Question

If $\overline{r}\times \overline{b}=\overline{c}\times \overline{b},\overline{r}\cdot \overline{a}=0,\overline{a}=2\overline{i}+3\overline{j}-\overline{k},\overline{b}=3\overline{i}-\overline{j}+\overline{k},\overline{c}=\overline{i}+\overline{j}+3\overline{k}$ then $\overline{r}=$

• A. $\frac{1}{2}(\overline{i}+\overline{j}+\overline{k})$
• B. $2(\overline{i}+\overline{j}+\overline{k})$
• C. $2(-\overline{i}+\overline{j}+\overline{k})$
• D. $\frac{1}{2}(\overline{i}-\overline{j}+\overline{k})$

Answer 1

The correct option is C $2(-\overline{i}+\overline{j}+\overline{k})$

Given

$\overrightarrow{r}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{b}$

$\overrightarrow{r}\times \overrightarrow{b}-\overrightarrow{c}\times \overrightarrow{b}=\overrightarrow{0}$

$(\overrightarrow{r}-\overrightarrow{c})\times \overrightarrow{b}=\overrightarrow{0}$

Hence, $\overrightarrow{r}-\overrightarrow{c}$ and $\overrightarrow{b}$ are parallel vector

$\therefore \overrightarrow{r}-\overrightarrow{c}=\lambda \overrightarrow{b}$

$\Rightarrow \overrightarrow{r}+\overrightarrow{c}=\lambda \overrightarrow{b}$ ___ (1)

Taking both the sides Dot product with $\overrightarrow{a}$

$\overrightarrow{r}.\overrightarrow{a}=\overrightarrow{c}.\overrightarrow{a}+\lambda \overrightarrow{b}.\overrightarrow{a}$

$\Rightarrow 0=(\overrightarrow{l}+\overrightarrow{j}+3\overrightarrow{k}).(2\overrightarrow{i}+3\overrightarrow{j}-\overrightarrow{k})+\lambda (2\overrightarrow{l}+3\overrightarrow{j}-\overrightarrow{k})(3\overrightarrow{l}-\overrightarrow{j}+\overrightarrow{k})$

$\Rightarrow 0=(2+3-3)+\lambda (6-3-1)$

$\Rightarrow 0=2+2\lambda$

$\therefore \lambda =-1$ ___ (ii)

using (ii) in (i)

$\overrightarrow{r}=\overrightarrow{c}-\overrightarrow{b}$

$=\hat{l}+\hat{j}+3\hat{k}-3\hat{l}+\hat{j}-\hat{k}$

$\overrightarrow{r}=-2\hat{l}+2\hat{j}+2\hat{k}$

Hence, $\overrightarrow{r}=2(-\hat{l}+\hat{j}+\hat{k})$