If are mutually perpendicular vectors of equal magnitudes, show that the vector

Solution:

Since $\vec{a}, \vec{b}$, and $\vec{c}$ are mutually perpendicular vectors, we have

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

It is given that:

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

Let vector $\vec{a}+\vec{b}+\vec{c}$ be inclined to $\vec{a}, \vec{b}$, and $\vec{c}$ at angles $\theta_{1}, \theta_{2}$, and $\theta_{3}$ respectively.

Then, we have:

$\cos \theta_{1}=\frac{(\vec{a}+\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}+\vec{b}+\vec{c}||\vec{a}|}=\frac{\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{a}+\vec{c} \cdot \vec{a}}{|\vec{a}+\vec{b}+\vec{c}||\vec{a}|}$

$=\frac{|\vec{a}|^{2}}{|\vec{a}+\vec{b}+\vec{c}||\vec{a}|}$                $[\vec{b} \cdot \vec{a}=\vec{c} \cdot \vec{a}=0]$

$=\frac{|\vec{a}|}{|\vec{a}+\vec{b}+\vec{c}|}$

$\cos \theta_{2}=\frac{(\vec{a}+\vec{b}+\vec{c}) \cdot \vec{b}}{|\vec{a}+\vec{b}+\vec{c}||\vec{b}|}=\frac{\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{b}}{|\vec{a}+\vec{b}+\vec{c}| \cdot|\vec{b}|}$

$=\frac{|\vec{b}|^{2}}{|\vec{a}+\vec{b}+\vec{c}| \cdot|\vec{b}|}$                             $[\vec{a} \cdot \vec{b}=\vec{c} \cdot \vec{b}=0]$

$=\frac{|\vec{b}|}{|\vec{a}+\vec{b}+\vec{c}|}$

Now, as $|\vec{a}|=|\vec{b}|=|\vec{c}|, \cos \theta_{1}=\cos \theta_{2}=\cos \theta_{3}$.

$\therefore \theta_{1}=\theta_{2}=\theta_{3}$

Hence, the vector $(\vec{a}+\vec{b}+\vec{c})$ is equally inclined to $\vec{a}, \vec{b}$, and $\vec{c}$.