Question

Show that the equation of the plane passing through the points with position vectors $3\overline{i}-5\overline{j}-\overline{k},-\overline{i}+5\overline{j}+7\overline{k}$ and parallel to the vector $3\overline{i}-\overline{j}-7\overline{k}$ is $3x+2y-z=0$.

Answer 1

$\overrightarrow{a}=3\hat{i}-5\hat{j}-\hat{k},\overrightarrow{b}=-\hat{i}+5\hat{j}+7\hat{k}\therefore \overrightarrow{AB}=4\hat{i}-10\hat{j}-8\hat{k},\overrightarrow{r}=3\hat{i}-\hat{j}+7\hat{k}$

$\therefore$ Let $\overrightarrow{n}$ be the normal vector to the plane $\Rightarrow \overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{r}$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& -10& -8\\ 3& -1& +7\end{array}\right|=-78\hat{i}-52\hat{j}+26\hat{k}$

$\therefore$ directions of normal $=(3,2,-1)$

$\therefore$ Equation of plane through $(3,2,-1)$ and directions of normal as $(3,2,-1)$ is,

$3(x-3)+2(y+5)-(z+1)=0$

$3x+2y-z=0$