A line with direction ratios 2,1,2 meets the lines x=y+2=z and x+2=2y=2z respectively at the points P and Q. If the length of the perpendicular from the point (1,2,12) to the line PQ is l, then l2 is .
Answer
65
Solution
P(t, t-2, t); Q(2s-2, s, s) (put x+2=2y=2z=2s)
QP=(t-2s+2, t-s-2, t-s) (2,1,2)
1st = 3rd component: t-2s+2=t-s s=2; then t-2=2(t-4) t=6
⇒ P(6, 4, 6), Q(2, 2, 2); A(1, 2, 12): PA=(-5,-2,6)
PA (2,1,2)=-10-2+12=0 ⇒ PA ⊥ PQ, so the foot of the perpendicular is P
[figure]
∴ l2=PA2=25+4+36=65
Answer: 65