The direction cosines of the line of intersection of two planes $x - y + z - 5 = 0, x - 3 y - 6 = 0$ are

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$⟨ \frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, - \frac{2}{\sqrt{14}} ⟩$

$⟨ 1, 0, 1 ⟩$

$⟨ 0, 1, 0 ⟩$

$⟨ 3, 1, - 2 ⟩$

answer is B.

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Detailed Solution

The direction ratios of line of intersection of two planes are proportional to the vector which is perpendicular to both normal vectors of the planes

The given planes are $x - y + z - 5 = 0, x - 3 y - 6 = 0$

Its normal vectors are $i - j + k, i - 3 j$

vector perpendicular to both the vectors is $|\begin{matrix} i & j & k \\ 1 & - 1 & 1 \\ 1 & - 3 & 0 \end{matrix}| = i (3) - j (- 1) + k (- 2) = 3 i - j - 2 k$

The line of intersection of two planes having direction ratios $⟨ 3, - 1, - 2 ⟩$ and its direction cosines are $⟨ \frac{3}{\sqrt{14}}, - \frac{1}{\sqrt{14}}, - \frac{2}{\sqrt{14}} ⟩$