Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes $OX, OY,$ and $OZ$.

(A) Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$.
The magnitude of the vector $\vec{a}$ is given by $|\vec{a}| = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}$.
The direction cosines $(l, m, n)$ of the vector $\vec{a}$ are given by the components of the unit vector $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.
Thus,$l = \frac{1}{\sqrt{3}}$,$m = \frac{1}{\sqrt{3}}$,and $n = \frac{1}{\sqrt{3}}$.
Let $\alpha, \beta,$ and $\gamma$ be the angles that the vector $\vec{a}$ makes with the positive directions of the $OX, OY,$ and $OZ$ axes respectively.
Then,$\cos \alpha = l = \frac{1}{\sqrt{3}}$,$\cos \beta = m = \frac{1}{\sqrt{3}}$,and $\cos \gamma = n = \frac{1}{\sqrt{3}}$.
Since $\cos \alpha = \cos \beta = \cos \gamma$,it follows that $\alpha = \beta = \gamma$.
Therefore,the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes $OX, OY,$ and $OZ$.