Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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

The direction vector $\vec{v}$ of the line of intersection $L$ is perpendicular to both normal vectors, given by their cross product:

$\vec{v}=\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 1\\ 1& 3& 2\end{array}\right|$

$\vec{v}=\hat{i}(6-3)-\hat{j}(4-1)+\hat{k}(6-3)=3\hat{i}-3\hat{j}+3\hat{k}$

The direction ratios of the line $L$ are proportional to $(3,-3,3)$, which simplifies to $(1,-1,1)$.

The angle $\alpha$ is the angle between the line $L$ and the positive X-axis (which has the direction vector $\hat{i}$).

$\cos \alpha =\frac{\vec{v}\cdot \hat{i}}{|\vec{v}||\hat{i}|}=\frac{3}{\sqrt{3^2+(-3{)}^2+3^2}\sqrt{1^2}}$

$\cos \alpha =\frac{3}{\sqrt{27}}=\frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}}$

Answer: $\frac{1}{\sqrt{3}}$