# Solve this

Question:

Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$. If $\overrightarrow{\mathrm{c}}$ is a vectorsuch that $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3$, then $\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})$ is equal to :

1. $-2$

2. –6

3. 2

Correct Option: 1

Solution:

$|\overrightarrow{\mathrm{a}}|=\sqrt{3} ; \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3 ; \quad \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$

Cross with $\vec{a}$.

$\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$

$\Rightarrow(\vec{a} \cdot \vec{c}) \overrightarrow{\mathrm{a}}-\mathrm{a}^{2} \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$

$\Rightarrow 3 \overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{c}}=-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\Rightarrow 3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}-3 \overrightarrow{\mathrm{c}}=-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\Rightarrow \overrightarrow{\mathrm{c}}=\frac{5 \hat{\mathrm{i}}}{3}+\frac{2 \hat{\mathrm{j}}}{3}+\frac{2 \hat{\mathrm{k}}}{3}$

$\therefore \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}=\frac{-10}{3}+\frac{2}{3}+\frac{2}{3}=-2$