If the line x-31=y+2-1=z+-2 lies in the plane 2x-4y+3z=2, then the shortest distance betwe…

Question

If the line $\dfrac{x-3}1=\dfrac{y+2}{-1}=\dfrac{z+\lambda}{-2}$ lies in the plane $2x-4y+3z=2$, then the shortest distance between this line and the line $\dfrac{x-1}{12}=\dfrac y9=\dfrac z4$ is:

Accepted Answer

Point $(3,-2,-\lambda)$ lies on the plane: $6+8-3\lambda=2\Rightarrow\lambda=4$.
Line 1: $(k_1+3,-k_1-2,-2k_1-4)$; Line 2: $(12k_2+1,9k_2,4k_2)$. Equating:
$$k_1+3=12k_2+1,\quad -k_1-2=9k_2,\quad -2k_1-4=4k_2.$$
Solving gives $k_2=0,\ k_1=-2$, and both give the common point $(1,0,0)$. Since the lines intersect, the shortest distance is $0$.

Solutions · 1