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 .