Question

Find the image of the point $(2,-1,5)$ in the line $\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}$
Also, find the length of the perpendicular from the point $(2,-1,5)$ to the line.

Answer 1

Let $P(2,-1,5)$ be the given point and $AB$ the given line
$\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}=r$ (say)
Draw $PM\perp AB$. Produce $PM$ to $P^{\prime }$ such that $PM=M{P}^{\prime }$. Then $P^{\prime }$ is image of $P$ in $AB$.
Any point on $AB$ is $M(10r+11,-4r-2,-11r-8)$
Then, dr's of $MP$ are $10r+11-2,-4r-2-(-1),-11r-8-5$
i.e., $10r+9,-4r-1,-11r-13$
and the d.c's of $AB$ are proportional to $10,-4,-11$
$\therefore$ $MP\perp AB$
$\therefore$ $10(10r+9)-4(-4r-1)-11(-11r-13)=0$
$\Rightarrow$ $r=-1$
From equation $\left(i\right)$, $M\equiv (1,2,3)$
Let $P^{\prime }$ be the point $(\alpha ,\beta ,\gamma )$
Simce $M(1,2,3)$ is the mid point of $P{P}^{\prime }$
$\therefore$ $\therefore 1=\frac{\alpha +2}{2},2=\frac{\beta -1}{2},3=\frac{\gamma +5}{2}\Rightarrow \alpha =0,\beta =5,\gamma =1$
$\therefore$ $P^{\prime }\equiv (0,5,1)$
thus $P^{\prime }\equiv (0,5,1)$ is the image of $P(2,-1,5)$ in the line $AB$.
Also, $MP=\sqrt{{(1-2)}^2+{(2+1)}^2+{(3-5)}^2}=\sqrt{1+9+4}=\sqrt{14}$