Vector Algebra Question 11
Question 11 - 29 January - Shift 2
If $\vec{a}=\hat{i}+2 \hat{k}, \vec{b}=\hat{i}+\hat{j}+\hat{k}, \vec{c}=7 \hat{i}-3 \hat{k}+4 \hat{k}$, $\overrightarrow{{}r} \times \overrightarrow{{}b}+\overrightarrow{{}b} \times \overrightarrow{{}c}=\overrightarrow{{}0}$ and $\overrightarrow{{}r} . \overrightarrow{{}a}=0$ then $\overrightarrow{{}r} . \overrightarrow{{}c}$ is equal to
(1) 34
(2) 12
(3) 36
(4) 30
Show Answer
Answer: (1)
Solution:
Formula: Vector product of two vectors when two vectors are parallel, Properties of Scalar product of vectors
$r \times b-\overrightarrow{{}c} \times b=0$
$\Rightarrow(\overrightarrow{{}r}-\overrightarrow{{}c}) \times \overrightarrow{{}b}=0$
$\Rightarrow \overrightarrow{{}r}-\overrightarrow{{}c}=\lambda \overrightarrow{{}b}$
$\Rightarrow \overrightarrow{{}r}=\overrightarrow{{}c}+\lambda \overrightarrow{{}b}$
And given that $\overrightarrow{{}r} \cdot \overrightarrow{{}a}=0$
$\Rightarrow(\overrightarrow{{}c}+\lambda \overrightarrow{{}b}) \cdot \overrightarrow{{}a}=0$
$\Rightarrow \vec{c} \cdot \vec{a}+\lambda \vec{b} \cdot \vec{a}=0$
$\Rightarrow \lambda=\frac{-\overrightarrow{{}c} \cdot \overrightarrow{{}a}}{\overrightarrow{{}b} \cdot \overrightarrow{{}a}}$
Now $\overrightarrow{{}r} \cdot \overrightarrow{{}c}=(\overrightarrow{{}c}+\lambda \overrightarrow{{}b}) \cdot \overrightarrow{{}c}$
$ \begin{aligned} & =(\overrightarrow{{}c}-\frac{\overrightarrow{{}c} \cdot \overrightarrow{{}a}}{\overrightarrow{{}b} \cdot \overrightarrow{{}a}}) \cdot \overrightarrow{{}c} \\ & =|\overrightarrow{{}c}|-(\frac{\overrightarrow{{}c} \cdot \overrightarrow{{}a}}{\overrightarrow{{}b} \cdot \overrightarrow{{}a}})(\overrightarrow{{}b} \cdot \overrightarrow{{}c}) \\ & =74-[\frac{15}{3}] 8 \\ & =74-40=34 \end{aligned} $
