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 }=\hat{5 i }-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$