JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(\mathrm{L}_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}\) and \(\mathrm{L}_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in \mathbf{R}\), be two lines, which intersect at the point \(B\). If \(P\) is the foot of perpendicular from the point \(A(1,1,-1)\) on \(L_2\), then the value of \(26 \alpha(\mathrm{~PB})^2\) is _________
- A 212
- B 214
- C 216
- D 218
Answer & Solution
Correct Answer
(C) 216
Step-by-step Solution
Detailed explanation
Point \(B\) \(\begin{aligned} & (3 \lambda+1,-\lambda+1,-1) \equiv(2 \mu+2,0, \alpha \mu-4) \\ & 3 \lambda+1=2 \mu+2 \\ & -\lambda+1=0 \\ & -1=\alpha \mu-4 \\ & \lambda=1, \mu=1, \alpha=3 \\ & B(4,0,-1) \end{aligned}\)…
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