A vector $\mathbf{r}$ is inclined at equal angles to the three axes. If the magnitude of $\mathbf{r}$ is $2\sqrt{3}$ units, then find the value of $\mathbf{r}$.

$\pm 2(\hat{i} + \hat{j} + \hat{k})$

The correct answer is Option (2) → $\pm 2(\hat{i} + \hat{j} + \hat{k})$ ##

We have, $|\mathbf{r}| = 2\sqrt{3}$

Since, $\mathbf{r}$ is equally inclined to the three axes, so direction cosines of the unit vector of $\mathbf{r}$ will be same. i.e., $l = m = n$

We know that,

$l^2 + m^2 + n^2 = 1 \quad [∵l=m=n]$

$\Rightarrow l^2 + l^2 + l^2 = 1$

$\Rightarrow l^2 = \frac{1}{3}$

$\Rightarrow l = \pm \left( \frac{1}{\sqrt{3}} \right)$

So, $\hat{r} = \pm \frac{1}{\sqrt{3}}\hat{i} \pm \frac{1}{\sqrt{3}}\hat{j} \pm \frac{1}{\sqrt{3}}\hat{k}$

$∴\mathbf{r} = \hat{r}|\mathbf{r}| \quad \left[ ∵\hat{r} = \frac{\mathbf{r}}{|\mathbf{r}|} \right]$

$= \left[ \pm \frac{1}{\sqrt{3}}\hat{i} \pm \frac{1}{\sqrt{3}}\hat{j} \pm \frac{1}{\sqrt{3}}\hat{k} \right] 2\sqrt{3} \quad [∵|\mathbf{r}| = 2\sqrt{3}]$

$= \pm 2\hat{i} \pm 2\hat{j} \pm 2\hat{k} = \pm 2(\hat{i} + \hat{j} + \hat{k})$