Consider a line L passing through the points P(1,2,1) and Q(2,1,-1).

Question

Consider a line $L$ passing through the points $P(1,2,1)$ and $Q(2,1,-1)$. If the mirror image of the point $A(2,2,2)$ in the line $L$ is $(\alpha ,\beta ,\gamma )$, then $\alpha +\beta +6\gamma $ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 6

Accepted Answer

$\begin{gathered} A(2,2,2) \ PQ:\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{-2} \end{gathered}$

General point,

$(k+1,-k+2,-2k+1)$

$\begin{gathered} \overset{	o \leftarrow }{OA}=(k+1-2)\hat{i}+(-k+2-2)\hat{j}+(-2k+1-2)\hat{k} \ \overset{	o \leftarrow }{OA}=(k-1)\hat{i}-k\hat{j}+(-2k-1)\hat{k} \ \overset{	o \leftarrow }{PQ}=1\hat{i}-\hat{j}-2\hat{k} \ \overset{	o \leftarrow }{OA}\cdot \overset{	o \leftarrow }{PQ}=0 \ (k-1)+k+2(2k+1)=0 \ k-1+k+4k+2=0 \ 6k+1=0 \ k=\frac{-1}{6} \ 0(\frac{-1}{6}+1,\frac{+1}{6}+2,-2(\frac{-1}{6})+1) \ 0(\frac{5}{6},\frac{13}{6},\frac{-8}{6}) \ 0(\frac{5}{6},\frac{13}{6},\frac{8}{6})=(\frac{\alpha +2}{2},\frac{\beta +2}{2},\frac{\gamma +2}{2}) \end{gathered}$

$\begin{gathered} \alpha +2=\frac{10}{6} \beta +2=\frac{26}{6} \ \alpha =\frac{10}{6}-2 \beta =\frac{26-12}{6} \ \alpha =\frac{-2}{6} \beta =\frac{14}{6} \ \alpha =-\frac{1}{3} \beta =\frac{7}{3} \end{gathered}$

$\begin{gathered} \gamma +2=\frac{16}{6} \ \gamma =\frac{16-12}{6} \ \gamma =\frac{4}{6} \ \Rightarrow \alpha +\beta +6\gamma \ \Rightarrow \frac{-1}{3}+\frac{7}{3}+6	imes \frac{4}{6} \ \Rightarrow 2+4=6 \end{gathered}$

Solutions · 1