If (a,b,c) is the image of the point (1,2,-3) in the line x+12=y-3-2=z-1, then a+b+c is eq…

Question

If $(a,b,c)$ is the image of the point $(1,2,-3)$ in the line $\dfrac{x+1}{2}=\dfrac{y-3}{-2}=\dfrac{z}{-1}$, then $a+b+c$ is equal to:

Accepted Answer

Point on line: $Q(2\lambda-1,-2\lambda+3,-\lambda)$; $\overrightarrow{PQ}=(2\lambda-2,-2\lambda+1,-\lambda+3)$ where $P=(1,2,-3)$. Perpendicular to direction $(2,-2,-1)$:
$$2(2\lambda-2)-2(-2\lambda+1)-(-\lambda+3)=0\implies 9\lambda-9=0\implies\lambda=1.$$
$Q$ (midpoint of $P$ and image) $=(1,1,-1)$. Image $=2Q-P=(1,0,1)=(a,b,c)$.
$a+b+c=1+0+1=2.$

Solutions · 1