Question

Let four vectors $\overrightarrow{r}=3\hat{i}+2\hat{j}-5\hat{k},\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}+3\hat{j}-2\hat{k}$ and $\overrightarrow{c}=-2\hat{i}+\hat{j}-3\hat{k}$ are such that $\overrightarrow{r}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}+v\overrightarrow{c},$ then $\lambda +\mu +v$ is

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is D $6$

$3\hat{i}+2\hat{j}-5\hat{k}=\lambda (2\hat{i}-\hat{j}+\hat{k})+\mu (\hat{i}+3\hat{j}-2\hat{k})+v(-2\hat{i}+\hat{j}-3\hat{k})$
$=(2\lambda +\mu -2v)\hat{i}+(-\lambda +3\mu +v)\hat{j}+(\lambda -2\mu -3v)\hat{k}$
Equating components of equal vectros
$2\lambda +\mu -2v=3\cdots \left(1\right)$
$-\lambda +3\mu +v=2\cdots \left(2\right)$
$\lambda -2\mu -3v=-5\cdots \left(3\right)$
On solving $\left(1\right),\left(2\right),\left(3\right),$
we get $\lambda =3,\mu =1,v=2$
So , $\lambda +\mu +v=6$