# Get Answers to all your Questions

#### Let  be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle , with the vector . Then  is equal to ________.

Given, $|\vec{a}|=|\vec{b}|=|\vec{c}|$

and these are mutually perpendicular,

So, $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0$

Angle of $\vec{a}$ with $(\vec{a}+\vec{b}+\vec{c})=\theta$

\begin{aligned} &\therefore \quad \vec{a} \cdot(\vec{a}+\vec{b}+\vec{c})=|\vec{a}||\vec{a}+\vec{b}+\vec{c}| \cdot \cos \theta \\ &\Rightarrow \quad \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=|\vec{a}||\vec{a}+\vec{b}+\vec{c}| \cos \theta \\ &\Rightarrow \quad|\vec{a}|^{2}+0+0=|\vec{a}||\vec{a}+\vec{b}+\vec{c}| \cos \theta \\ &\Rightarrow \quad \frac{|\vec{a}|}{|\vec{a}+\vec{b}+\vec{c}|}=\cos \theta\; \; \; \; \; \; \; \; \; \; \; \; \; ...........(i) \end{aligned}

Now,

\begin{aligned} &|\vec{a}+\vec{b}+\vec{c}|^{2}=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c}) \\ &|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\ \Rightarrow &|\vec{a}+\vec{b}+\vec{c}|^{2}=3|\vec{a}|^{2}+2(0) \\ \Rightarrow &|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3}|\vec{a}| \end{aligned}

Using this in (i)

\begin{aligned} &\Rightarrow \quad \cos \theta=\frac{1}{\sqrt{3}} \\ &\Rightarrow \quad \cos 2 \theta=2 \cos ^{2} \theta-1=\frac{2}{3}-1=-\frac{1}{3} . \\ &\Rightarrow \quad 36 \cos ^{2} 2 \theta=\quad 36 \cdot\left(\frac{1}{9}\right)=4 . \end{aligned}

Hence, the correct answer is 4