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

The Correct Option is B

The correct answer is (B) : 93
P₁: x + 3y – z = 6
P₂: –6x + 5y – z = 7
Family of planes passing through line of intersection of P₁ and P₂ 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
–5x + 8y – 2z – 13 = 0
Or
\(\stackrel{→}{r} . ( -5\hat{i} + 8\hat{j} - 2\hat{k} ) = 13\)
\(\frac{| 13a→ |²}{ |d|²} = \frac{13²}{(13)²} . |\stackrel{→}{a}|^2 = 93\)