JEE Main 12 Jan 2019 Morning Question 1
Question: The sum of the distinct real values of $ \mu $ , for which the vectors, $ \mu \hat{i}+\hat{j}+\hat{k},\hat{i}+\mu \hat{j}+\hat{k},\hat{i}+\hat{j}+\mu ,\hat{k} $ are co-planar, is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 2
B) 0
C) -1
D) 1
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ \vec{a}=\mu \hat{i}+\hat{j}+\hat{k}, $ $ \hat{b}=\hat{i}+\mu \hat{j}+\hat{k} $ and $ \vec{c}=\hat{i}+\hat{j}+\mu \hat{k} $
$ \therefore $ $ \vec{a},\vec{b},\vec{c}= \begin{vmatrix} \mu & 1 & 1 \\ 1 & \mu & 1 \\ 1 & 1 & \mu \\ \end{vmatrix} $
$ =\mu ({{\mu }^{2}}-1)-1(\mu -1)+1(1-\mu ) $ $ ={{\mu }^{3}}-\mu -\mu +1+1-\mu ={{\mu }^{3}}-3\mu +2 $ $ \vec{a},\vec{b} $ and $ \vec{c}
$ are coplanar, so $ [\vec{a},\vec{b},\vec{c}]=0 $
$ \therefore $ $ {{\mu }^{3}}-3\mu +2=0\Rightarrow {{(\mu -1)}^{2}}(\mu +2)=0 $
$ \Rightarrow \mu =1,,1,-2 $
$ \therefore $ The sum of the distinct real values of $ \mu =1-2=-1. $
