Vectors Ques 58
- The sum of the distinct real values of $\mu$, for which the vectors, $\mu \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+\mu \hat{\mathbf{j}}+\hat{\mathbf{k}}, \quad \hat{\mathbf{i}}+\hat{\mathbf{j}}+\mu \hat{\mathbf{k}}$ are coplanar, is
(2019 Main, 12 Jan I)
(a) 2
(b) 0
(c) 1
(d) -1
Show Answer
Answer:
Correct Answer: 58.(d)
Solution:
Formula:
Scalar Product Of Two Vectors:
- Given vectors, $\mu \hat{i}+\hat{j}+\hat{k}, \hat{i}+\mu \hat{j}+\hat{k}, \hat{i}+\hat{j}+\mu \hat{k}$ will be coplanar, if
$$ \begin{array}{rlrl} & \left|\begin{array}{ccc} \mu & 1 & 1 \\ 1 & \mu & 1 \\ 1 & 1 & \mu \end{array}\right|=0 \\ \Rightarrow & \mu\left(\mu^{2}-1\right)-1(\mu-1)+1(1-\mu)=0 \\ \Rightarrow & (\mu-1)[\mu(\mu+1)-1-1]=0 \\ \Rightarrow & (\mu-1)\left[\mu^{2}+\mu-2\right]=0 \\ \Rightarrow & (\mu-1)[(\mu+2)(\mu-1)]=0 \\ \Rightarrow & \mu=1 \text { or }-2 \end{array} $$
So, sum of the distinct real values of $\mu=1-2=-1$.
BETA
