Question

A vector \[\vec{r}\] is inclined at equal angles to the three axes. If the magnitude of \[\vec{r}\] is \[2\sqrt{3}\], find \[\vec{r}\].

Answer 1

Let l, m, n be the direction cosines of \[\vec{r}\]
Now, \[\vec{r}\] is inclined at equal angles to the three axes.
\[\therefore l = m = n \left[ \alpha = \beta = \gamma \Rightarrow \cos\alpha = \cos\beta = \cos\gamma \right]\]
\[\text { So }, l^2 + m^2 + n^2 = 1\]
\[ \Rightarrow 3 l^2 = 1\]
\[ \Rightarrow l = \pm \frac{1}{\sqrt{3}}\]
\[ \Rightarrow l = m = n = \pm \frac{1}{\sqrt{3}}\]

We know that

\[\vec{r} = \left| \vec{r} \right|\left( l \hat{i} + m \hat{j} + n \hat{k} \right)\]\[ \Rightarrow \vec{r} = 2\sqrt{3}\left( \pm \frac{1}{\sqrt{3}} \hat{i} \pm \frac{1}{\sqrt{3}} \hat{j} \pm \frac{1}{\sqrt{3}} \hat{k} \right)\]
\[ \Rightarrow \vec{r} = 2\left( \pm \hat{i} \pm \hat{j} \pm \hat{k} \right)\]