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

Question

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

Professor PC Thomas Classes · Answer

(594) : Let $Q(\alpha, \beta, \gamma)$ be the image of $P$, in the plane $2 x-y+z=9$\ $\frac{\alpha-1}{2}=\frac{\beta-2}{-1}=\frac{\gamma-3}{1}=2 \quad 	herefore \alpha=5, \beta=0, \gamma=5$\ So, point $Q$ is $(5,0,5)$. Also, $P(1,2,3)$ and $R(6,10,7)$.\ $\overrightarrow{P Q}=4 \hat{i}-2 \hat{j}+2 \hat{k} ; \overrightarrow{P R}=5 \hat{i}+8 \hat{j}+4 \hat{k}$\ Area of triangle $P Q R=\frac{1}{2}|\overrightarrow{P Q} 	imes \overrightarrow{P R}|$\ $=\left|\frac{1}{2}ight| \begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \ 4 & -2 & 2 \ 5 & 8 & 4\end{array}| |$\ $=\frac{1}{2}|[\hat{i}(-8-16)-\hat{j}(16-10)+\hat{k}(32+10)]|$\ $=\frac{1}{2}|-24 \hat{i}-6 \hat{j}+42 \hat{k}|=|-12 \hat{i}-3 \hat{j}+21 \hat{k}|$\ $=\sqrt{12^{2}+3^{2}+21^{2}}=\sqrt{594}$\ $	herefore \quad$ Square of area of triangle $P Q R=594$\