Question

Find the vector and cartesian equation of the plane which passes through the point $(2,-3,4)$ and perpendicular to the line with direction ratios $3,-5,4$.

Answer 1

We have the position vector of point $(2,-3,4)$ as $2\hat{i}-3\hat{j}+4\hat{k}$ and the normal vector $\overrightarrow{n}$ perpendicular to the plane as $\overrightarrow{n}=3\hat{i}-5\hat{j}+4\hat{k}$.
Therefore, the vector equation of the plane is given by $(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$.
or $[\overrightarrow{r}-(2\hat{i}-3\hat{j}+4\hat{k}\left)\right].(3\hat{i}-5\hat{j}+4\hat{k})=0$
transforming into cartesian form by replacing $\overrightarrow{r}$ with $x\hat{i}+y\hat{j}+z\hat{k}$
$\left[\right(x-2)\hat{i}+(y+3)\hat{j}+(z-3\left)\hat{k}\right].(3\hat{i}-5\hat{j}+4\hat{k})=0$
$3(x-2)-5(y+3)+4(z-4)=0$
$3x-5y+4z=37$