If $\vec{a}, \vec{b}$ and $\vec{c}$ 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 the 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:
$\cos \theta_{1} = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a} + \vec{c} \cdot \vec{a}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{|\vec{a}|^2}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{|\vec{a}|}{|\vec{a} + \vec{b} + \vec{c}|}$.
Similarly,$\cos \theta_{2} = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{b}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{b}|} = \frac{\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{b}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{b}|} = \frac{|\vec{b}|^2}{|\vec{a} + \vec{b} + \vec{c}| |\vec{b}|} = \frac{|\vec{b}|}{|\vec{a} + \vec{b} + \vec{c}|}$.
And $\cos \theta_{3} = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{c}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{c}|} = \frac{\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{c}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{c}|} = \frac{|\vec{c}|^2}{|\vec{a} + \vec{b} + \vec{c}| |\vec{c}|} = \frac{|\vec{c}|}{|\vec{a} + \vec{b} + \vec{c}|}$.
Since $|\vec{a}| = |\vec{b}| = |\vec{c}|$,we have $\cos \theta_{1} = \cos \theta_{2} = \cos \theta_{3}$,which implies $\theta_{1} = \theta_{2} = \theta_{3}$.
Hence,the vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}, \vec{b}$ and $\vec{c}$.
It is given that $|\vec{a}| = |\vec{b}| = |\vec{c}|$.
Let the 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:
$\cos \theta_{1} = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a} + \vec{c} \cdot \vec{a}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{|\vec{a}|^2}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{|\vec{a}|}{|\vec{a} + \vec{b} + \vec{c}|}$.
Similarly,$\cos \theta_{2} = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{b}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{b}|} = \frac{\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{b}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{b}|} = \frac{|\vec{b}|^2}{|\vec{a} + \vec{b} + \vec{c}| |\vec{b}|} = \frac{|\vec{b}|}{|\vec{a} + \vec{b} + \vec{c}|}$.
And $\cos \theta_{3} = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{c}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{c}|} = \frac{\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{c}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{c}|} = \frac{|\vec{c}|^2}{|\vec{a} + \vec{b} + \vec{c}| |\vec{c}|} = \frac{|\vec{c}|}{|\vec{a} + \vec{b} + \vec{c}|}$.
Since $|\vec{a}| = |\vec{b}| = |\vec{c}|$,we have $\cos \theta_{1} = \cos \theta_{2} = \cos \theta_{3}$,which implies $\theta_{1} = \theta_{2} = \theta_{3}$.
Hence,the vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}, \vec{b}$ and $\vec{c}$.
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