JEE Main 12 Jan 2019 Evening Question 6
Question: Let S be the set of all real values of $ \lambda $ such that a plane passing through the points $ (-{{\lambda }^{2}},1,1), $ $ (1,-{{\lambda }^{2}},1) $ and $ (1,1,-{{\lambda }^{2}}) $ also passes through the point ( $ -1,-1,1 $ ). Then S is equal to
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ {1,-1} $
B) $ {3,-3} $
C) $ {\sqrt{3}} $
D) $ {\sqrt{3},-\sqrt{3}} $
Show Answer
Answer:
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 $ \begin{vmatrix} x-{{\lambda }^{2}} & y-1 & z-1 \\ 1+{{\lambda }^{2}} & -{{\lambda }^{2}}-1 & 0 \\ 1+{{\lambda }^{2}} & 0 & -{{\lambda }^{2}}-1 \\ \end{vmatrix} =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={ \sqrt{3},-\sqrt{3} } $
