Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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

The vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ lying in the plane are:

$\overrightarrow{AB}=(-1-4)\hat{i}+(4-2)\hat{j}+(2-3)\hat{k}=-5\hat{i}+2\hat{j}-\hat{k}$

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

The normal vector $\vec{n}$ to the plane is given by the cross product $\overrightarrow{AB}\times \overrightarrow{AC}$:

$\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -5& 2& -1\\ -1& 0& -2\end{array}\right|$

$\vec{n}=\hat{i}(-4-0)-\hat{j}(10-1)+\hat{k}(0-(-2))$

$\vec{n}=-4\hat{i}-9\hat{j}+2\hat{k}$

The direction ratios of the normal are $-4,-9,2$.

The magnitude of the normal vector is:

$|\vec{n}|=\sqrt{(-4{)}^2+(-9{)}^2+2^2}=\sqrt{16+81+4}=\sqrt{101}$

The direction cosines of the normal are $\frac{-4}{\sqrt{101}},\frac{-9}{\sqrt{101}},\frac{2}{\sqrt{101}}$.

Answer: $\frac{-4}{\sqrt{101}},\frac{-9}{\sqrt{101}},\frac{2}{\sqrt{101}}$