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