JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the line L: \(\frac{ x -1}{2}=\frac{ y +1}{-1}=\frac{ z -3}{1}\) intersect the plane \(2 x+y+3 z=16\) at the point \(P.\) Let the point \(Q\) be the foot of perpendicular from the point \(R(1,-1,-3)\) on the line \(L\). If \(\alpha\) is the area of triangle \(PQR.\) then \(\alpha^2\) is equal to \(...........\).
- A \(180\)
- B \(90\)
- C \(45\)
- D \(62\)
Answer & Solution
Correct Answer
(A) \(180\)
Step-by-step Solution
Detailed explanation
Any point on \(L ((2 \lambda+1),(-\lambda-1),(\lambda+3))\) \(2(2 \lambda+1)+(-\lambda-1)+3(\lambda+3)=16\) \(6 \lambda+10=16 \Rightarrow \lambda=1\) \(\therefore P=(3,-2,4)\) \(DR\) of \(QR =\langle 2 \lambda,-\lambda, \lambda+6\rangle\) \(DR\) of \(L =\langle 2,-1,1\rangle\)…
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