Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the given points be $A(2,1,0)$, $B(4,1,1)$ and $C(5,0,1)$.

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

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

$\overrightarrow{AC}=(5-2)\hat{i}+(0-1)\hat{j}+(1-0)\hat{k}=3\hat{i}-\hat{j}+\hat{k}$

The normal vector $\vec{n}$ to the plane is:

$\vec{n}=\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 0& 1\\ 3& -1& 1\end{array}\right|=\hat{i}(0-(-1))-\hat{j}(2-3)+\hat{k}(-2-0)=\hat{i}+\hat{j}-2\hat{k}$

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

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

$x+y-2z-3=0$

Let the image of the point $R(2,1,6)$ in the plane be $(x^{\prime },y^{\prime },z^{\prime })$. Using the formula for the image of a point in a plane:

$\frac{x^{\prime }-2}{1}=\frac{y^{\prime }-1}{1}=\frac{z^{\prime }-6}{-2}=-2\frac{1(2)+1(1)-2(6)-3}{1^2+1^2+(-2{)}^2}$

$\frac{x^{\prime }-2}{1}=\frac{y^{\prime }-1}{1}=\frac{z^{\prime }-6}{-2}=-2\left(\frac{2+1-12-3}{6}\right)$

$\frac{x^{\prime }-2}{1}=\frac{y^{\prime }-1}{1}=\frac{z^{\prime }-6}{-2}=-2\left(\frac{-12}{6}\right)=4$

Solving for $x^{\prime },y^{\prime },z^{\prime }$:

$x^{\prime }=2+4=6$

$y^{\prime }=1+4=5$

$z^{\prime }=6-8=-2$

The image of $R$ in the plane is $(6,5,-2)$.

Answer: $(6,5,-2)$