Question

Let R be the point \( (13, -8, 10) \) and P\( (\alpha, \beta, \lambda) \) be the image of the point Q \( (5, 2, 1) \) in the line \( x-2=3-y=z+1 \). Then the square of the area of the triangle PQR is ________.

Hint:

Find the foot of the perpendicular from Q to the line, get P as the reflection of Q, then use the cross product for the area.

Answer 1

Correct Answer: 486

The line \( x-2=3-y=z+1=t \) has points \( (2+t,\,3-t,\,-1+t) \) and direction \( (1,-1,1) \).
Foot of perpendicular F from Q(5,2,1): the vector from \( (2,3,-1) \) to Q is \( (3,-1,2) \); its dot product with \( (1,-1,1) \) is 6, so \( t=6/3=2 \) and \( F=(4,1,1) \).
P is the image of Q, so \( P=2F-Q=(3,0,1) \).
\( \vec{PQ}=(2,2,0) \), \( \vec{PR}=(10,-8,9) \).
\( \vec{PQ}\times\vec{PR}=(2\cdot9-0,\ 0\cdot10-2\cdot9,\ 2(-8)-2\cdot10)=(18,-18,-36) \).
Area \( =\tfrac12\sqrt{18^2+18^2+36^2}=\tfrac12\sqrt{1944} \), so (Area)\( ^2=\dfrac{1944}{4}=486 \).
Answer: 486