Question

Find the equation of the plane passing through $(2,0,1)$ and $(3,-3,4)$ and perpendicular to $x-2y+z=6$.

Answer 1

As second plane is perpendicular to first plane, its normal will be parallel to required plane, also $\overline{AB}$ lies in the plane and therefore parallel to it $(A=2,0,1;B=3,-3,4)$

$\Rightarrow \overrightarrow{A_1}=\overrightarrow{n_2}\times \overline{AB}$ $(\overrightarrow{n_1},\overrightarrow{n_2}=normals)$

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

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

$=(-6+3)\hat{i}+(1-3)\hat{j}+(-3+2)\hat{k}$

$=-3\hat{i}-2\hat{j}-\hat{k}$

$\Rightarrow$ Equation plane : $(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$

$\Rightarrow (\overrightarrow{r}-(2\hat{i}+\hat{k}\left)\right).(-3\hat{i}-2\hat{j}-\hat{k})=0$

$\Rightarrow [\overrightarrow{r}-(2\hat{i}+\hat{k}\left)\right].(3\hat{i}+2\hat{j}+\hat{k})=0$