Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the given points be $A(1,-2,1)$, $B(2,-1,-3)$ and $C(0,1,5)$.

The vectors in the plane are:
$\overrightarrow{AB}=(2-1)\hat{i}+(-1-(-2))\hat{j}+(-3-1)\hat{k}=\hat{i}+\hat{j}-4\hat{k}$
$\overrightarrow{AC}=(0-1)\hat{i}+(1-(-2))\hat{j}+(5-1)\hat{k}=-\hat{i}+3\hat{j}+4\hat{k}$

The normal vector to the plane is given by the cross product $\vec{n}=\overrightarrow{AB}\times \overrightarrow{AC}$:
$\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& -4\\ -1& 3& 4\end{array}\right|=\hat{i}(4-(-12))-\hat{j}(4-4)+\hat{k}(3-(-1))=16\hat{i}+4\hat{k}$

A simpler normal vector is ${\vec{n}}^{\prime }=4\hat{i}+\hat{k}$.

The equation of the plane passing through $A(1,-2,1)$ with normal vector $4\hat{i}+\hat{k}$ is:
$4(x-1)+0(y-(-2))+1(z-1)=0$
$4x+z-5=0$

The perpendicular distance from the origin $(0,0,0)$ to the plane $4x+z-5=0$ is:
$d=\frac{|4(0)+0(0)+1(0)-5|}{\sqrt{4^2+0^2+1^2}}=\frac{5}{\sqrt{17}}$

Answer: $\frac{5}{\sqrt{17}}$