Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the given points be $A(2,1,0)$, $B(3,-2,4)$ and $C(1,-3,3)$.

The vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are given by:

$\overrightarrow{AB}=(3-2)\hat{i}+(-2-1)\hat{j}+(4-0)\hat{k}=\hat{i}-3\hat{j}+4\hat{k}$

$\overrightarrow{AC}=(1-2)\hat{i}+(-3-1)\hat{j}+(3-0)\hat{k}=-\hat{i}-4\hat{j}+3\hat{k}$

A normal vector $\vec{n}$ to the plane is the cross product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$:

$\vec{n}=\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 4\\ -1& -4& 3\end{array}\right|$

$\vec{n}=\hat{i}(-9+16)-\hat{j}(3+4)+\hat{k}(-4-3)=7\hat{i}-7\hat{j}-7\hat{k}$

Dividing by $7$, a simpler normal vector is $\hat{i}-\hat{j}-\hat{k}$.

The equation of the plane passing through $A(2,1,0)$ with normal vector $\hat{i}-\hat{j}-\hat{k}$ is:

$1(x-2)-1(y-1)-1(z-0)=0$

$x-y-z-1=0$

The perpendicular distance $d$ of the point $P(5,3,-1)$ from the plane $x-y-z-1=0$ is:

$d=\frac{|1(5)-1(3)-1(-1)-1|}{\sqrt{1^2+(-1{)}^2+(-1{)}^2}}$

$d=\frac{|5-3+1-1|}{\sqrt{3}}$

$d=\frac{2}{\sqrt{3}}$ units

Answer: $\frac{2}{\sqrt{3}}$ units