Question

Find the image of $(1,5,1)$ in the plane $x-2y+z+5=0$

Answer 1

$m=(1,5,1)+k_1(1,-2,1)=(1+k_{1}5-2k_1-1+k_1)$
$\therefore 1+k_1-2(5-2k_1)+1+k_1=0$
$\therefore 1+k_1-10+4k_1+1+k_1=0$
$\therefore 6k_1=3$
$\therefore k_1=\frac{1}{2}$
Position vector of the foot of perpendicular
$\overline{m}=(1+\frac{1}{2},5-1,1+\frac{1}{2})=(\frac{3}{2},4,\frac{3}{2})$
Now from $M$ is mid point
$\therefore \frac{\overline{a}+\overline{b}}{2}=\overline{m}$
$\therefore \frac{a+1}{2}=\frac{3}{2},\frac{b+5}{2}=4,\frac{c+1}{2}=\frac{3}{2}$
$\therefore a=2,b=3,c=2$
Required image is at the point A is $B\left(\overline{b}\right)=(2,3,2)$