Question

Find the image of the point $P(2,-3,1)$ about the line
$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+2}{-1}$

• A. $(1,-3,2)$
• B. $(\frac{-58}{7},\frac{36}{14},\frac{-40}{14})$
• C. $(\frac{58}{7},\frac{36}{14},\frac{-40}{14})$
• D. $(\frac{-58}{7},\frac{36}{14},\frac{40}{14})$

Answer 1

The correct option is B

$(\frac{-58}{7},\frac{36}{14},\frac{-40}{14})$

We know that midpoint of P and I (image) will be the foot of the perpendicular Q. We can find the coordinates of I if we know the coordinates of P and Q.

Finding foot of the perpendicular :

Given line is
$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+2}{-1}=r$...(1)

Point $P\equiv (2,-3,1)$

Co-ordinates of foot of the perpendicular on line (1) may be taken as
$Q\equiv (2r–1,3r+3,-r–2)$

We get direction ratios of $PQ\equiv 2r–3,3r+6,-r–3$

Direction ratios of line segment are $2,3,-1$ (from (1))

Since PQ perpendicular to AB

$\therefore 2(2r–3)+3(3r+6)–1(-r–3)=0$

or, $14r+15=0$

$\therefore r=\frac{-15}{14}$

$\therefore Q\equiv (2r–1,3r+3,-r–2$
$\Rightarrow Q\equiv (\frac{-22}{7},\frac{-3}{14},\frac{-13}{14})$

Let the coordinates of I be $(x,y,z)$.
Midpoint of P and I$Q\equiv (\frac{x+2}{2},\frac{y-3}{2},\frac{z+1}{2})=\frac{-22}{7},\frac{-3}{14},\frac{-13}{14}\Rightarrow \frac{x+2}{2}=\frac{-22}{7},\frac{y-3}{2}=\frac{-3}{14},\frac{z+1}{2}=\frac{-13}{14}\Rightarrow x=\frac{-58}{7},y=\frac{36}{14},z=\frac{-40}{14}$