Question

A vector $\overrightarrow{r}$ is inclined at equal angles to the three axes.If the magnitude or $\overrightarrow{r}is2\sqrt{3}$ units, then find the value of $\overrightarrow{r}$ .

Answer 1

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

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

We know that,

$l^2+m^2+n^2=1\Rightarrow l^2+l^2+l^2=1\Rightarrow l^2=\frac{1}{3}\Rightarrow l=\pm \left(\frac{1}{\sqrt{3}}\right)So,\hat{r}=\pm \frac{1}{\sqrt{3}}\hat{i}\pm \frac{1}{\sqrt{3}}\hat{j}\pm \frac{1}{\sqrt{3}}\hat{k}\therefore \overrightarrow{r}=\hat{r}|\overrightarrow{r}|[\because \hat{r}=\frac{\overrightarrow{r}}{|\overrightarrow{r}|}]=[\pm \frac{1}{\sqrt{3}}\hat{i}\pm \frac{1}{\sqrt{3}}\hat{j}\pm \frac{1}{\sqrt{3}}\hat{k}]2\sqrt{3}[\because |r|=2\sqrt{3}]=\pm 2\hat{i}\pm 2\hat{j}\pm 2\hat{k}=\pm 2(\hat{i}+\hat{j}+\hat{k})$