# Solve the Following Questions

Question:

Let $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$. If $\overrightarrow{\mathrm{c}}$ is a vector such that $\vec{a} \cdot \vec{c}=|\vec{c}|,|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $(\vec{a} \times \vec{b})$ and $\vec{c}$ is $\frac{\pi}{6}$, then the value of $|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}|$ is :

1. $\frac{2}{3}$

2. 4

3. 3

4. $\frac{3}{2}$

Correct Option: , 4

Solution:

$|\overrightarrow{\mathrm{a}}|=3=\mathrm{a} ; \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=\mathrm{c}$

Now $|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2}$

$\Rightarrow \mathrm{c}^{2}+\mathrm{a}^{2}-2 \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=8$

$\Rightarrow \mathrm{c}^{2}+9-2(\mathrm{c})=8$

$\Rightarrow \mathrm{c}^{2}-2 \mathrm{c}+1=0 \Rightarrow \mathrm{c}=1=|\overrightarrow{\mathrm{c}}|$

Also, $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

Given $(\vec{a} \times \vec{b})=|\vec{a} \times \vec{b}||\vec{c}| \sin \frac{\pi}{6}$

$=(3)(1)(1 / 2)$

$=3 / 2$