Question

Let $▽.\left(f\overrightarrow{v}\right)=x^{2}y+y^{2}z+z^{2}x,$ where $f$ and $v$ are scalar and vector fields respectively. If $\overrightarrow{v}=y\hat{i}+z\hat{j}+x\hat{k},$ then $\overrightarrow{v}.▽f$ is

• A. $x^{2}y+y^{2}z+z^{2}x$
• B. $2xy+2yz+2zx$
• C. $x+y+z$
• D. $0$

Answer 1

The correct option is A $x^{2}y+y^{2}z+z^{2}x$

Given $▽.\left(f\overrightarrow{v}\right)=x^{2}y+y^{2}z+z^{2}x$
and $\overrightarrow{v}=y\hat{i}+z\hat{j}+x\hat{k}\Rightarrow ▽.\overrightarrow{v}=0$
Now, using vector identity,
$▽.\left(f\overrightarrow{v}\right)=f(▽.\overrightarrow{v})+▽f.\overrightarrow{v}$
$=0+\overrightarrow{v}.▽f$
(as dot product is commutative)
$\overrightarrow{v}.▽f=▽.\left(f\overrightarrow{v}\right)$
$=x^{2}y+y^{2}z+z^{2}x$