Prove the following

Question:

Let $\vec{b}, \vec{b}$ and $\vec{c}$ be three vectors such that $|\vec{a}|=\sqrt{3}$, $|\vec{b}|=5, \vec{b} \cdot \vec{c}=10$ and the angle between $\vec{b}$ and $\vec{c}$

is $\frac{\pi}{3}$. If $\vec{b}$ is perpendicular to the vector $\vec{b} \times \vec{c}$, then $|\vec{a} \times(\vec{b} \times \vec{c})|$ is equal to_______.

Solution:

$\vec{b} \cdot \vec{c}=10 \Rightarrow|\vec{b}||\vec{c}| \cos \left(\frac{\pi}{3}\right)=10$

$\Rightarrow 5 \cdot|\vec{c}| \cdot \frac{1}{2}=10 \Rightarrow|\vec{c}|=4$

Since, $\vec{a}$ is perpendicular to the vector $\vec{b} \times \vec{c}$, then

$\vec{a} \cdot(\vec{b} \times \vec{c})=0$

Now, $|\vec{a} \times(\vec{b} \times \vec{c})|=|\vec{a}||\vec{b} \times \vec{c}| \sin \left(\frac{\pi}{2}\right)$

$=\sqrt{3} \times|\vec{b}||\vec{c}| \sin \frac{\pi}{3} \times 1=\sqrt{3} \times 5 \times 4 \times \frac{\sqrt{3}}{2}=30$

Hence, $|\vec{a} \times(\vec{b} \times \vec{c})|=30$