Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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

The line of intersection of the two planes is perpendicular to the normal vectors of both planes. Therefore, its direction vector $\vec{d}$ is parallel to the cross product of $\overrightarrow{n_1}$ and $\overrightarrow{n_2}$.

$\vec{d}=\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ 1& -3& 0\end{array}\right|$

$\vec{d}=\hat{i}(0-(-3))-\hat{j}(0-1)+\hat{k}(-3-(-1))$

$\vec{d}=3\hat{i}+\hat{j}-2\hat{k}$

The direction ratios of the line of intersection are $3,1,-2$.

Answer: $3,1,-2$