Question

Let a line with direction ratios a, – 4a, –7 be perpendicular to the lines with direction ratios 3, – 1, 2b and b, a, – 2. If the point of intersection of the line
\(\begin{array}{l} \frac{x+1}{a^2+b^2}=\frac{y-2}{a^2-b^2}=\frac{z}{1} \end{array}\)
and the plane x – y + z = 0 is (\(α, β, γ\)), then \(α + β + γ\) is equal to ______.

Answer 1

Correct Answer: 10

Given a.3 + (– 4a)(–1) + (–7) 2b = 0 …(1)

and ab –4a² + 14 = 0 …(2)

⇒ a² = 4 and b² = 1
\(\begin{array}{l} \therefore\ \text{Line }L\equiv\frac{x+1}{5}=\frac{y-2}{3}=\frac{z}{1}=\lambda(\text{ say})\end{array}\)
⇒ General point on line is (5λ – 1, 3λ + 2, λ)
for finding point of intersection with x – y + z = 0
we get (5λ – 1) – (3λ + 2) + (λ) = 0
⇒ 3λ – 3 = 0 ⇒λ = 1
∴ Point at intersection (4, 5, 1)
∴ \(α + β + γ\) = 4 + 5 + 1 = 10