Question

The acute angle between the planes P 1 and P 2, when P 1 and P 2 are the planes passing through the intersection of the planes 5 x + 8 y + 13 z - 29 = 0 and 8 x - 7 y + z - 20 = 0 and the points (2, 1, 3) and (0, 1, 2), respectively, is

• A. $\frac{π}{3}$
• B. $\frac{π}{4}$
• C. $\frac{π}{6}$
• D. $\frac{π}{12}$

Answer 1

Answer: (A)

$\frac{π}{3}$

Answer 2

The correct answer is (A) : $\frac{π}{3}$
Family of Plane’s equation can be given by
(5 + 8λ)x + (8 – 7λ)y + (13 + λ) z – (29 + 20λ) = 0
P 1 passes through (2, 1, 3)
⇒ (10 + 16λ) + (8 – 7λ) + (39 + 3λ) – (29 + 20λ) = 0
⇒ –8λ + 28 = 0
$⇒ λ =\frac{7}{2}$
d.r,s of normal to P 1
$⟨33, \frac{−33}{2} ,\frac{33}{2}⟩ or ⟨1, \frac{−1}{2}, \frac{1}{2}⟩$
P 2 passes through (0, 1, 2)
8 – 7λ + 26 + 2λ – (29 + 20λ) = 0
⇒ 5 – 25λ = 0
$⇒ λ = \frac{1}{5}$
d.r, s of normal to P 2
$⟨ \frac{33}{5} ,\frac{33}{5}, \frac{66}{5}⟩ or ⟨1, 1, 2⟩$
Angle between normals
$= \frac{( \hat{i} -\frac{1}{2}\hat{j} + \frac{1}{2}\hat{k} ) . ( \hat{i} + \hat{j} + 2\hat{k} )}{ \frac{\sqrt{3}}{2} \hspace{1cm}\sqrt6}$
$cosθ = \frac{1 -\frac{1}{2} + 1}{3} = \frac{1}{2}$
$θ = \frac{π}{3}$