Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The line of intersection of planes $\vec{r}\cdot \overrightarrow{n_1}=0$ and $\vec{r}\cdot \overrightarrow{n_2}=0$ is parallel to $\overrightarrow{n_1}\times \overrightarrow{n_2}$.

Here, $\overrightarrow{n_1}=\hat{i}+2\hat{j}+3\hat{k}$ and $\overrightarrow{n_2}=3\hat{i}+2\hat{j}+\hat{k}$

The direction vector of the line is $\vec{d}=\overrightarrow{n_1}\times \overrightarrow{n_2}=-4\hat{i}+8\hat{j}-4\hat{k}$

The angle $\theta$ with $\hat{j}$ is given by $\cos \theta =\frac{|\vec{d}\cdot \hat{j}|}{|\vec{d}||\hat{j}|}=\frac{8}{\sqrt{16+64+16}}=\frac{8}{\sqrt{96}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}$

Given $\cos \theta =\sqrt{\frac{\lambda }{3}}$, so $\frac{2}{3}=\frac{\lambda }{3}$

Thus, $\lambda =\frac{2\times 3}{3}=2$