JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the image of the point \(P(0, -5, 0)\) in the line \(\dfrac{x-1}{2} = \dfrac{y}{1} = \dfrac{z+1}{-2}\) be the point \(R\) and the image of the point \(Q\left(0, \dfrac{-1}{2}, 0\right)\) in the line \(\dfrac{x-1}{-1} = \dfrac{y+9}{4} = \dfrac{z+1}{1}\) be the point \(S\). Then the square of the area of the parallelogram \(PQRS\) is __________.
- A 160
- B 161
- C 162
- D 163
Answer & Solution
Correct Answer
(C) 162
Step-by-step Solution
Detailed explanation
Let \(M_1\) be the foot of the perpendicular from \(P(0, -5, 0)\) to the first line \(L_1: \dfrac{x-1}{2} = \dfrac{y}{1} = \dfrac{z+1}{-2}\). A general point on \(L_1\) is \(M_1(2\lambda + 1, \lambda, -2\lambda - 1)\). The direction ratios of \(PM_1\) are…
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