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 $\overrightarrow{n_2}=3\hat{i}+3\hat{j}+\hat{k}$.

The direction vector of the line is $\vec{d}=\overrightarrow{n_1}\times \overrightarrow{n_2}=-7\hat{i}+8\hat{j}-3\hat{k}$.

The angle $\theta$ with $\hat{j}$ is given by $\cos \theta =\frac{|\vec{d}\cdot \hat{j}|}{|\vec{d}||\hat{j}|}=\frac{|8|}{\sqrt{(-7{)}^2+8^2+(-3{)}^2}\sqrt{1^2}}=\frac{8}{\sqrt{49+64+9}}=\frac{8}{\sqrt{122}}$.

Given $\cos \theta =\sqrt{\frac{\lambda }{3}}$, so $\frac{64}{122}=\frac{\lambda }{3}$.

Thus, $\lambda =\frac{64\times 3}{122}=\frac{192}{122}=\frac{96}{61}$.