JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Self Evaluation Test - Three Dimensional Geometry

From a point \[P(\lambda ,\lambda ,\lambda ),\] perpendiculars PQ and PR are drawn, respectively, on the lines \[y=x,\text{ }z=1\] and \[y=-x,\text{ }z=-1\]. If \[\angle QPR\] is a right angle, then the possible value(s) of \[\lambda \] is/are

A) 2

B) 1

C) -1

D) \[-\,\sqrt{2}\]

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. \[\therefore \lambda =-1\]