Skip to content
dronacharya.ai
← JEE Main 2026 · 6 Apr · Shift 2
MathematicsInteger+4 / -1Question 24 of 75

JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 24

Let the image of the point P(0,−5,0)P(0,-5,0) in the line x−12=y1=z+1−2\dfrac{x-1}2=\dfrac y1=\dfrac{z+1}{-2} be the point R and the image of the point Q(0,−12,0)Q\left(0,\dfrac{-1}2,0\right) in the line x−1−1=y+94=z+11\dfrac{x-1}{-1}=\dfrac{y+9}4=\dfrac{z+1}1 be the point S. Then the square of the area of the parallelogram PQRS is ________.

Answer and worked solution

Official answer: 162

Image of P. Line through (1,0,−1)(1,0,-1), direction (2,1,−2)(2,1,-2). The foot is (−1,−1,1)(-1,-1,1), so R=(−2,3,2)R=(-2,3,2).

Image of Q. Line through (1,−9,−1)(1,-9,-1), direction (−1,4,1)(-1,4,1). The foot is again (−1,−1,1)(-1,-1,1), so S=(−2,−32,2)S=\left(-2,-\tfrac32,2\right).

Area. The diagonals PR and QS bisect each other at (−1,−1,1)(-1,-1,1), so PQRS is a parallelogram with area 12∣PR⃗×QS⃗∣\tfrac12|\vec{PR}\times\vec{QS}|:
PR⃗=(−2,8,2),QS⃗=(−2,−1,2),PR⃗×QS⃗=(18,0,18).\vec{PR}=(-2,8,2),\quad\vec{QS}=(-2,-1,2),\quad\vec{PR}\times\vec{QS}=(18,0,18).
Area =92=9\sqrt2, and its square is 162.

Answer: 162

Topics:Equation of a line in spaceCross product
← Previous questionNext question →