Question

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

Answer 1

Let l, m, n be the direction cosines of $\overrightarrow{r}$.

Now, $\overrightarrow{r}$ is inclined at equal angles to the three axes.

$$\begin{gathered} \therefore l=m=n\left[\alpha =\beta =\gamma \Rightarrow \cos \alpha =\cos \beta =\cos \gamma \right] \\ \mathrm{So},l^2+m^2+n^2=1 \\ \Rightarrow 3l^2=1 \\ \Rightarrow l=\pm \frac{1}{\sqrt{3}} \\ \Rightarrow l=m=n=\pm \frac{1}{\sqrt{3}} \end{gathered}$$
We know that

$$\begin{gathered} \overrightarrow{r}=\left|\overrightarrow{r}\right|\left(l\hat{i}+m\hat{j}+n\hat{k}\right) \\ \Rightarrow \overrightarrow{r}=2\sqrt{3}\left(\pm \frac{1}{\sqrt{3}}\hat{i}\pm \frac{1}{\sqrt{3}}\hat{j}\pm \frac{1}{\sqrt{3}}\hat{k}\right) \\ \Rightarrow \overrightarrow{r}=2\left(\pm \hat{i}\pm \hat{j}\pm \hat{k}\right) \end{gathered}$$