Vector Algebra Question 17
Question 17 - 31 January - Shift 2
Let $\vec{a}=\hat{i}+2 j+3 k, \overrightarrow{{}b}=i-j+2 k$ and $\vec{c}=5 \hat{i}-3 \hat{j}+3 \hat{k}$ be three vectors. If $\vec{r}$ is a vector such that, $\overrightarrow{{}r} \times \overrightarrow{{}b}=\overrightarrow{{}c} \times \overrightarrow{{}b}$ and $\overrightarrow{{}r} \cdot \overrightarrow{{}a}=0$. Then $25|\overrightarrow{{}r}|^{2}$ is equal to
(1) 449
(2) 336
(3) 339
(4) 560
Show Answer
Answer: (3)
Solution:
Formula: Vector product of two vectors when two vectors are parallel, Properties of Scalar product of vectors
$ \begin{aligned} & \overrightarrow{{}a}=\hat{i}+2 \hat{j}+3 \hat{k} \\ & \overrightarrow{{}b}=\hat{i}-\hat{j}+2 \hat{k} \\ & \overrightarrow{{}c}=5 \hat{i}-3 \hat{j}+3 \hat{k} \\ & (\overrightarrow{{}r}-\overrightarrow{{}c}) \times \overrightarrow{{}b}=0, \overrightarrow{{}r} \cdot \overrightarrow{{}a}=0 \\ & \Rightarrow \overrightarrow{{}r}-\overrightarrow{{}c}=\lambda \overrightarrow{{}b} \end{aligned} $
Also, $(\overrightarrow{{}c}+\lambda \overrightarrow{{}b}) \cdot \overrightarrow{{}a}=0$
$\Rightarrow \vec{a} \cdot \vec{c}+\lambda(\vec{a} \cdot \vec{b})=0$
$\therefore \lambda=\frac{\overrightarrow{{}a} \cdot \overrightarrow{{}c}}{\overrightarrow{{}a} \cdot \overrightarrow{{}b}}=\frac{-8}{5}$
$\overrightarrow{{}r}=\frac{5(5 \hat{i}-3 \hat{i}+3 \hat{k})-8(\hat{i}-\hat{j}+2 \hat{k})}{5}$
$\overrightarrow{{}r}=\frac{17 \hat{i}-7 \hat{j}+\hat{k}}{5}$
$|\overrightarrow{{}r}|^{2}=\frac{1}{25}(289+50)$
$25|\overrightarrow{{}r}|^{2}=339$
