The square of the distance of the point P(-2,-8,6) from the line L_1: x-11=y-12=z-1, measu…

Question

The square of the distance of the point $P(-2,-8,6)$ from the line $L_1:\ \dfrac{x-1}{1}=\dfrac{y-1}{2}=\dfrac{z}{-1}$, measured along the line $L_2:\ \dfrac{x+5}{1}=\dfrac{y+5}{-1}=\dfrac{z}{2}$ (i.e. parallel to $L_2$), is equal to:

Accepted Answer

Line $L_3$ through $P(-2,-8,6)$ parallel to $L_2$'s direction $(1,-1,2)$:
$$\frac{x+2}{1}=\frac{y+8}{-1}=\frac{z-6}{2}.$$
General point on $L_1$: $Q(\lambda+1,\ 2\lambda+1,\ -\lambda)$. Requiring $Q$ on $L_3$:
$$\frac{\lambda+3}{1}=\frac{2\lambda+9}{-1} \implies \lambda+3=-2\lambda-9 \implies \lambda=-4.$$
$Q=(-3,-7,4).$
$PQ^2=(-3+2)^2+(-7+8)^2+(4-6)^2=1+1+4=6.$

Solutions · 1