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

Answer 1

The correct option is C $\left(D\right)\to \left(S\right)$

$\left(C\right)$ If vectors $2\hat{i}-\hat{j}+\hat{k}$, $\hat{i}+2\hat{j}+(1+a)\hat{k}$ and $3\hat{i}+a\hat{j}+5\hat{k}$ are coplanar, then
$\left|\begin{array}{ccc}2& -1& 1\\ 1& 2& 1+a\\ 3& a& 5\end{array}\right|=0$

$\Rightarrow a^2+2a-8=0$
$\Rightarrow (a+4)(a-2)=0$
$\Rightarrow a=-4,2$
$\left(C\right)\to \left(P\right),\left(S\right)$

$\left(D\right)\overrightarrow{A}=2\hat{i}+\lambda \hat{j}+3\hat{k}$
$B=2\hat{i}+\lambda \hat{j}+\hat{k}$
$C=3\hat{i}+\hat{j}+0\hat{k}$
$\overrightarrow{A}+\lambda \overrightarrow{B}=2(1+\lambda )\hat{i}+(\lambda +{\lambda }^2)\hat{j}+(3+\lambda )\hat{k}$

Now, $(\overrightarrow{A}+\lambda \overrightarrow{B})\perp \overrightarrow{C}$
$\therefore (\overrightarrow{A}+\lambda \overrightarrow{B})\cdot \overrightarrow{C}=0$
$\Rightarrow 6(1+\lambda )+(\lambda +{\lambda }^2)+0=0$
$\Rightarrow {\lambda }^2+7\lambda +6=0$
$\Rightarrow (\lambda +6)(\lambda +1)=0$
$\Rightarrow \lambda =-6,-1$
$\Rightarrow |2\lambda |=12,2$
$\left(D\right)\to \left(S\right)$