Question

The sum of distinct real values of $\mu$, for which the vectors, $\mu \hat{i}+\hat{j}+\hat{k}$, $\hat{i}+\mu \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+\mu \hat{k}$ are coplanar is
$\begin{aligned} & \left( A \right)2 \\\ & \left( B \right)0 \\\ & \left( C \right)-1 \\\ & \left( D \right)3 \\\ \end{aligned}$

Answer 1

We solve this question by going through the concept of scalar triple product and then find the scalar triple product of the given vectors. Then we equate the obtained scalar triple product and equate it to zero to find the value of $\mu$, for which the given vectors are coplanar. Then we add them to find the required value.

Complete step-by-step solution:
First, let us go through the concept of the scalar triple product.
For any three vectors $\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$, $\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ and $\overrightarrow{c}={{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ scalar triple product is defined as
$\left[ \begin{matrix} \overrightarrow{a} & \overrightarrow{b} & \overrightarrow{c} \\\ \end{matrix} \right]=\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\ \end{matrix} \right|$
Now, let us go through the concept of coplanar. Any three vectors are said to be coplanar if they lie on the same plane. The three vectors $\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$, $\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ and $\overrightarrow{c}={{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ are said to be coplanar if their scalar triple product is equal to zero, that is $\left[ \begin{matrix} \overrightarrow{a} & \overrightarrow{b} & \overrightarrow{c} \\\ \end{matrix} \right]=0$, because if $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar then the cross product of any two of them, say $\overrightarrow{b}\times \overrightarrow{c}$ is perpendicular to the plane formed by them. Then the dot product of any vector on the plane and the normal is zero as they are perpendicular. So, we get that $\overrightarrow{a}.\left( \overrightarrow{b}\times \overrightarrow{c} \right)=0$, that is scalar product of $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ is zero.
$\left[ \begin{matrix} \overrightarrow{a} & \overrightarrow{b} & \overrightarrow{c} \\\ \end{matrix} \right]=0$

The vectors we were given are $\mu \hat{i}+\hat{j}+\hat{k}$, $\hat{i}+\mu \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+\mu \hat{k}$.
Now, let us find the scalar triple product of given vectors.
$\begin{aligned} & \left| \begin{matrix} \mu & 1 & 1 \\\ 1 & \mu & 1 \\\ 1 & 1 & \mu \\\ \end{matrix} \right|=\mu \left| \begin{matrix} \mu & 1 \\\ 1 & \mu \\\ \end{matrix} \right|-\left| \begin{matrix} 1 & 1 \\\ 1 & \mu \\\ \end{matrix} \right|+\left| \begin{matrix} 1 & \mu \\\ 1 & 1 \\\ \end{matrix} \right| \\\ & \Rightarrow \mu \left( {{\mu }^{2}}-1 \right)-\left( \mu -1 \right)+\left( 1-\mu \right) \\\ & \Rightarrow {{\mu }^{3}}-\mu -\mu +1+1-\mu \\\ & \Rightarrow {{\mu }^{3}}-3\mu +2 \\\ \end{aligned}$
Now, let us equate the obtained scalar triple product to zero as they are coplanar.
$\begin{aligned} & \Rightarrow {{\mu }^{3}}-3\mu +2=0 \\\ & \Rightarrow \left( \mu -1 \right)\left( {{\mu }^{2}}+\mu -2 \right)=0 \\\ & \Rightarrow \left( \mu -1 \right)\left( \mu -1 \right)\left( \mu +2 \right)=0 \\\ & \Rightarrow {{\left( \mu -1 \right)}^{2}}\left( \mu +2 \right)=0 \\\ & \Rightarrow \mu =1,-2 \\\ \end{aligned}$
So, the distinct values of $\mu$ are 1 and -2.
We need to find the sum of those values. S, by adding them we get
$\begin{aligned} & \Rightarrow 1+\left( -2 \right) \\\ & \Rightarrow -1 \\\ \end{aligned}$
So, sum of distinct values of $\mu$ is -1
Hence answer is option C.

Note: Here while solving this problem one might confuse and add 1 two times at the end while finding the sum of the roots, then we will get the answer as $1+1-2=0$, which is Option B, but we are asked to find the sum of distinct values of $\mu$. So, one must read the question carefully and solve the problem.