Question:
Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes $O X, O Y$, and $O Z$.
Solution:
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$
Then,
$|\vec{a}|=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$
Therefore, the direction cosines of $\vec{a}$ are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$
Now, let $\alpha, \beta$, and $\gamma$ be the angles formed by $\vec{a}$ with the positive directions of $x, y$, and $z$ axes.
Then, we have $\cos \alpha=\frac{1}{\sqrt{3}}, \cos \beta=\frac{1}{\sqrt{3}}, \cos \gamma=\frac{1}{\sqrt{3}}$.
Hence, the given vector is equally inclined to axes OX, OY, and OZ.