Question

Find the equation of the plane passing through the point $(5,-1,3)$ and parallel to the vectors $\hat{i}+2\hat{j}+3\hat{k}$ and $3\hat{i}-\hat{j}+4\hat{k}$

Answer 1

The vectors $\hat{i}+2\hat{j}+3\hat{k}$ and $3\hat{i}-\hat{j}+4\hat{k}$ lie in the plane. Hence, perpendicular vector to the plane is

$\overrightarrow{n}=(\hat{i}+2\hat{j}+3\hat{k})\times (3\hat{i}-\hat{j}+4\hat{k})$

$\therefore \overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 3& -1& 4\end{array}\right|$

$\therefore \overrightarrow{n}=(8+3)\hat{i}-(4-9)\hat{j}+(-1-6)\hat{k}$

$\therefore \overrightarrow{n}=11\hat{i}+5\hat{j}-7\hat{k}$

Now, equation of plane passing through $(5,-1,3)$ and $\overrightarrow{n}$ is

$11(x-5)+5(y+1)-7(z-3)=0$

$\therefore 11(x-5)+5(y+1)-7(z-3)=0$

$\therefore 11x+5y-7z=29$

This is the required equation of the plane.