Vector Algebra Question 23
Question 23 - 01 February - Shift 2
Let $\quad a=21-7 J+5 k \quad, \quad b=1+k \quad$ and $\overrightarrow{{}c}=\hat{i}+2 \hat{j}-3 \hat{k}$ be three given vectors. If $\overrightarrow{{}r}$ is a vector such that $\overrightarrow{{}r} \times \overrightarrow{{}a}=\overrightarrow{{}c} \times \overrightarrow{{}a}$ and $\overrightarrow{{}r} \cdot \overrightarrow{{}b}=0$, then $|\overrightarrow{{}r}|$ is equal to :
(1) $\frac{11}{7} \sqrt{2}$
(2) $\frac{11}{7}$
(3) $\frac{11}{5} \sqrt{2}$
(4) $\frac{\sqrt{914}}{7}$
Show Answer
Answer: (1)
Solution:
Formula: Vector product of two vectors when two vectors are parallel, Scalar product of vectors
$\overrightarrow{{}a}=2 \hat{i}-7 \hat{j}+5 \hat{k}$
$\overrightarrow{{}b}=\hat{i}+\hat{k}$
$\overrightarrow{{}c}=\hat{i}+2 \hat{j}-3 \hat{k}$
$\overrightarrow{{}r} \times \overrightarrow{{}a}=\overrightarrow{{}c} \times \overrightarrow{{}a} \Rightarrow(\overrightarrow{{}r}-\overrightarrow{{}c}) \times \overrightarrow{{}a}=0$
$\therefore \overrightarrow{{}r}=\overrightarrow{{}c}+\lambda \overrightarrow{{}a}$
$\overrightarrow{{}r} \cdot \overrightarrow{{}b}=0 \Rightarrow \overrightarrow{{}c} \cdot \overrightarrow{{}b}+\lambda \overrightarrow{{}b} \cdot \overrightarrow{{}a}=0$
$-2+\lambda(7)=0 \Rightarrow \lambda=\frac{2}{7}$
$\therefore \overrightarrow{{}r}=\overrightarrow{{}c}+\frac{2 \overrightarrow{{}a}}{7}=\frac{1}{7}(11 \hat{i}-11 \hat{k})$
$\therefore |\overrightarrow{{}r}| = \frac{11}{7} \sqrt{2}$
