Let Q(a,b,c) be the image of the point P(3,2,1) in the line L_1: x-11=y2=z-11.

Question

Let $Q(a,b,c)$ be the image of the point $P(3,2,1)$ in the line $L_1:\ \dfrac{x-1}{1}=\dfrac{y}{2}=\dfrac{z-1}{1}$. Then the distance of $Q$ from the line $L_2:\ \dfrac{x-9}{3}=\dfrac{y-9}{2}=\dfrac{z-5}{-2}$ is:

Accepted Answer

General point on $L_1$: $(\lambda+1,\ 2\lambda,\ \lambda+1)$. Foot of perpendicular $M$ from $P(3,2,1)$: $\vec{PM}=(\lambda-2,\ 2\lambda-2,\ \lambda)$, perpendicular to $(1,2,1)$:
$$(\lambda-2)+2(2\lambda-2)+\lambda=0\implies 6\lambda-6=0\implies\lambda=1.$$
$M=(2,2,2).$
Since $M$ is the midpoint of $PQ$: $Q=2M-P=(1,2,3)$, so $(a,b,c)=(1,2,3)$.

Distance of $Q(1,2,3)$ from $L_2$ (point $A(9,9,5)$, direction $\vec d=(3,2,-2)$):
$$\vec{AQ}=(-8,-7,-2).$$
$$\vec{AQ}\times\vec d=\hat i[(-7)(-2)-(-2)(2)]-\hat j[(-8)(-2)-(-2)(3)]+\hat k[(-8)(2)-(-7)(3)]=(18,-22,5).$$
$$|\vec{AQ}\times\vec d|=\sqrt{324+484+25}=\sqrt{833},\qquad |\vec d|=\sqrt{17}.$$
$$D=\sqrt{\frac{833}{17}}=\sqrt{49}=7.$$

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