Question:
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3 \hat{i}+5 \hat{j}-6 \hat{k}$.
Solution:
The normal vector is, $\vec{n}=3 \hat{i}+5 \hat{j}-6 \hat{k}$
$\therefore \hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{3 \hat{i}+5 \hat{j}-6 \hat{k}}{\sqrt{(3)^{2}+(5)^{2}+(6)^{2}}}=\frac{3 \hat{i}+5 \hat{j}-6 \hat{k}}{\sqrt{70}}$
It is known that the equation of the plane with position vector $\vec{r}$ is given by, $\vec{r} \cdot \hat{n}=d$
$\Rightarrow \hat{r} \cdot\left(\frac{3 \hat{i}+5 \hat{j}-6 \hat{k}}{\sqrt{70}}\right)=7$
This is the vector equation of the required plane.