{\mid(\overline{\mathrm{a}}-\overline{\mathrm{c}}) \times(\overline{\mathrm{c}}-\overline{\mathrm{b}})}{|\overline{\mathrm{a}}-\overline{\mathrm{b}}|}\)

→ P, Q, R are non collinear points. Then distance of P to the plane OQR is OP. \(\left|\frac{\overline{\mathrm{OP}} \cdot(\overline{\mathrm{OQ}} \times \overline{\mathrm{OR}})}{|\overline{\mathrm{OQ}} \times \overline{\mathrm{OR}}|}\right|\)

→ Perpendicular distance from P(α̅ ) to the plane passing through A(a̅) and parallel to the vectors b and c is

→ Length of the perpendicular from the point c̅ to the line \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\mathrm{tb}\) is \(\frac{|(\overline{\mathrm{c}}-\overline{\mathrm{a}}) \times \overline{\mathrm{b}}|}{|\overline{\mathrm{b}}|}\)

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 4 Addition of Vectors to solve questions creatively.

Intermediate 1st Year Maths 1A Addition of Vectors Formulas

Scalar :
A physical quantity which has only magnitude is called a scalar quantity. All the real numbers will be taken as scalars.

Vector :
A physical quantity which has both magnitude and direction.
e.g. : Velocity acceleration, force, momentum.

Position vector :
Let ‘O’ and ‘P be any points in space. Then OP is called the position vector of the point ‘P w.r.t. origin ‘O’.
Note : \(\overline{A B}\) = Position vector of B- position vector of A’
= \(\overline{O B}-\overline{O A}\)

Coinitial vector:
Vectors having the same initial point are called coinitial vectors.
e.g. : \(\overline{O A}, \overline{O B}, \overline{O C}\) etc.

Unit vector:
A vector whose magnitude is one-unit is called unit vector.
Unit vector in the direction of a is denoted by â = \(\frac{\bar{a}}{|a|}\).
For any non-zero vector \(\bar{a}=|\bar{a}|\) â.

Like vectors :
If two vectors are parallel and having the same direction then they are called like vectors.

Unlike vectors :
If two vectors are parallel and having Opposite direction then they are called unlike vectors.
The position vector of any point C on \(\overline{A B}\) can be taken as λ\(\bar{a}\) + µ\(\bar{a}\), where λ + µ = 1

Angle between vectors:
Let \(\overline{O A}=\bar{a}\) = a, \(\overline{O B}=\bar{a}\) = b be any two non – zero vectors, then angle AOB is defined as angle between vectors \(\bar{a}, \bar{b}\) and is denoted by \((\bar{a}, \bar{b})\) where 0 ≤ \((\bar{a}, \bar{b})\) ≤ 180°.

Addition of vectors (or) Parallelogram Law:
If \(\bar{a}, \bar{b}\) are the adjacent sides of a parallelogram the diagonals which is coinitial with \(\bar{a}, \bar{b}\) is given by \(\bar{a}+ \bar{b}\) and its magnitude is given by
\(|\bar{a}+\bar{b}|=\sqrt{|\bar{a}|^{2}+|\bar{b}|^{2}+2|\bar{a}||\bar{b}| \cos (\bar{a}, \bar{b})}\)

Triangle law:
If two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their sum is represented by the third side taken in the reverse order.

Note : Addition of vectors is of 2 types :

1. Commutative
2. Associative.

(ie) (i) \(\bar{a}+\bar{b}=\bar{b}+\bar{a}\) and (ii) \((\bar{a}+\bar{b})+\bar{c}=\bar{a}+(\bar{b}+\bar{c})\)
Rule: \(|\bar{a}| \sim|\bar{b}| \leq|\bar{a}-\bar{b}| \leq|\bar{a}+\bar{b}| \leq|\bar{a}|+|\bar{b}|\)

Parallel (or) Collinear vector:

1. Two vectors a, b are parallel or collinear, then a = λ. b, where ‘λ’ is a scalar.
2. If \(\bar{a}\) = (a₁, a₂, a₃), b = (b₁, b₂, b₃) are parallel or collinear, then
   \(\frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=\frac{a_{3}}{b_{3}}\)
   Note : Zero vector is parallel to any vector.
3. If three points with position vectors a, b, c are to be collinear. The necessary and sufficient condition is that there exists scalars \(\bar{a}, \bar{b}, \bar{c}\) not all zero, such that
   l\(\bar{a}\) + m\(\bar{b}\) + n\(\bar{c}\) = 0, l + m + n = 0.
4. If \(\bar{a}, \bar{b}, \bar{c}\) are non-zero, non-collinear vectors such that l\(\bar{a}\) + m\(\bar{b}\) + n\(\bar{c}\) = 0, then l = 0, m = 0, n = 0.

Linear combination of vectors :
A linear combination of the system of vectors \(\bar{a}_{1}, \bar{a}_{2}, \ldots, \bar{a}_{n}\) is a vector.
r = x₁\(\bar{a}_{1}\) + x₂\(\bar{a}_{2}\) + x₃\(\bar{a}_{3}\) + ………………. + x_n\(\bar{a}_{n}\)
where x₁ x₂, x₃, ………………., x_n are scalars.

Coplanar vectors:
If three or more vectors lie in the same plane (or) parallel to the same plane then they are called coplanar vectors. If one vector can be expressed as a linear combination of the remaining vectors, then the vectors are coplanar vectors.
If \(\bar{a}\) = x\(\bar{b}\) + y\(\bar{c}\), where x, y, are scalars, then \(\bar{a}, \bar{b}, \bar{c}\) are coplanar.

Linearly dependent system of vectors :
A system of vectors \(\bar{a}_{1}, \bar{a}_{2}, \bar{a}_{3}, \ldots \ldots, \bar{a}_{n}\) is said to
be linearly dependent, if there exists a system of scalars x₁, x₂, x₃, ……………….., x_n not all zero
such that
x₁\(\bar{a}_{1}\) + x₂\(\bar{a}_{2}\) + x₃\(\bar{a}_{3}\) + ………………. + x_n\(\bar{a}_{n}\) = \(\bar{0}\)

• The null vector is linearly dependent.
• Two collinear vectors are linearly dependent.
  \(\bar{a}\) = λ\(\bar{b}\) ⇒ (1)\(\bar{a}\) + (-λ)\(\bar{b}\) = 0
• Any three coplanar vectors are linearly dependent.
  \(\bar{a}\) = x\(\bar{b}\) + y\(\bar{c}\) ⇒ (1)\(\bar{a}\) + (-x)\(\bar{b}\) + (-y)\(\bar{c}\) = 0
• Any four vectors in space from a linearly dependent set of vectors.

Linearly independent system of vectors:
A system of vectors \(\bar{a}_{1}, \bar{a}_{2}, \bar{a}_{3}, \ldots \ldots, \bar{a}_{n}\) is said to be linearly independent, if x₁\(\bar{a}_{1}\) + x₂\(\bar{a}_{2}\) + x₃\(\bar{a}_{3}\) + ………………. + x_n\(\bar{a}_{n}\) = \(\bar{0}\) implies

• Two non-zero, non-collinear vectors are linearly independent.
• Any three non-coplanar vectors are linearly independent. If a, b, c are three non-coplanar vectors and \(\bar{r}\) be any other vector. Then there exists unique scalars x, y, z such that
  \(\bar{r}\) = x\(\bar{a}\) + y\(\bar{b}\) + z\(\bar{c}\).

Right handed system of vectors:
Three non-coplanar vectors \(\bar{a}, \bar{b}, \bar{c}\) are said to form a – right-handed system, if the rotation is form \(\bar{a}\) to \(\bar{b}\) in anti-clockwise direction, through an angle less than 180° as seen from the terminal point of c.
If \(\bar{a}, \bar{b}, \bar{c}\) from RHS and \(\bar{b}, \overline{,}, \bar{a}\) and \(\bar{c}, \bar{a}, \bar{b}\) also from RHS.

Left-handed system of vectors:
Three non-coplanar vectors \(\bar{a}, \bar{b}, \bar{c}\) are said to form a left handed system. If the rotation is from \(\bar{a}\) to \(\bar{b}\) in clockwise direction through an angle less than 180° as seen from terminal point of c.

If \(\bar{a}, \bar{b}, \bar{c}\) are in RHS, then
\(\bar{b}, \overline{,}, \bar{a}\); \(\bar{c}, \bar{a}, \bar{b}\) also from R.H.S.
and \(-\bar{a}, \bar{b}, \bar{c} ; \bar{a},-\bar{b}, \bar{c} ; \bar{a}, \bar{b},-\bar{c}\) form L.H.S.

Direction cosines of a vector and direction ratios of a vector:
If α, β, γ are the angles made by a line with +ve directions of x, y, z axes respectively, then cos α, cos β, cos γ are known as the direction cosines of that line.
Generally the direction cosines are denoted by l, m, n and l² + m² + n² = 1.
Any numbers proportional to the direction cosines are known as direction ratios.

→ Unit vector in the direction of \(\bar{a}\) = \(\frac{\bar{a}}{|\bar{a}|}\)

→ Unit vector parallel to the resultant of the vectors \(\frac{\bar{a}+\bar{b}+\bar{c}}{|\bar{a}+\bar{b}+\bar{c}|}\) is ± \(\frac{\bar{a}+\bar{b}+\bar{c}}{|\bar{a}+\bar{b}+\bar{c}|}\)

→ The vector parallel to the resultant of the vectors \(\frac{\bar{a}+\bar{b}+\bar{c}}{|\bar{a}+\bar{b}+\bar{c}|}\) and having magnitude λ is ±λ \(\frac{\bar{a}+\bar{b}+\bar{c}}{|\bar{a}+\bar{b}+\bar{c}|}\)

→ The position vector of the point which divides the line segment joining the points A and B whose position vectors are \(\bar{a}, \bar{b}\) respectively, internally in the ratio, l:m is \(\frac{m \overline{\mathrm{a}}+l \bar{b}}{l+m}\) externally in the ratio l: m is \(\frac{m \bar{a}-l \bar{b}}{m-l}\), m ≠ l.

→ The ratio in which the line joining the points A(x₁ y₁, z₁) and B(x₂, y₂, z₂) is divided by

• xy – plane is – z₁: z₂
• yz – plane is -x₁ : x₂
• zx – plane is -y₁: y₂

→ If ‘c’ is the mid-point of line AB, the position vector of ‘c’ is \(\overline{O C}=\frac{\overline{O A}+\overline{O B}}{2}\)
= \(\frac{\bar{a}+\bar{b}}{2}\)

→ If \(\bar{a}, \bar{b}\) are two unit vectors, then the unit vector along the bisector of the angle between \(\bar{a}, \bar{b}\) is given by
\(\bar{c}=\frac{\bar{a}+\bar{b}}{|\bar{a}+\bar{b}|}\)
or
\(\bar{c}=\frac{\bar{a}-\bar{b}}{|\bar{a}-\bar{b}|}\)

→ The Vector \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃), \(\bar{c}\) = (c₁, c₂, c₃) are coplanar or linearly dependent iff \(\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) = 0

→ The Vector \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃), \(\bar{c}\) = (c₁, c₂, c₃) are non-coplanar or linearly independent iff \(\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) ≠ 0

→ The necessary and sufficient condition for four points with position vectors \(\bar{a}, \bar{b}, \bar{c}, \bar{d}\) are coplanar is that there exists scalars l, m, n, p not all zero such that
l\(\bar{a}\) + m\(\bar{b}\) + n\(\bar{c}\) + p\(\bar{da}\) = \(\bar{0}\), l + m + n + p = 0.

→ If \(\bar{a}, \bar{b}, \bar{c}\) are three non-zero, non – coplanar vectors and x, y, z are three scalars such that x\(\bar{a}\) + y\(\bar{b}\) + z\(\bar{c}\) = 0, then x = 0, y = 0, z = 0.
Note : Collinearity implies coplanarity but coplanarity does not imply collinearity.

→ If \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃), \(\bar{c}\) = (c₁, c₂, c₃) and if \(\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) > 0 then \(\bar{a}, \bar{b}, \bar{c}\) are in R.H.S

→ If \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃), \(\bar{c}\) = (c₁, c₂, c₃) and if \(\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) < 0 then \(\bar{a}, \bar{b}, \bar{c}\) are in L.H.S

→ If \(\bar{r}\) = x\(\bar{i}\) + y\(\bar{j}\) + z\(\bar{k}\), then |\(\bar{r}\)| = r = \(\sqrt{x^{2}+y^{2}+z^{2}}\) .

→ If l, m, n are the direction cosines of a line, then l² + m² + n² = 1.

→ If a vector makes angles α, β, γ with co-ordinate axes, then

• cos² α + cos² β + cos² γ = 1
• sin² α + sin² β + sin² γ =2

→ The direction ratios of a vector \(\bar{r}\) = a\(\bar{i}\) + b\(\bar{j}\) + c\(\bar{k}\) are a, b, c.

→ If A = (x₁, y₁ z₁), B = (x₂, y₂, z₂) then the direction ratios of a vector
AB are x₂ – x₁, y₂ – y₁, z₂ – z₁

→ If a, b, c are the direction ratios of a vector then direction cosines of a vector are
\(\pm \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \pm \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, \pm \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

→ If a, b, c are the direction ratios of a line then λa, λb, λc also become the direction ratios of that line where ‘λ’ is a non-zero scalar.

→ If \(\bar{r}\) = x\(\bar{i}\) + y\(\bar{j}\) + z\(\bar{k}\), then the direction cosines of a vector are \(\frac{a}{|\vec{r}|}, \frac{b}{|\vec{r}|}, \frac{c}{|\vec{r}|}\)

→ If \(\bar{r}\) = x\(\bar{i}\) + y\(\bar{j}\) + z\(\bar{k}\) makes angles α, γ with co-ordinate axes respectively, then

cos α = \(\frac{a}{|\vec{r}|}\),
cos β = \(\frac{b}{|\vec{r}|}\)
cos γ = \(\frac{c}{|\vec{r}|}\).

• The direction cosines of x- axis are 1, 0, 0.
• The direction cosines of y – axis are 0, 1,0.
• The direction cosines of z- axis are 0, 0, 1.

→ If a vector make equal angles with co-ordinate axes then direction cosines of a vector are \(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}\)

→ If l, m, n are direction cosines of a vector OP and ‘O’ is the origin OP = r, then P = (lr, mr, nr).

→ If 1, m, n are direction cosines of a vector, then the maximum value of lmn is \(\frac{1}{3 \sqrt{3}}\)

→ The maximum value of l + m + n = \(\frac{1}{3 \sqrt{3}}\)

→ The vector equation of a straight-line passing through the point whose position vector is a and parallel to the vector \(\bar{b}\) is \(\bar{r}=\bar{a}+t \bar{b}\), where ‘t’ is parameter.

→ In cartesian form its equation is \(\frac{x-a_{1}}{b_{1}}=\frac{y-a_{2}}{b_{2}}=\frac{z-a_{3}}{b_{3}}\)
where \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃).

→ The vector equation of a straight-line passing through the point whose position vectors are
a and b is \(\bar{r}\) = (1 – t)\(\bar{a}\) (or) \(\bar{r}=\bar{a}+t(\bar{b}-\bar{a})\); where’t’ is a parameter.

→ In cartesian form \(\frac{x-a_{1}}{b_{1}-a_{1}}=\frac{y-a_{2}}{b_{2}-a_{2}}=\frac{z-a_{3}}{b_{3}-a_{3}}\)
where \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃).

→ Vector equation of the plane through the points whose position vectors are \(\bar{a}, \bar{b}, \bar{c}\) is \(\bar{r}=(1-s-t) \bar{a}+s \bar{b}+t \bar{c}\), where s and t are parameters (or)
\(\bar{r}=\bar{a}+s(\bar{b}-\bar{a})+t(\bar{c}-\bar{a})\).
where \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃), \(\bar{c}\) = (c₁, c₂, c₃)

→ The vector equation of the plane through the point whose position vector is \(\bar{a}\) and parallel to the vectors \(\bar{b}\) and \(\bar{c}\) is \(\bar{r}=\bar{a}+s \bar{b}+t \bar{c}\), where ‘s’ and ‘t’ are parameter.
In cartesian form it is \(\left|\begin{array}{ccc}
x-a_{1} & y-a_{2} & z-a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) = 0

→ Vector equation of the plane through the point’s whose position vectors are a, b and parallel to vector c is
\(\bar{r}=(1-s) \bar{a}+s \bar{b}+t \bar{c}\)
or
\(\bar{r}=\bar{a}+s(\bar{b}-\bar{a})+t \bar{c}\)
In cartesian form it is \(\left|\begin{array}{ccc}
x-a_{1} & y-a_{2} & z-a_{3} \\
b_{1}-a_{1} & b_{2}-a_{2} & b_{3}-a_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) = 0

Scalar:
A quantity which has only magnitude but no directions is called scalar quantity.
Ex: Length, mass, time ………….

Vector:
A quantity which has both magnitude and direction is called a vector quantity
Ex: Displacement, Velocity, Force ………………
A vector can also be denoted by a single letter \(\vec{a}, \vec{b}, \vec{c}\) …….. or bold letter a, b, c
Length of \(\vec{a}\) is dinded by |\(\vec{a}\)|. Length of \(\vec{a}\) is called magnitude of \(\vec{a}\).

Zero vector (Null Vector):
The vector of length O and having any direction is called null vector. It is denoted by \(\bar{O}\)

Note:

• If A is any point in the space then \(\overrightarrow{A A}=\vec{O}\)
• A non zero vector is called a proper vector.

Free vector:
A vector which is independent of its position is called free vector

Localised vector:
If \(\vec{a}\) is a vector P is a point then the ordere of pair, (P,a) is called localized vector at P

Multiplication of a vector by a scalar:

• Let m be any scalar and \(\vec{a}\) be any vector then the vector m \(\vec{a}\) is defined as
• Length of m \(\vec{a}\) is |m| times of length of \(\vec{a}\) i.e. |m\(\vec{a}\)| = |m||a|
• The line of support of m\(\vec{a}\) is same or parallel to that of \(\vec{a}\)

The sense the direction of m\(\vec{a}\) is same as that of \(\vec{a}\) if m is positive, the direction of m\(\vec{a}\) is opposite to that \(\vec{a}\) if m is negative

Note:

• o\(\vec{a}\) = o
• m\(\vec{0}\) = \(\vec{0}\)
• (mn)\(\vec{a}\) = m(n\(\vec{a}\)) = n(m\(\vec{a}\))
• (-1)\(\vec{a}\)

Negative of a vector:
\(\vec{a}\) ,\(\vec{b}\) are two vectors having same length but their directions are opposite to each other then each vector is called the negative of the other vector.
Here \(\vec{a}\) = –\(\vec{b}\) and \(\vec{b}\) = –\(\vec{a}\)

Unit vector:
A vector whose magnitude is unity is called unit vector.
Note: If \(\vec{a}\) is any vector then unit vector is \(\frac{\vec{a}}{|\vec{a}|}\) this is denoted by â {read as ‘a’ cap}

Collinear or parallel vectors:
Two or more vectors are said to the collinear vectors if the have same line of support. The vectors are said to be parallel if they have parallel lines of support.

Like parallel vectors:
Vectors having same direction are called like parallel vectors.

Unlike parallel vectors:
Vectors having different direction are called unlike parallel vectors.

Note:

• If \(\vec{a}\), \(\vec{b}\) are two non-zero collinear or parallel vectors then there exists a non zero scalar m such that \(\vec{a}\) = m\(\vec{b}\)
• Conversly if there exists a relation of the type \(\vec{a}\) = m\(\vec{b}\) between two non zero two non zero vectors a, b then a, b must be parallel or collinear.

Cointialvectors:
The vectors having the same initial point are called co-initial vectors.

Co planar vectors:
Three or more vectors are said to be coplanar if they lie on a plane parallel to same plane. Other wise the vectors are non coplanar vectors.

Angle between two non-zero vectors:
Let \(\overline{O A}=\vec{a}\) and \(\overline{O B}=\vec{b}\) be two non-zero vectors. Then the angle between \(\vec{a}\) and \(\vec{b}\) is defined as that angle AOB where O ≤ ∠AOB ≤ 180°

The angle between \(\vec{a}\) and \(\vec{b}\) is denoted by (\(\vec{a}\), \(\vec{b}\))

Note:

• If \(\vec{a}\) and \(\vec{b}\)are any two vectors such that (\(\vec{a}\), \(\vec{b}\)) = 0 then o ≤ 0 ≤ 180° {this is the range of θ i.e. angle beween vectors}
• \((\vec{a}, \vec{b})=(\vec{b}, \vec{a})\)
• \((\vec{a},-\vec{b})=(-\vec{a}, \vec{b})\) = 180° – \((\vec{a}, \vec{b})\)
• \((-\vec{a},-\vec{b})=(\vec{a}, \vec{b})\)

• If k > 0; 1 > 0 then \((k \vec{a}, l \vec{b})=(\vec{a}, \vec{b})\)
• If \((\vec{a}, \vec{b})\) = 0 then \(\vec{a}, \vec{b}\) are like parallel vectors
• If \((\vec{a}, \vec{b})\) = 180° then \(\vec{a}, \vec{b}\) are unlike parallel vectors
• If \((\vec{a}, \vec{b})\) = 90° then \(\vec{a}, \vec{b}\) are orthogonal or perpendicular vectors.

Addition of vectors:
Let \(\vec{a}\) and \(\vec{b}\) be any two given vectors. If three points O, A, B are taken such that \(\overline{O A}=\vec{a}\), \(\overrightarrow{A B}=\vec{b}\) then vectors \(\) is called the vector sum or resultant of the given vectors a and b we write \(\overline{O B}=\overline{O A}+\overline{A B}=\vec{a}+\vec{b}\)

Triangular law of vectors:
Trianglular law states that if two vectors are represented in magnitude and direction by two sides of a triangle taken in order, then their sum or resultant is represented in magnitude and direction by the third side of the triangle taken in opposite direction.

Position vector:
Let O be a fixed point in the space called origin. If P is any point in the space then \(\) is called position vector of P relative to O.

Note:
If \(\vec{a}\) and \(\vec{b}\) be two non collinear vectors, then there exists a unique plane through a,b this plane is called plane generated by a, b. If \(\overline{O A}=\vec{a} ; \overline{O B}=\vec{b}\) then the plane generated by \(\vec{a}\), \(\vec{b}\) is denoted by \(\overline{A O B}\).

• Two non zero vectors a, b are collinear if m\(\vec{a}\) + n\(\vec{b}\) = 0 for some scalars m,n not both zero
• Let \(\vec{a}, \vec{b}, \vec{c}\) be the position vectors of the points A, B, C respectively. Then A, B, C are collinear iff m\(\vec{a}\) + n\(\vec{b}\) + p\(\vec{c}\) = 0 for some scalars m, n, p not all zero such that m + n + p = 0
• Let \(\vec{a}, \vec{b}\) be two non collinear vectors, If \(\vec{r}\) is any vector in the plane generated by a,b then there exist a unique pair of real numbers x, y such that \(\vec{r}\) = x\(\vec{a}\) + y\(\vec{b}\).
• Let \(\vec{a}, \vec{b}\) be two non collinear vectors. If \(\vec{r}\) is any vector such that \(\vec{r}\) = xa + yb for some real numbers then \(\vec{r}\) lies in the plane generated by a, b.
• Three vectors \(\vec{a}, \vec{b}, \vec{c}\) coplanar iff x\(\vec{a}\) + y\(\vec{b}\) + z\(\vec{c}\) = 0 for some scalars x, y, z not all zero.
• Let \(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) be the position vectors of the points A, B, C, D in which no three of them are collinear. Then A, B, C, D are coplanar iff m\(\vec{a}\) + n\(\vec{b}\) + p\(\vec{c}\) + q\(\vec{d}\) = 0 for some scalars m,n,p,q not all zero such that m + n + p + q = 0
• If \(\vec{a}, \vec{b}, \vec{c}\) are three non coplanar vectors and \(\vec{r}\) is any vector then there exist a unique traid of real numbers. x, y, z such that \(\vec{r}\) = x\(\vec{a}\) + y\(\vec{b}\) + z\(\vec{c}\).

Right handed system of orthonormal vectors
A triced of three non-coplanar vectors\(\vec{i}, \vec{j}, \vec{k}\) is said to be a right hand system of orthonormal triad of vectors if

• \(\vec{i}, \vec{j}, \vec{k}\) from a right handed system
• \(\vec{i}, \vec{j}, \vec{k}\) are unit vectors
• (i, j) = 900 = \((\vec{j}, \vec{k})=(\vec{k}, i)\)

Right handed system:
Let \(\overline{O A}=\vec{a}, \overline{O B}=\vec{b}\) and \(\overrightarrow{O C}=\vec{c}\) be three non coplanar vectors. If we observe from the point C that a rotation from OA to OB through an angle not greater than 1800 is in the anti clock wise direction then the vectors a,b, c are said to form ‘Right handed system’.

Left handed system:
If we observe from C that a rotation from OA to OB through an angle not greater than 1800 is in the clock -wise direction then the vectors a,b, c are said to form a “Left handed system”.
If \(\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}\) then \(|\vec{r}|=\sqrt{x_{2}+y_{2}+z_{2}}\).

Direction Cosines:
If a given directed line makes angles α, β, γ, with positive direction of axes of x,y, and z respectively then cos α,cos β,cos γ are called direction cosines of the line and these are denoted by l,m,n.

Direction ratios:
Thre real numbers a,b,c are said to be direction ratios of a line if a:b:c = l:m:n where l,m,n are the direction cosines of the line.

Linear combination:
Let \(\vec{a}_{1}, \bar{a}_{2}, \overline{a_{3}} \ldots \ldots \overline{a_{n}}\) be n vectors and l₁,l₂,l₃….l_n be n scalars then \(l_{1} \bar{a}_{1}+l_{2} \overline{a_{2}}+l_{3} \overline{a_{3}}+\ldots+l_{n} \overline{a_{n}}\) is called a. linear combination of \(\vec{a}_{1}, \bar{a}_{2} \ldots \ldots . \overline{a_{n}}\)

Linear dependent vectors:
The vectors \(\bar{a}_{1}, \overline{a_{2}} \ldots \ldots \overline{a_{n}}\) are said to be linearly dependent if there exist scalars l₁,l₂,l₃….l_n not all zero such that \(_{1} \vec{a}_{1}+l_{2} \overline{a_{2}}+\ldots+l_{n} \overline{a_{n}}]\) = 0

Linear independent The vectors a₁, a₂, a₃ ……………. a_n are said to be linearly independt if l₁, l₂, l₃ ……………. l_n are scalars, \(l_{1} \vec{a}_{1}+l_{2} \overline{a_{2}}+\ldots l_{n} \overline{a_{n}}\) = 0
⇒ l₁ = 0 l₂ = o…. l_n = o

• Let \(\vec{a}, \vec{b}\) be the position vectors of A, B respectively the position vector of the point P which divides \(\overline{A B}\) in the ratio m:n is \(\frac{m \vec{b}+n \vec{a}}{m+n}\). Conversely the point P with position vector \(\frac{m \vec{b}+n \vec{a}}{m+n}\) lies on the lines \(\overline{A B}\) and divides \(\overline{A B}\) in the ratio m:n.
• The medians of a triangle are concurrent. The point of concurrence divides each median in the ratio 2:1
• Let ABC be a triangle and D be a point which is not in the plane of \(\overline{A B C}\) the lines joining O, A, B,C with the centroids of triangle ABC, triangle BCD, triangle CDA and triangle DAB respectively are concurrent and the point of concurrence divides each line segment in the ratio 3:1
• The equation of the line passing through the point A = (x₁, y₁, z₁) and parallel to the vector b = (l, m, n) is \(\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}\) = t
• The equation of the line having through the points A(x₁, y₁, z₁), B(x₂, y₂, z₂) is \(\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\) = t
• The unit vector bisecting the angle between the vectors \(\vec{a}, \vec{b}\) is \(\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}\)
• The internal bisector the angle between \(\vec{a}, \vec{b}\) is \(\frac{\overline{O P}}{O P}=\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}\) are concurrent

Theorem 1:
The vector equation of a line parallel to the vector \(\vec{b}\) and passing through the point A with position vector \(\vec{a}\) is \(\vec{r}=\vec{a}+t \vec{b}\) where t is a scalar.
Proof:
Let \(\overline{O A}=t \vec{a}\) be the given point and \(\overline{O P}=\vec{r}\) be any point on the line
\(\overline{A P}=t \vec{b}\) where ‘t’ is a scalar
\(\vec{r}-\vec{a}=t \vec{b} \Rightarrow \vec{r}=\vec{a}+t \vec{b}\)

Theorem 2:
The vector equation of the line passing through the points A, B whose position vectors \(\vec{a}, \vec{b}\) respectively is \(\vec{r}=\vec{a}(1-t)+t \vec{b}\) where t is a scalar.
Proof:
let P a a point on the line joining of A, B

Theorem 3:
The vector equation of the plane passing through the point A with position vector \(\vec{b}, \vec{c}\) and parallel to the vectors \(\vec{a}\) is \(\vec{r}=\vec{a}+s \vec{b}+t \vec{c}\) where s,t are scalars
Proof:
Given that \(\overline{O A}=\vec{a}\)
Let \(\overline{O P}=\vec{r}\) be the position vector of P
\(\overrightarrow{A P}, \vec{b}, \vec{c}\) are coplanar
∴ \(\overrightarrow{A P}=s \vec{b}+t \vec{c} \Rightarrow \vec{r}-\vec{a}=s \vec{b}+t \vec{c}\)
∴ \(\vec{r}=\vec{a}+s \vec{b}+t \vec{c}\)

Theorem 4:
The vector equation of the plane passing through the points A, B with position vectors \(\vec{a}, \vec{b}\) and parallel to the vector \(\vec{c}\) is \(\vec{r}=(1-s) \vec{a}+s \vec{b}+t \vec{c}\)
Proof:
Let P be a point on the plane and \(\overline{O P}=\vec{r}\)
\(\overline{O A}=\vec{a} \overline{O B}=\vec{b}\) be the given points \(\overline{A P}, \overrightarrow{A B}, \vec{C}\) are coplanar
\(\overrightarrow{A P}=s \overrightarrow{A B}+t \vec{C}\)
\(\vec{r}-\vec{a}=s(\vec{b}-\vec{a})+t \vec{c}\)
\(\vec{r}=(1-s) \vec{a}+s \vec{b}+t \vec{c}\)

Theorem 5:
The vector equation of the plane passing through the points A, B, C having position vectors \(\vec{a}, \vec{b}, \vec{c}\) is \(\vec{r}=(1-s-t) \vec{a}+s \vec{b}+t \vec{c}\) where s,t are scalars
Proof:
Let P be a point on the plane and \(\overline{O P}=\vec{r}\)
\(\overline{O A}=\vec{a} \overline{O B}=\vec{b}, \overline{O C}=\bar{c}\)
Now the vectors \(\overline{A B}, \overline{A C}, \overline{A P}\) are coplanar.
Therefore, \(\overline{A P}=s \overline{A B}+t \overline{A C}\), t,s are scalars.
\(\bar{r}-\bar{a}=s(\bar{b}-\bar{a})+t(\bar{c}-\bar{a})\)
\(\vec{r}=(1-s-t) \vec{a}+s \vec{b}+t \vec{c}\)

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 2 Mathematical Induction to solve questions creatively.

Intermediate 1st Year Maths 1A Mathematical Induction Formulas

Principle of finite mathematical induction:
Let S be a subset of N such that

• 1 ∈ S
• For any k ∈ N, k ∈S ⇒ (k + 1) ∈ S

Then S = N

Principle of complete mathematical induction:
Let S be a subset of N such that

• 1 ∈ S
• For any k ∈ N {1, 2, 3 … k} ⊆ S
  ⇒ (k + 1) ∈ S

Then S = N

Steps to prove a statement using the principle of mathematical induction :

• Basis of induction : Show that P(1) is true
• Inductive hypothesis : For k > 1, assume that P(k) is true
• Inductive Step : Show that P(k + 1) is true on the basis of the inductive hypothesis.

Principle of finite Mathematical Induction:
Let {P(n) / n ∈ N} be a set of statements. If

• p(1) is true
• p (m) is true ⇒ p (m+1) is true ; then p (n) is true for every n ∈ N.

Principle of complete induction:
Let {P (n) / n N} be a set of statements. If p (1) is true and p(2), p(3) …. p (m-1) are true ⇒ p(m) is true, then p (n) is true for every n e N.

Note:

• The principle of mathematical induction is a method of proof of a statement.
• We often use the finite mathematical induction, hence or otherwise specified the mathematical induction is the finite mathematical induction.

Some important formula:

• Σn = \(\frac{n(n+1)}{2}\)
• Σn² = \(\frac{n(n+1)(2 n+1)}{6}\)
• Σn³ = \(\frac{n^{2}(n+1)^{2}}{4}\)
• a, (a + d), (a + 2d), ……….. are in a.p
  n th term t_n = a + (n – 1)d, sum of n terms S_n = \(\frac{n}{2}\)[ 2a + (n – 1)d] = \(\frac{n}{2}\)[a + l]
  a = first term, l= last term.
• a, ar, ar², ………… is a g.p
  Nth terms t_n = a.r^n-1 a = 1st term, r = common ratio
• Sum of n terms s_n = a\(\frac{\left(r^{n}-1\right)}{r-1}\); r > 1 = a\(\left(\frac{1-r^{n}}{1-r}\right)\); r < 1

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b)

All problems in this exercise have reference to ΔABC.

I.

Question 1.
Express \(\Sigma r_{1} \cot \frac{A}{2}\) in terms of s.
Solution:
\(\Sigma r_{1} \cot \frac{A}{2}\) = \(\Sigma\left(s \tan \frac{A}{2}\right) \cot \frac{A}{2}\)
= Σs
= s + s + s
= 3s

Question 2.
Show that Σa cot A = 2(R + r).
Solution:
L.H.S = Σa . cot A
= Σ2R sin A \(\frac{\cos A}{\sin A}\)
= 2R Σ cos A
= \(2 R\left(1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)\) (From transformants)
= \(2\left(R+4 R \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right)\)
= 2(R + r)
= R.H.S

Question 3.
In ∆ABC, prove that r₁ + r₂ + r₃ – r = 4R.
Solution:

Question 4.
In ∆ABC, prove that r + r₁ + r₂ – r₃ = 4R cos C.
Solution:

Question 5.
If r + r₁ + r₂ + r₃ then show that C = 90°.
Solution:

II.

Question 1.
Prove that 4(r₁r₂ + r₂r₃ + r₃r₁) = (a + b + c)²
Solution:

Question 2.
Prove that \(\left(\frac{1}{r}-\frac{1}{r_{1}}\right)\left(\frac{1}{r}-\frac{1}{r_{2}}\right)\left(\frac{1}{r}-\frac{1}{r_{3}}\right)=\frac{a b c}{\Delta^{3}}=\frac{4 R}{r^{2} s^{2}}\)
Solution:

Question 3.
Prove that r(r₁ + r₂ + r₃) = ab + bc + ca – s².
Solution:

Question 4.
Show that \(\sum \frac{r_{1}}{(s-b)(s-c)}=\frac{3}{r}\)
Solution:

Question 5.
Show that \(\left(r_{1}+r_{2}\right) \tan \frac{C}{2}=\left(r_{3}-r\right) \cot \frac{C}{2}=c\)
Solution:

Question 6.
Show that r₁r₂r₃ = \(r^{3} \cot ^{2} \frac{A}{2} \cdot \cot ^{2} \frac{B}{2} \cdot \cot ^{2} \frac{C}{2}\)
Solution:

III.

Question 1.
Show that cos A + cos B + cos C = 1 + \(\frac{r}{R}\)
Solution:
L.H.S = cos A + cos B + cos C
= 2 cos(\(\frac{A+B}{2}\)) cos(\(\frac{A-B}{2}\)) + cos C

Question 2.
Show that \(\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}+\cos ^{2} \frac{C}{2}=2+\frac{r}{2 R}\)
Solution:

Question 3.
Show that \(\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=1-\frac{r}{2 R}\)
Solution:

Question 4.
Show that
(i) a = (r₂ + r₃) \(\sqrt{\frac{r r_{1}}{r_{2} r_{3}}}\)
(ii) ∆ = r₁r₂ \(\sqrt{\frac{4 R-r_{1}-r_{2}}{r_{1}+r_{2}}}\)
Solution:

Question 5.
Prove that \(r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r^{2}\) = 16R² – (a² + b² + c²).
Solution:

Question 6.
If p₁, p₂, p₃ are altitudes drawn from vertices A, B, C to the opposite sides of a triangle respectively, then show that
(i) \(\frac{1}{p_{1}}+\frac{1}{p_{2}}+\frac{1}{p_{3}}=\frac{1}{r}\)
(ii) \(\frac{1}{p_{1}}+\frac{1}{p_{2}}-\frac{1}{p_{3}}=\frac{1}{r_{3}}\)
(iii) p₁ . p₂ . p₃ = \(\frac{(a b c)^{2}}{8 R^{3}}=\frac{8 \Delta^{3}}{a b c}\)
Solution:

Question 7.
If a = 13, b = 14, c = 15, show that R = \(\frac{65}{8}\), r = 4, r₁ = \(\frac{21}{2}\), r₂ = 12 and r₃ = 14.
Solution:
a = 13, b = 14, c = 15
s = \(\frac{a+b+c}{2}\)
= \(\frac{13+14+15}{2}\)
= 21
s – a = 21 – 13 = 8
s – b = 21 – 14 = 7
s – c = 21 – 15 = 6

Question 8.
If r₁ = 2, r₂ = 3, r₃ = 6 and r = 1, prove that a = 3, b = 4 and c = 5.
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(a)

All problems in this exercise refer to ΔABC

I.

Question 1.
Show that Σa(sin B – sin C) = 0
Solution:

Question 2.
If a = √3 + 1 cms, ∠B = 30°, ∠C = 45°, then find c.
Solution:
∠B = 30°, ∠C = 45° and a = (√3 + 1) cms
A = 180° – (B + C)
= 180° – (30° + 45°)
= 180° – 75°
= 105°

Question 3.
If a = 2 cms, b = 3 cms, c = 4 cms, then find cos A.
Solution:
a = 2 cms, b = 3 cms and c = 4 cms

Question 4.
If a = 26 cms, b = 30 cms and cos C = \(\frac{63}{65}\), then find c.
Solution:
a = 26 cms, b = 30 cms and cos C = \(\frac{63}{65}\)
c² = a² + b² – 2ab cos C
⇒ c² = 676 + 900 – 2 × 26 × 30 × \(\frac{63}{65}\)
⇒ c² = 1576 – 1512
⇒ c² = 64
⇒ c = 8

Question 5.
If the angles are in the ratio 1 : 5 : 6, then find the ratio of its sides.
Solution:
Given \(\frac{A}{1}=\frac{B}{5}=\frac{C}{6}\), B = 5A, C = 6A
A + B + C = 180°
⇒ A + 5A + 6A = 180°
⇒ 12A = 180°
⇒ A = 15°
∴ B = 5A = 75°
∴ C = 6A = 90°
a : b : c = sin A : sin B : sin C
= sin 15° : sin 75° : sin 90°
= \(\frac{\sqrt{3}-1}{2 \sqrt{2}}: \frac{\sqrt{3}+1}{2 \sqrt{2}}: 1\)
= (√3 – 1) : (√3 + 1) : 2√2

Question 6.
Prove that 2(bc cos A + ca cos B + ab cos C) = a² + b² + c².
Solution:
L.H.S = Σ2bc cos A
= Σ2bc \(\frac{\left(b^{2}+c^{2}-a^{2}\right)}{2 b c}\)
= Σ(b² + c² – a²)
= b² + c² – a² + c² + a² – b² + a² + b² – c²
= a² + b² + c²
= R.H.S

Question 7.
Prove that \(\frac{a^{2}+b^{2}-c^{2}}{c^{2}+a^{2}-b^{2}}=\frac{\tan B}{\tan C}\)
Solution:
L.H.S = \(\frac{a^{2}+b^{2}-c^{2}}{c^{2}+a^{2}-b^{2}}=\frac{\tan B}{\tan C}\)
[∵ c² = a² + b² – 2ab cos C and b² = a² + c² – 2ac cos B]

Question 8.
Prove that (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c
Solution:
L.H.S = (b + c) cos A + (c + a) cos B + (a + b) cos C
= (b cos A + c cos A) + (c cos B + a cos B) + (a cos C + b cos C)
= (b cos C + c cos B) + (a cos C + c cos A) + (a cos B + b cos A)
= a + b + c
= R.H.S

Question 9.
Prove that (b – a cos C) sin A = a cos A sin C
Solution:
LHS = (b – a cos C) sin A
= (a cos C + c cos A – a cos C) sin A
= c cos A sin A [∵ b = a cos C + c cos A]
= (2R sin C) cos A sin A
= a cos A sin C (∵ 2R sin A = a)

Question 10.
If 4, 5 are two sides of a triangle and the included angle is 60°, find its area.
Solution:
Let a = 4, b = 5 are two sides and included angle is C = 60° then
Area of ∆ = \(\frac{1}{2}\) ab sin C
= \(\frac{1}{2}\) × 4 × 5 × sin 60°
= 2 × 5 × \(\frac{\sqrt{3}}{2}\)
= 5√3 sq.cm

Question 11.
Show that \(b \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{B}{2}=s\)
Solution:

Question 12.
If \(\frac{a}{\cos A}=\frac{b}{\cos B}=\frac{c}{\cos C}\), then show that ∆ABC is equilateral.
Solution:
Given that \(\frac{a}{\cos A}=\frac{b}{\cos B}=\frac{c}{\cos C}\)
⇒ \(\frac{2 R \sin A}{\cos A}=\frac{2 R \sin B}{\cos B}=\frac{2 R \sin C}{\cos C}\)
⇒ \(\frac{\sin A}{\cos A}=\frac{\sin B}{\cos B}=\frac{\sin C}{\cos C}\)
⇒ tan A = tan B = tan C
⇒ A = B = C
⇒ ∆ABC is an equilateral triangle.

II.

Question 1.
Prove that a cos A + b cos B + c cos C = 4R sin A sin B sin C.
Solution:
L.H.S = (2R sin A) cos A + (2R sin B) cos B + (2R sin C) cos C
= R (sin 2A + sin 2B + sin 2C)
= R (2 sin (A + B) cos (A – B) + sin 2C)
= R [2 sin (180° – C) cos (A – B) + sin 2C]
= R (2 sin C . cos (A – B) + 2 sin C . cos C)
= 2R sin C (cos (A – B)) + cos C)
= 2R sin C (cos (A – B) + cos (180° – \(\overline{\mathrm{A}+\mathrm{B}}\))
= 2R sin C [cos (A – B) – cos (A + B)]
= 2R sin C (2 sin A sin B)
= 4R sin A sin B sin C
= R.H.S.

Question 2.
Prove that Σa³ sin(B – C) = 0.
Solution:
L.H.S = Σa² [a sin (B – C)]
= Σa² [2R . sin A sin (B – C)]
= R Σ a² (2 sin (180° – \(\overline{\mathrm{B}+\mathrm{C}}\)) sin (B – C))
= R Σ a² [2 sin (B + C) . sin (B – C)]
= R Σ a² (sin² B – sin² C)
= R Σ a² \(\left(\frac{b^{2}}{4 R^{2}}-\frac{c^{2}}{4 R^{2}}\right)\)
= \(\frac{1}{2 R}\) Σ[a² (b² – c²)]
= \(\frac{1}{2 R}\) a² (b² – c²) + b² (c² – a²) + c² (a² – b²)]
= \(\frac{1}{2 R}\) (a²b² – a²c² + b²c² – a²b² + a²c² – b²c²)
= \(\frac{1}{2 R}\) × 0
= 0
= R.H.S

Question 3.
Prove that \(\frac{a \sin (B-C)}{b^{2}-c^{2}}=\frac{b \sin (C-A)}{c^{2}-a^{2}}=\frac{c \sin (A-B)}{a^{2}-b^{2}}\)
Solution:

Question 4.
Prove that \(\sum a^{2} \frac{a^{2} \sin (B-C)}{\sin B+\sin C}=0\)
Solution:

Question 5.
Prove that \(\frac{a}{b c}+\frac{\cos A}{a}=\frac{b}{c a}+\frac{\cos B}{b}=\frac{c}{a b}+\frac{\cos C}{c}\)
Solution:

Question 6.
Prove that \(\frac{1+\cos (A-B) \cos C}{1+\cos (A-C) \cos B}=\frac{a^{2}+b^{2}}{a^{2}+c^{2}}\)
Solution:

Question 7.
If C = 60°, then show that
(i) \(\frac{\mathbf{a}}{\mathbf{b}+\mathbf{c}}+\frac{\mathbf{b}}{\mathbf{c}+\mathbf{a}}=1\)
(ii) \(\frac{b}{c^{2}-a^{2}}+\frac{a}{c^{2}-b^{2}}=0\)
Solution:
∠C = 60°
⇒ c² = a² + b² – 2ab cos C
⇒ c² = a² + b² – 2ab cos 60°
⇒ c² = a² + b² – 2ab (\(\frac{1}{2}\))
⇒ c² = a² + b² – ab

Question 8.
If a : b : c = 7 : 8 : 9, find cos A : cos B : cos C.
Solution:

Question 9.
Show that \(\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}\)
Solution:

Question 10.
Prove that (b – a) cos C + c (cos B – cos A) = c . sin(\(\frac{A-B}{2}\)) cosec(\(\frac{A+B}{2}\))
Solution:
L.H.S = b cos C – a cos C + c cos B – c cos A
= (b cos C + c cos B) – (a cos C + c cos A)
= a – b
= 2R (sin A – sin B)

Question 11.
Express \(a \sin ^{2} \frac{C}{2}+c \cdot \sin ^{2} \frac{A}{2}\) interms of s, a, b, c.
Solution:

Question 12.
If b + c = 3a, then rind the value of \(\cot \frac{\mathrm{B}}{2} \cot \frac{\mathrm{C}}{2}\)
Solution:

Question 13.
Prove that (b + c) cos (\(\frac{B+C}{2}\)) = a cos (\(\frac{B-C}{2}\))
Solution:

Question 14.
In ΔABC, show that \(\frac{b^{2}-c^{2}}{a^{2}}=\sin \frac{(B-C)}{(B+C)}\)
Solution:

III.

Question 1.
Prove that
(i) \(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\frac{s^{2}}{\Delta}\)
(ii) \(\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}\) = \(\frac{b c+c a+a b-s^{2}}{\Delta}\)
(iii) \(\frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}{\cot A+\cot B+\cot C}\) = \(\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}\)
Solution:

Question 2.
Show that
(i) Σ(a + b) tan(\(\frac{A-B}{2}\)) = 0
(ii) \(\frac{\mathbf{b}-\boldsymbol{c}}{\mathbf{b}+\mathbf{c}} \cot \frac{A}{2}+\frac{b+c}{b-c} \tan \frac{A}{2}\) = 2 cosec(B – C)
Solution:

Question 3.
(i) If sin θ = \(\frac{a}{b+c}\), then show that cos θ = \(\frac{2 \sqrt{b c}}{b+c} \cdot \cos \frac{A}{2}\)
Solution:

(ii) If a = (b + c) cos θ, then prove that sin θ = \(\frac{2 \sqrt{b c}}{b+c} \cos \left(\frac{A}{2}\right)\)
Solution:

(iii) For any anlge θ show that a cos θ = b cos (C + θ) + c cos (B – θ).
Solution:
b cos (C + θ) + c cos (B – θ)
= b (cos C . cos θ – sin C sin θ) + c (cos B cos θ + sin B sin θ)
= (b cos C + c cos B) cos θ + (-b sin C + C sin B) sin θ
= a cos θ + (-2R sin B sin C + 2R sin B sin C) sin θ
= a cos θ

Question 4.
If the angles of ∆ABC are in A.P and b : c = √3 : √2 , then show that A = 75°.
Solution:
∵ The angles A, B, C of a triangle are in A.P.
⇒ 2B = A + C
⇒ 3B = A + B + C
⇒ 3B = 180°
⇒ B = 60°

Question 5.
If \(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}=\frac{\sin C}{\sin (A-B)}\), prove that ∆ABC is either isosceles or right-angled.
Solution:

⇒ sin 2A = sin 2B
⇒ A = B
⇒ ∆ABC is isosceles
or 2A = 180° – 2B
or A = 90° – B
or A + B = 90°
so A ≠ B ⇒ ∆ABC is a right-angled triangle
∴ ∆ABC is either isosceles (or) right-angled.

Question 6.
If cos A + cos B + cos C = \(\frac{3}{2}\), then show that the triangle is equilateral.
Solution:

Question 7.
If cos² A + cos² B + cos² C = 1, then show that ∆ABC is right-angled.
Solution:
Given cos² A + cos² B + cos² C = 1 …….(1)
Now cos² A + cos² B + cos² C
= cos² A + 1 – sin² B + cos² C
= 1 + (cos² A – sin² B) + cos² C
= 1 + cos (A + B) cos (A – B) + cos² C
= 1 + cos (180° – C) cos (A – B) + cos² C
= 1 – cos C . cos (A – B) + cos² C
= 1 – cos C (cos (A – B) – cos C)
= 1 – cos C (cos (A – B) – cos (180° – \(\overline{A+B}\)))
= 1 – cos C (cos (A – B) + cos (A + B))
= 1 – cos C (2 cos A cos B)
= 1 – 2 cos A cos B cos C
Substituting in (1) we get
1 – 2 cos A cos B cos C = 1
∴ 2 cos A cos B cos C = 0
∴ cos A = 0 or cos B = 0 or cos C = 0
i.e., A = 90° or B = 90° or C = 90°
∴ ∆ABC is right-angled.

Question 8.
If a² + b² + c² = 8R², then prove that the triangle is right angled.
Solution:
Given a² + b² + c² = 8R²
⇒ 4R² (sin² A + sin² B + sin² C) = 8R²
⇒ sin² A + sin² B + sin² C = 2 ……(1)
Now sin² A + sin² B + sin² C
= 1 – cos² A + sin² B + sin² C
= 1 – (cos² A – sin² B) + sin² C
= 1 – cos (A + B) . cos (A – B) + sin² C
= 1 – cos (180° – C) cos (A – B) + sin² C
= 1 + cos C cos (A – B) + 1 – cos² C
= 2 + cos C (cos (A – B) – cos C)
= 2 + cos C (cos (A – B) – cos (180° – \(\overline{A+B}\)))
= 2 + cos C (cos (A – B) + cos (A + B))
= 2 + cos C (2 cos A cos B)
= 2 + 2 cos A cos B cos C
Substituting in (1), we get
2 + 2 cos A cos B cos C = 2
2 cos A cos B cos C = 0
⇒ cos A = 0 or cos B = 0 or cos C = 0
∴ A = 90° or B = 90° or C = 90°
∴ ΔABC is right-angled.

Question 9.
If cot \(\frac{A}{2}\), cot \(\frac{B}{2}\), cot \(\frac{C}{2}\) are in A.P., then prove that a, b, c are in A.P.
Solution:
∵ cot \(\frac{A}{2}\), cot \(\frac{B}{2}\), cot \(\frac{C}{2}\) are in A.P
⇒ \(\frac{(s)(s-a)}{\Delta}, \frac{(s)(s-b)}{\Delta}, \frac{(s)(s-c)}{\Delta}\) are in A.P.
⇒ s – a, s – b, s – c are in A.P
⇒ -a, -b, -c are in A.P
⇒ a, b, c are in A.P

Question 10.
If \(\sin ^{2} \frac{A}{2}, \sin ^{2} \frac{B}{2}, \sin ^{2} \frac{C}{2}\) are in H.P., show that a, b, c are in H.P.
Solution:

Question 11.
If C = 90° then prove that \(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\) sin (A – B) = 1
Solution:

Question 12.
Show that \(\frac{a^{2}}{4} \sin 2 C+\frac{c^{2}}{4} \sin 2 A\) = ∆
Solution:
LHS = \(\frac{a^{2}}{4} \sin 2 C+\frac{c^{2}}{4} \sin 2 A\)
= \(\frac{4 R^{2} \sin ^{2} A}{4}\) 2 sin C cos C + \(\frac{4 R^{2} \sin ^{2} C}{4}\) 2 sin A cos A
= 2R² sin²A sin C cos C + 2R² sin²C sin A cos A
= 2R² sin A sin C (sin A cos C + cos A sin C)
= 2R² sin A sin C sin(A + C)
= 2R² sin A sin C sin(180 – B)
= 2R² sin A sin B sin C
= ∆
= RHS

Question 13.
A lamp post is situated at the middle point I the side AC of a triangular plot A with BC = 7 meters, CA = 8 meters, and AB = 9 meters, lamp post subtends an angle of 15° at point B. find H height of the lamp post.
Solution:
Let MP be the height of the lamp post

Question 14.
Two ships leave a port at the same time. One goes 24 km. per hour in the direction N45°E and other travels 32 km per hour in the direction S75°E. Find the distance between the ships at the end of 3 hours.
Solution:
Given the first ship goes 24 km per hour After 3 hours the first ship goes 72 km.

Given the second ship goes 32 km per hour After 3 hours, the second ship goes 96 km.
Let AB = x
∠AOB = 180° – (75 + 45) = 60°
Apply cosine rule for ∆AOB,
cos 60° = \(\frac{(72)^{2}+(96)^{2}-x^{2}}{2(72)(96)}\)
⇒ \(\frac{1}{2}=\frac{5184+9216-x^{2}}{13824}\)
⇒ 13824 = 28800 – 2x²
⇒ 2x² = 14976
⇒ x² = 7488
⇒ x = 86.4 (Appx)
At the end of 3 hours the difference between ships was 86.4 km.

Question 15.
A tree stands vertically on the slant of the hill. From A point on the ground 35 meters down the hill from the base of the tree, the angle, the elevation of the top of the tree is 60° if the angle of elevation of the foot of the tree from A is 15°, then find the height of the tree.
Solution:
Let BC be the height of the tree
BC = h
Let BD = x, AD = y
Given AB = 35 m

Question 16.
The upper \(\frac{3}{4}\)th portion of a vertical pole subtends an angle tan-1(\(\frac{3}{5}\)) at a point in the horizontal plane through its foot and at a distance 40 m from the foot. Given that the vertical pole is at a height less than 100 m from the ground, find its height.
Solution:

⇒ 6400 + h² = 200h
⇒ h² – 200h + 6400 = 0
⇒ h² – 160h – 40h + 6400 = 0
⇒ h(h – 160) – 40(h – 160) = 0
⇒ (h – 160) (h – 40) = 0
⇒ h = 40 or 160
but the height of the pole should be less than 100m
∴ h = 40 m

Question 17.
AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of point A from a certain point C on the ground is 60°. He moves away from the pole along the line BC to a point D such that CD = 7 m. From D, the angle of elevation of point A is 45°. Find the height of the pole.
Solution:
Let ‘h’ be the height of the pole
AB = h
given CD = 7 m
∠ACB = 60°, ∠ADB = 45°, Let BC = x

Question 18.
Let an object be placed at some height h cm and let P and Q two points of observation which are at a distance of 10 cm apart on a line inclined at an angle of 15° to the horizontal. If the angles of elevation of the object from P and Q are 30° and 60° respectively then find h.
Solution:
Let AB = Height of the object from ‘A’ = h m
Given that P & Q are two observations,
PQ = 10 cms

From ∆APB,
∠P = 30; ∠A = 90; ∠B = ?
A + P + B = 180°
⇒ 30° + 30° + B = 180°
⇒ B = 180° – 120°
⇒ B = 60°
From ∆BQC,
∠Q = 60°; ∠C = 90°; ∠B = ?
Q + C + B = 180°
⇒ 60° + 90° + B = 180°
⇒ B = 180° – 150°
⇒ B = 30°
From ∆BQP,
∠P = 15; ∠B = 30; ∠Q = ?
P + B + Q = 180°
⇒ 15° + 30° + Q = 180°
⇒ Q = 180° – 45°
⇒ Q = 135°
Applying the ‘sin’ rule for ∆BQP

From ∆PAB,
sin 30° = \(\frac{B A}{B P}\)
BP . sin 30° = AB = h
√2 × 10 × \(\frac{1}{2}\) = AB = h
5√2 = AB = h
∴ h = 5√2

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Hyperbolic Functions Solutions Exercise 9(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Hyperbolic Functions Solutions Exercise 9(a)

Question 1.
If sinh x = \(\frac{3}{4}\), find cosh (2x) and sinh (2x).
Solution:

Question 2.
If sinh x = 3, then show that x = log_e(3 + √10).
Solution:

Question 3.
Prove that
(i) tanh (x – y) = \(\frac{\tanh x-\tanh y}{1-\tanh x \tanh y}\)
Solution:

(ii) coth (x – y) = \(\frac{{coth} x \cdot {coth} y-1}{{coth} y-{coth} x}\)
Solution:

Question 4.
Prove that
(i) (cosh x – sinh x)ⁿ = cosh (nx) – sinh (nx), for any n ∈ R.
Solution:

(ii) (cosh x + sinh x)ⁿ = cosh (nx) + sinh (nx), for any n ∈ R.
Solution:

Question 5.
Prove that \(\frac{\tanh x}{{sech} x-1}+\frac{\tanh x}{{sech} x+1}\) = -2 cosech x, for x ≠ 0
Solution:

Question 6.
Prove that \(\frac{\cosh x}{1-\tanh x}+\frac{\sinh x}{1-{coth} x}\) = sinh x + cosh x, for x ≠ 0
Solution:

Question 7.
For any x ∈ R, prove that cosh⁴x – sinh⁴x = cosh (2x)
Solution:
L.H.S = cosh⁴x – sinh⁴x
= (cosh²x)² – (sinh²x)²
= [cosh²x – sinh²x] [cosh²x + sinh²x]
= (1) cosh (2x)
= cosh (2x)
∴ cosh⁴x – sinh⁴x = cosh (2x)

Question 8.
If u = \(\log _{e}\left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right)\) and if cos θ > 0,then prove that cosh u = sec θ.
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(e) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(e)

I.

Question 1.
Find the adjoint and inverse of the following matrices.
(i) \(\left[\begin{array}{cc}
2 & -3 \\
4 & 6
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right]\)
Solution:

Question 2.
If A = \(\left[\begin{array}{cc}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right]\), a² + b² + c² + d² = 1, then find the inverse of A.
Solution:

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
0 & -1 & 4 \\
-2 & 2 & 1
\end{array}\right]\), then find A^-1
Solution:

Question 4.
If A = \(\left|\begin{array}{ccc}
-1 & -2 & -2 \\
2 & 1 & -2 \\
2 & -2 & 1
\end{array}\right|\), then show that the adjoint of A = 3A’ find A^-1.
Solution:

Question 5.
If abc ≠ 0, find the inverse of \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right]\)
Solution:

II.

Question 1.
If A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\) and B = \(\frac{1}{2}\left[\begin{array}{lll}
b+c & c-a & b-a \\
c-b & c+a & a-b \\
b-c & a-c & a+b
\end{array}\right]\) then show that ABA^-1 is a diagonal matrix.
Solution:

Question 2.
If 3A = \(\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
-2 & 2 & -1
\end{array}\right]\) then show that A^-1 = A’
Solution:

Question 3.
If A = \(\left[\begin{array}{rrr}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right]\), then show that A^-1 = A³
Solution:

∴ A⁴ = I
det A = 3(1) – 3(-2) + 4(-2) = 1
∵ A ≠ 0 ⇒ A^-1 exists
∵ A⁴ = I
Multiply with A^-1
A⁴ (A^-1) = I (A^-1)
⇒ A³ (AA^-1) = A^-1
⇒ A³ (I) = A^-1
∴ A^-1 = A³

Question 4.
If AB = I or BA = I, then prove that A is invertible and B = A^-1
Solution:
Given AB = I
⇒ AB| = |1|
⇒ |A| |B| = 1
⇒ |A| ≠ 0
∴ A is a non-singular matrix and BA = I
⇒ |BA| = |I|
⇒ |B| |A| = 1
⇒ |A| ≠ 0
∴ A is a non-singular matrix.
AB = I or BA = I, A is invertible.
∴ A^-1 exists.
AB = I
⇒ A^-1AB = A^-1I
⇒ IB = A^-1
⇒ B = A^-1
∴ B = A^-1