If $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{a}+\vec{b}+\vec{c}$ is equally inclined to $\vec{a}, \vec{b}$ and $\vec{c}$.
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 $\vec{c}$.