Question

Let: $\vec{a}=\hat{i}+2\hat{j}+3\hat{k},\vec{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\vec{c}=5\hat{i}-3\hat{j}+3\hat{k}$ be there vectors. If $\vec{r}$ is a vector such that, $\vec{r}\times \vec{b}=\vec{c}\times \vec{b}$ and $\vec{r}\cdot \vec{a}=0$, then $25|\vec{r}{|}^2$ is equal to

• A. 560
• B. 339
• C. 449
• D. 336

Answer 1

Answer: (B)

$\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$
$\vec{b}=\hat{i}-\hat{j}+2\hat{k}$
$\vec{c}=\widehat{5i}-3\hat{j}+3\hat{k}$
$(\vec{r}-\vec{c})\times \vec{b}=0,\vec{r}\cdot \vec{a}=0$
$\Rightarrow \vec{r}-\vec{c}=\lambda \vec{b}$
$\text{ Also, }(\vec{c}+\lambda \vec{b})\cdot \vec{a}=0$
$\Rightarrow \vec{a}\cdot \vec{c}+\lambda (\vec{a}\cdot \vec{b})=0$
$\therefore \lambda =\frac{\vec{a}\cdot \vec{c}}{\vec{a}\cdot \vec{b}}=\frac{-8}{5}$
$\vec{r}=\frac{5(5\hat{i}-3\hat{i}+3\hat{k})-8(\hat{i}-\hat{j}+2\hat{k})}{5}$
$\vec{r}=\frac{17\hat{i}-7\hat{j}+\hat{k}}{5}$
$|\vec{r}{|}^2=\frac{1}{25}(289+50)$
$25|\vec{r}{|}^2=339$