A) \[\{1,-1\}\]
B) \[\{3,-3\}\]
C) \[\{\sqrt{3}\}\]
D) \[\{\sqrt{3},-\sqrt{3}\}\]
Correct Answer: D
Solution :
Equation of plane passing through the points \[(-{{\lambda }^{2}},1,1),(1,-{{\lambda }^{2}},1)\]and \[(1,1,-{{\lambda }^{2}})\]is \[\left| \begin{matrix} x-{{\lambda }^{2}} & y-1 & z-1 \\ 1+{{\lambda }^{2}} & -{{\lambda }^{2}}-1 & 0 \\ 1+{{\lambda }^{2}} & 0 & -{{\lambda }^{2}}-1 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[(x+{{\lambda }^{2}}){{({{\lambda }^{2}}+1)}^{2}}+(y-1){{(1+{{\lambda }^{2}})}^{2}}\]\[+(z-1){{(1+{{\lambda }^{2}})}^{2}}=0\] \[\Rightarrow \]\[{{(1+{{\lambda }^{2}})}^{2}}[x+{{\lambda }^{2}})+(y-1+(z-1)]=0\] Since, this plane is also passes through the point (\[-1,\,-1,\,\,1\]). \[\therefore \]\[{{(1+{{\lambda }^{2}})}^{2}}[(-1+{{\lambda }^{2}})+(-1-1)]=0\] \[\Rightarrow \]\[{{(1+{{\lambda }^{2}})}^{2}}({{\lambda }^{2}}-3)=0\] So, real values of\[\lambda \] are\[~\pm \sqrt{3}.\] \[\therefore \]\[S=\left\{ \sqrt{3},-\sqrt{3} \right\}\]
You need to login to perform this action.
You will be redirected in
3 sec
