Question

Find the equation of the line passing through the point (3,0,1) and parallel ti the planes x+2y=0 and 3y-z=0.

Answer 1

Equation of the two planes are x+2y =0 and 3y-z=0.
Let $\overrightarrow{n_1}and\overrightarrow{n_2}$ are the normals to the two planes, respectively.

$\therefore \overrightarrow{n_1}=\hat{i}+2\hat{j}and\overrightarrow{n_2}=3\hat{j}-\hat{k}$
Since, required line is parallel to the given two planes.
Therefore, $\overrightarrow{b}=\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 0\\ 0& 3& -1\end{array}\right|$
=$\hat{i}(-2)-\hat{j}(-1)\hat{k}\left(3\right)=-2\hat{i}+\hat{j}+3\hat{k}$
So, the equation of the lines through the point (3,0,1) and parallel to the given two planes are
$(x-3)i+(y-0)j+(z-1)k+\lambda (-2\hat{i}+\hat{j}+3\hat{k})$
$\Rightarrow (x-3)\hat{i}+y\hat{j}+(z-1)\hat{k}+\lambda (-2\hat{i}+\hat{j}+3\hat{k})$ = 0