Let the image of the point P(1,2,3) in the plane 2x-y+z=9 be Q.

Question

Let the image of the point $P(1,2,3)$ in the plane $2x-y+z=9$ be $Q$. If the coordinates of the point $R$ are $(6,10,7)$, then the square of the area of the triangle $PQR$ is \_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 594

Accepted Answer

Let $Q(x,y,z)$ be the image of $P(1,2,3)$ in the plane

$\begin{gathered} 2x-y+z=9 \ 	herefore \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{1}=\frac{-2(2	imes 1+(-1)(2)+(1)(3)(-9)}{(2{)}^{2}+(-1{)}^{2}+(1{)}^{2}} \end{gathered}$

$\begin{gathered} \Rightarrow \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{1}=\frac{-2(-6)}{6}=2 \ \Rightarrow x=5,y=0,z=5 \end{gathered}$

Now, $\overset{	o \leftarrow }{PQ}=4\hat{i}-2\hat{j}+2\hat{k}$ and

$\overset{	o \leftarrow }{PR}=5\hat{i}+8\hat{j}+4\hat{k}$

$	herefore $ Area of the $	rianglePQR=\frac{1}{2}|\overset{	o \leftarrow }{PQ}	imes \overset{	o \leftarrow }{PR}|$

Now, $\overset{	o \leftarrow }{PQ}	imes \overset{	o \leftarrow }{PR}$

$\begin{gathered} =|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 4 \quad  -2 \quad  2 \ 5 \quad  8 \quad  4\end{array}| \ =\hat{i}(-8-16)-\hat{j}(16-10)+\hat{k}(32+10) \ =-24\hat{i}-6\hat{j}+42\hat{k} \end{gathered}$

$\begin{gathered} 	herefore \frac{1}{2}|(-24\hat{i} -6\hat{j}+42\hat{k})|=|-12\hat{i}-3\hat{j}+21\hat{k}| \ \ =\sqrt{(-12{)}^{2}+(-3{)}^{2}+(21{)}^{2}} \ \ =\sqrt{144+9+441}=\sqrt{594} \end{gathered}$

$	herefore $ The square of the area of the $	rianglePQR=594$

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