Question

If $\overline{a}=2\overline{i}+\overline{k},\overline{b}=\overline{i}+\overline{j}+\overline{k},\overline{c}=4\overline{i}-3\overline{j}+7\overline{k}$, then the vector $\overline{r}$ satisfying $\overline{r}\times \overline{b}=\overline{c}\times \overline{b}$ and $\overline{r}.\overline{a}=0$ is

• A. $\overline{i}+8\overline{j}+2\overline{k}$
• B. $\overline{i}-8\overline{j}+2\overline{k}$
• C. $\overline{i}-8\overline{j}-2\overline{k}$
• D. $-\overline{i}-8\overline{j}+2\overline{k}$

Answer 1

The correct option is D $-\overline{i}-8\overline{j}+2\overline{k}$

$\overline{r}\times \overline{b}=\overline{c}\times \overline{b}$
$\overline{r}\times \overline{b}-\overline{c}\times \overline{b}=\overline{0}$
$(\overline{r}-\overline{c})\times \overline{b}=\overline{0}$
$\therefore (\overline{r}-\overline{c})\parallel \overline{b}$
Thus $\overline{r}-\overline{c}=t\overline{b}$
$\overline{r}=\overline{c}+t\overline{b}$
Now $\overline{r}\cdot \overline{a}=\overline{c}.\overline{a}+t(\overline{b}.\overline{a})=0$
$\frac{-(\overline{c}.\overline{a})}{\overline{b}.\overline{a}}=t$

And, $\frac{(\overline{c}.\overline{a})}{\overline{b}.\overline{a}}=\frac{8-0+7}{2+0+1}=5$
$\therefore \overrightarrow{r}=\overrightarrow{c}-\left(\frac{\overrightarrow{c}.\overrightarrow{a}}{\overrightarrow{b}.\overrightarrow{a}}\right)\overrightarrow{b}=\overrightarrow{c}-5\overrightarrow{b}$
$\therefore \overrightarrow{r}=(4\overrightarrow{i}-3\overrightarrow{j}+7\overrightarrow{k})-5(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})=-\overrightarrow{i}-8\overrightarrow{j}+2\overrightarrow{k}$