Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The position vector of the given point is $\vec{a}=2\hat{i}+\hat{j}-\hat{k}$.

The equation of the plane is $\vec{r}\cdot (\hat{i}-2\hat{j}+4\hat{k})=9$.

The normal vector to the plane is $\vec{n}=\hat{i}-2\hat{j}+4\hat{k}$ and $d=9$.

The perpendicular distance of the point from the plane is given by $D=\frac{|\vec{a}\cdot \vec{n}-d|}{|\vec{n}|}$.

Substituting the values, we get:

$\vec{a}\cdot \vec{n}=(2)(1)+(1)(-2)+(-1)(4)=2-2-4=-4$

$|\vec{n}|=\sqrt{1^2+(-2{)}^2+4^2}=\sqrt{1+4+16}=\sqrt{21}$

$D=\frac{|-4-9|}{\sqrt{21}}=\frac{13}{\sqrt{21}}$ units.

Answer: $\frac{13}{\sqrt{21}}$ units