Let a line L pass through the point P(2,3,1) and be parallel to the line of intersection of x+3y−2z−2=0 and x−y+2z=0. If the distance of L from the point (5,3,8) is α, then 3α2 is equal to:
Answer: 158
Solutions · 1
Official solution
Direction =(1,3,−2)×(1,−1,2)=(4,−4,−4)∥(1,−1,−1).
L:1x−2=−1y−3=−1z−1. Foot of perpendicular R from Q(5,3,8): R=(k+2,−k+3,−k+1), QR=(k−3,−k,−k−7), perpendicular to (1,−1,−1):
\[(k-3)+k+(k+7)=0 \implies 3k+4=0 \implies k=-\frac43.\]
\[\vec{QR}=\left(-\frac{13}{3},\frac43,-\frac{17}{3}\right).\]
\[\alpha^2=\frac{169+16+289}{9}=\frac{474}{9} \implies 3\alpha^2=158.\]