Question

Find the vector equation of a plane which is at a distance of $7$ units from the origin and normal to the vector $3i+5j-6\hat{k}$.

Answer 1

The normal vector is $\overrightarrow{n}=3\hat{i}+5\hat{j}-6\hat{k}$
$\therefore$ $\hat{n}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{3\hat{i}+5\hat{j}-6\hat{k}}{\sqrt{{\left(3\right)}^2+{\left(5\right)}^2+{\left(6\right)}^2}}=\frac{3\hat{i}+5\hat{j}-6\hat{k}}{\sqrt{70}}$
It is known that the equation of the plane with position vector $\overrightarrow{r}$ is given by, $\overrightarrow{r}.\hat{n}=d$
$\Rightarrow$ $\overrightarrow{r}\cdot \left(\frac{3\hat{i}+5\hat{j}-6\hat{k}}{\sqrt{70}}\right)=7$
This is the vector equation of the required plane.