Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The line of intersection of the planes $x-y+2z=5$ and $3x+y+z=6$ is perpendicular to the normal vectors of both planes.

The normal vectors of the given planes are $\overrightarrow{n_1}=\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{n_2}=3\hat{i}+\hat{j}+\hat{k}$.

The direction vector of the line is given by $\vec{b}=\overrightarrow{n_1}\times \overrightarrow{n_2}$.

$\vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ 3& 1& 1\end{array}\right|$

$\vec{b}=\hat{i}(-1-2)-\hat{j}(1-6)+\hat{k}(1+3)$

$\vec{b}=-3\hat{i}+5\hat{j}+4\hat{k}$

The direction ratios of the line are $-3,5,4$.

The magnitude of the direction vector is $\sqrt{(-3{)}^2+5^2+4^2}=\sqrt{9+25+16}=\sqrt{50}=5\sqrt{2}$.

Therefore, the direction cosines of the line are $\frac{-3}{5\sqrt{2}},\frac{5}{5\sqrt{2}},\frac{4}{5\sqrt{2}}$.

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