Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{The possible value of}a\text{if}\overrightarrow{r}=(\hat{i}+\hat{j})+\lambda (\hat{i}+2\hat{j}-\hat{k})& \text{(P)}& -4\\ & \text{and}\overrightarrow{r}=(\hat{i}+2\hat{j})+\mu (-\hat{i}+\hat{j}+a\hat{k})\text{are not consistent,}& & \\ & \text{where}\lambda \text{and}\mu \text{are scalars, is}& & \\ \text{(B)}& \text{The angle between vectors}\overrightarrow{a}=\lambda \hat{i}-3\hat{j}-\hat{k}\text{and}& \text{(Q)}& -2\\ & \overrightarrow{b}=2\lambda \hat{i}+\lambda \hat{j}-\hat{k}\text{is acute, whereas vector}\overrightarrow{b}& & \\ & \text{makes an obtuse angle with the axes of coordinates.}& & \\ & \text{Then}\lambda \text{can be}& & \\ \text{(C)}& \text{The possible value of}a\text{such that}2\hat{i}-\hat{j}+\hat{k},& \text{(R)}& 1\\ & \hat{i}+2\hat{j}+(1+a)\hat{k}\text{and}3\hat{i}+a\hat{j}+5\hat{k}\text{are coplanar, is}& & \\ \text{(D)}& \text{If}\overrightarrow{A}=2\hat{i}+\lambda \hat{j}+3\hat{k},\overrightarrow{B}=2\hat{i}+\lambda \hat{j}+\hat{k},\overrightarrow{C}=3\hat{i}+\hat{j}& \text{(S)}& 2\\ & \text{and}\overrightarrow{A}+\lambda \overrightarrow{B}\text{is perpendicular to}\overrightarrow{C},\text{then}|2\lambda |\text{is}& & \\ & & \text{(T)}& 3\end{array}$

Which of the following is the only CORRECT combination?

• A. $\left(A\right)\to \left(P\right),\left(Q\right),\left(R\right)$
• B. $\left(A\right)\to \left(P\right),\left(Q\right),\left(S\right),\left(T\right)$
• C. $\left(B\right)\to \left(P\right),\left(S\right)$
• D. $\left(B\right)\to \left(S\right),\left(T\right)$

Answer 1

The correct option is B $\left(A\right)\to \left(P\right),\left(Q\right),\left(S\right),\left(T\right)$

$\left(A\right)$ Given equations are consistent if
$(\hat{i}+\hat{j})+\lambda (\hat{i}+2\hat{j}-\hat{k})=(\hat{i}+2\hat{j})+\mu (-\hat{i}+\hat{j}+a\hat{k})$
$\Rightarrow 1+\lambda =1-\mu ,1+2\lambda =2+\mu ,-\lambda =a\mu$
$\Rightarrow \lambda =\frac{1}{3}$ and $\mu =-\frac{1}{3}$
$\Rightarrow a=1$
So, for inconsistency, $a\ne 1$
$\left(A\right)\to \left(P\right),\left(Q\right),\left(S\right),\left(T\right)$

$\left(B\right)\overrightarrow{a}=\lambda \hat{i}-3\hat{j}-\hat{k}$
$\overrightarrow{b}=2\lambda \hat{i}+\lambda \hat{j}-\hat{k}$
Angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is acute.
$\therefore \overrightarrow{a}\cdot \overrightarrow{b}>0$
$\Rightarrow 2{\lambda }^2-3\lambda +1>0$
$\Rightarrow (2\lambda -1)(\lambda -1)>0$
$\Rightarrow \lambda \in (-\infty ,\frac{1}{2})\cup (1,\infty )$

Also, $\overrightarrow{b}$ makes an obtuse angle with the axes. Therefore,
$\overrightarrow{b}.\hat{i}<0\Rightarrow \lambda <0$
$\overrightarrow{b}.\hat{j}<0\Rightarrow \lambda <0$
Hence, $\lambda$ can be $-4,-2$
$\left(B\right)\to \left(P\right),\left(Q\right)$