Question

For a position vector $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ the norm of the vector can be defined as $|\overrightarrow{r}|=\sqrt{x^2+y^2+z^2}.$ Given a function $\varphi =ln|\overrightarrow{r}|,$ its gradient $▽\varphi$ is

• A. $\overrightarrow{r}$
• B. $\frac{\overrightarrow{r}}{|\overrightarrow{r}|}$
• C. $\frac{\overrightarrow{r}}{\overrightarrow{r}.\overrightarrow{r}}$
• D. $\frac{\overrightarrow{r}}{|\overrightarrow{r}{|}^3}$

Answer 1

The correct option is C $\frac{\overrightarrow{r}}{\overrightarrow{r}.\overrightarrow{r}}$

$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ and $r=|\overrightarrow{r}|=\sqrt{x^2+y^2+z^2}$
We know that
$gradf\left(r\right)=f^{\prime }\left(r\right)\left(\frac{\overrightarrow{r}}{r}\right)$
So, $grad\phi =grad\left(logr\right)$
$=\frac{1}{r}\left(\frac{\overrightarrow{r}}{r}\right)=\frac{\overrightarrow{r}}{r^2}=\frac{\overrightarrow{r}}{\overrightarrow{r}.\overrightarrow{r}}$