Let the image of the point P(0,−5,0) in the line 2x−1=1y=−2z+1 be the point R, and the image of the point Q(0,−21,0) in the line −1x−1=4y+9=1z+1 be the point S. Then the square of the area of the parallelogram PQRS is \_\_\_\_.
Answer: 162
Solutions · 1
Official solution
Reflection of P(0,−5,0) in line 1: Parametrize as (1+2t,t,−1−2t), direction d1=(2,1,−2). Foot of perpendicular M: with PM=(1+2t,t+5,−1−2t),
\[\vec{PM}\cdot\vec d_1=0 \implies 2(1+2t)+(t+5)-2(-1-2t)=0 \implies 9+9t=0 \implies t=-1.\]
\[M=(-1,-1,1) \implies R=2M-P=(-2,3,2).\]
Reflection of Q(0,−21,0) in line 2: Parametrize as (1−u,−9+4u,−1+u), direction d2=(−1,4,1). With QN=(1−u,−217+4u,−1+u),
\[\vec{QN}\cdot\vec d_2=0 \implies -(1-u)+4\left(-\tfrac{17}{2}+4u\right)+(-1+u)=0 \implies 18u-36=0 \implies u=2.\]
\[N=(-1,-1,1) \implies S=2N-Q=\left(-2,\ -\tfrac32,\ 2\right).\]
Area of parallelogram with P(0,−5,0),Q(0,−21,0),R(−2,3,2),S(−2,−23,2): checking sides, PS=QR=(−2,27,2), so the correct cyclic order is P,Q,R,S with adjacent sides PQ=(0,29,0) and PS=(−2,27,2).
\[\vec{PQ}\times\vec{PS}=\hat i\left(\tfrac92\cdot2-0\right)-\hat j(0-0)+\hat k\left(0-\left(-9\right)\right)=(9,0,9).\]
Area =81+81=92. Square of area =(92)2=162.