The lines x-22=y-2=z-716 and x+34=y+23=z+21 intersect at the point P.

Question

The lines $\dfrac{x-2}{2}=\dfrac{y}{-2}=\dfrac{z-7}{16}$ and $\dfrac{x+3}{4}=\dfrac{y+2}{3}=\dfrac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\dfrac{x+1}{2}=\dfrac{y-1}{3}=\dfrac{z-1}{1}$ is $l$, then $14l^2$ is equal to:

Answer: 108

Accepted Answer

Parametrize and solve for the intersection: $t=-\tfrac12,\ s=1$, giving
$P=(1,1,-1).$
Distance from $P$ to the third line ($A(-1,1,1)$, direction $(2,3,1)$): $\vec{AP}=(2,0,-2)$.
$$\vec{AP}\times(2,3,1)=(6,-6,6),\qquad |\vec{AP}\times(2,3,1)|^2=108.$$
$$l^2=\frac{108}{|(2,3,1)|^2}=\frac{108}{14} \implies 14l^2=108.$$

Solutions · 1