Question

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

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

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

$\Rightarrow \overrightarrow{r}-\overrightarrow{c}$ is parallel to $\overrightarrow{b}\Rightarrow \overrightarrow{r}-\overrightarrow{c}=\lambda \overrightarrow{b}$

Now, $\overrightarrow{r}\cdot \overrightarrow{a}=0$

$\Rightarrow (\overrightarrow{c}+\lambda \overrightarrow{b})\cdot \overrightarrow{a}=0$

$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{c}+\lambda \overrightarrow{b}\cdot \overrightarrow{a}=0$

$\Rightarrow \lambda =\frac{-(\overrightarrow{a}-\overrightarrow{c})}{\overrightarrow{a}\cdot \overrightarrow{b}}$

$=-\frac{(2+3-3)}{6-3-1}=\frac{-2}{2}=-1$

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

$=-2\hat{i}+2\hat{j}+2\hat{k}=2(-i+j+k)$

$\Rightarrow \left(C\right)$.