Question

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

Answer 1

The normal vector is,

$\overrightarrow n= 3\hat i+5\hat j-6\hat k.$

∴$\hat n=\frac{\overrightarrow n}{\mid \overrightarrow n\mid}$

=$\frac{3\hat i+5\hat j-6\hat k.}{\sqrt{(3)^2+(5)^2+(6)^2}}$

=$\frac{3\hat i+5\hat j-6\hat k.}{\sqrt 70}$

It is known that the equation of the plane with position vector $\overrightarrow r$ is given by,

$\overrightarrow r.\hat n=d$
$\Rightarrow \hat r.\bigg(\frac{3\hat i+5\hat j-6\hat k.}{\sqrt {70}}\bigg)=7$

This is the vector equation of the required plane.