A line with direction ratios 2,1,2 meets the lines x=y+2=z and x+2=2y=2z at the points P and Q respectively. If the length of the perpendicular from the point (1,2,12) to the line PQ is l, then l2 is:
Answer: 65
Solutions · 1
Official solution
P=(t,t−2,t) on the first line; Q=(u−2,u/2,u/2) on the second. Requiring PQ∥(2,1,2) gives u=4,t=6:
\[P=(6,4,6),\qquad Q=(2,2,2).\]
With A=(1,2,12), QA=(−1,0,10):
\[\vec{QA}\times(2,1,2)=(-10,22,-1),\qquad |\vec{QA}\times(2,1,2)|^2=100+484+1=585.\]
\[l^2=\frac{585}{|(2,1,2)|^2}=\frac{585}{9}=65.\]