Three Dimensional Geometry Question 391
Question: From a point $ P(\lambda ,\lambda ,\lambda ), $ perpendiculars PQ and PR are drawn, respectively, on the lines $ y=x,z=1 $ and $ y=-x,z=-1 $ . If $ \angle QPR $ is a right angle, then the possible value(s) of $ \lambda $ is/are
Options:
A) 2
B) 1
C) -1
D) $ -,\sqrt{2} $
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Answer:
Correct Answer: C
Solution:
[c] Line 1 : $ \frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r,Q(r,r,1) $ Line 2 : $ \frac{x}{1}=\frac{x}{-1}=\frac{z+1}{0}=k,R(k,-k-1) $ $ \overrightarrow{PQ}=(\lambda -r)\hat{i}+(\lambda -r)\hat{j}+(\lambda -1)\hat{k} $ $ \overrightarrow{PQ} $ is perpendicular to line 1.
$ \Rightarrow \lambda -r+\lambda -r=0\Rightarrow \lambda =r $ $ \overrightarrow{PR}=(\lambda -k)\hat{i}+(\lambda +k)\hat{j}+(\lambda +1)\hat{k} $ $ \overrightarrow{PR} $ is perpendicular to line 2.
$ \Rightarrow \lambda -k-\lambda -k=0\Rightarrow k=0 $ Now, $ \overrightarrow{PQ}\bot \overrightarrow{PR} $
$ \Rightarrow (\lambda -r)(\lambda -k)+(\lambda -r)(\lambda +k)+(\lambda -1)(\lambda +1) $ $ =0 $
$ \Rightarrow \lambda =\pm 1 $
For $ \lambda =1, $ points P and Q coincide.
$ \therefore \lambda =-1 $
BETA
