Question

Let the plane
$P : \stackrel{→}{r} . \stackrel{→}{a} = d$
contain the line of intersection of two planes
$\stackrel{→}{r} . ( \hat{i} + 3\hat{j} - \hat{k} ) = 6$
and
$\stackrel{→}{r} . ( -6\hat{i} + 5\hat{j} - \hat{k} ) = 7$
. If the plane P passes through the point (2, 3, 1/2),
then the value of $\frac{| 13a→|² }{d²}$ is equal to

• A. 90
• B. 93
• C. 95
• D. 97

Answer 1

Answer: (B)

93

Answer 2

The correct answer is (B) : 93
P 1: x + 3 y – z = 6
P 2: –6 x + 5 y – z = 7
Family of planes passing through line of intersection of P 1 and P 2 is given by x(1 – 6λ) + y(3 + 5λ) + z (–1 – λ) – (6 + 7λ) = 0
It passes through (2, 3, 1/2)
So,
$2 ( 1 - 6λ ) + 3(3 + 5λ) + \frac{1}{2} (-1 - λ) - ( 6 + 7λ ) = 0$
$⇒ 2 - 12λ + 9 + 15λ - \frac{1}{2} - \frac{λ}{2} - 6 - 7λ = 0$
$⇒ \frac{9}{2} - \frac{9λ}{2} = 0 ⇒ λ = 1$
Required plane is
–5 x + 8 y – 2 z – 13 = 0
Or
$\stackrel{→}{r} . ( -5\hat{i} + 8\hat{j} - 2\hat{k} ) = 13$
$\frac{| 13a→ |²}{ |d|²} = \frac{13²}{(13)²} . |\stackrel{→}{a}|^2 = 93$