The direction cosines of the line formed by t | vedclass

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Question

(A) The line of intersection of two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c_2z = d_2$ has direction ratios proportional to the cross pr

Vedclass · Accepted Answer

$\frac{-3}{5\sqrt{2}}, \frac{5}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}$

Vedclass · Answer

(A) The line of intersection of two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c_2z = d_2$ has direction ratios proportional to the cross product of their normals $\vec{n_1} = (1, -1, 2)$ and $\vec{n_2} = (3, 1, 1)$.Let the direction ratios be $(a, b, c) = \vec{n_1} 	imes \vec{n_2}$.$(a, b, c) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ 3 & 1 & 1 \end{vmatrix} = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 + 3) = -3\hat{i} + 5\hat{j} + 4\hat{k}$.So,the direction ratios are $(-3, 5, 4)$.The magnitude is $\sqrt{(-3)^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}$.The direction cosines are $\left(\frac{-3}{5\sqrt{2}}, \frac{5}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}ight)$ or $\left(\frac{3}{5\sqrt{2}}, \frac{-5}{5\sqrt{2}}, \frac{-4}{5\sqrt{2}}ight)$.Comparing with the given options,option $A$ is correct.

The direction cosines of the line formed by the intersection of the planes $x - y + 2z = 5$ and $3x + y + z = 6$ are:

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