Question

If $a=2\hat{i}+2\hat{j}+3\hat{k},\hat{b}=-\hat{i}+2\hat{j}+\hat{k}$ and $c=3\hat{i}+\hat{j}$ such that $a+\lambda b$ is perpendicular to $c$, then the value of $\lambda$ is

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

Answer: (D)

The given vectors are $a=2\hat{i}+2\hat{j}+3\hat{k},b=-\hat{i}+2\hat{j}+\hat{k}$ and $c=3\hat{i}+\hat{j}$.
Now, $(a+\lambda b)\perp c$(given)
$\Rightarrow (a+\lambda b)\cdot c=0$
$(\because$ scalar product of two perpendicular vectors is zero )
$\Rightarrow [(2\hat{i}+2\hat{j}+3\hat{k})+\lambda (-\hat{i}+2\hat{j}+\hat{k})]\cdot (3\hat{i}+\hat{j})=0$
$\Rightarrow [(2-\lambda )\hat{i}+(2+2\lambda )\hat{j}+(3+\lambda )\hat{k}]\cdot (3\hat{i}+\hat{j})=0$
$\Rightarrow (2-\lambda )3+(2+2\lambda )1+(3+\lambda )0=0$
$\Rightarrow 6-3\lambda +2+2\lambda =0$
$\Rightarrow 8-\lambda =0$
$\Rightarrow \lambda =8$
Hence, the required value of $\lambda$ is 8 .