(a) If A and B are two vector functions, what does the expression $\left(\mathbf{A}·\nabla \right)\mathbf{B}$mean?(That is, what are its x, y, and z components, in terms of the Cartesian componentsof A, B, and V?)

(b) Compute $\left(\hat{\mathbf{r}}·\nabla \right)\hat{\mathbf{r}}$, where r is the unit vector defined in Eq. 1.21.

(c) For the functions in Prob. 1.15, evaluate $\left({\mathbf{v}}_{\mathbf{a}}·\nabla \right){\mathbf{v}}_{\mathbf{b}}$.

The seperation vector is obtained by subtracting the source vector ${\hat{\mathbf{r}}}_{\mathbf{2}}$from the destination${\hat{\mathbf{r}}}_{\mathbf{1}}$ . The expression of position vector is as follows:

${\hat{\mathbf{r}}}_{}=xi+yj+zk$

Where i, j, k are unit vectors along x, y, z coordintaes

Consider the expression is $\left(\mathrm{A}·\nabla \right)\mathrm{B}$.

Write the expression as:

Solve further as:

$$\begin{gathered} \begin{array}{rcl}\left(A·\nabla \right)B& =& \left(A_x\frac{\partial }{\partial x}+A_y\frac{\partial }{\partial y}+A_z\frac{\partial }{\partial z}\right)\overrightarrow{B}\\ & =& \left(A_x\frac{\partial }{\partial x}+A_y\frac{\partial }{\partial y}+A_z\frac{\partial }{\partial z}\right)\left(B_x\hat{x}+B_y\hat{y}+B_z\hat{z}\right)\\ & =& \left\{\left(A_x\frac{\partial B_x}{\partial x}+A_y\frac{\partial B_x}{\partial y}+A_z\frac{\partial B_x}{\partial z}\right)\hat{x}+\left(A_x\frac{\partial B_y}{\partial x}+A_y\frac{\partial B_y}{\partial y}+A_z\frac{\partial B_y}{\partial z}\right)\hat{y} \\ +\left(A_x\frac{\partial B_z}{\partial x}+A_y\frac{\partial B_z}{\partial y}+A_z\frac{\partial B_z}{\partial z}\right)\hat{z}\right\}\\ & & \end{array} \end{gathered}$$

Therefore, the required expression is $$\begin{gathered} \left(A·\nabla \right)B=\left\{\left(A_x\frac{\partial B_x}{\partial x}+A_y\frac{\partial B_x}{\partial y}+A_z\frac{\partial B_x}{\partial z}\right)\hat{x}+\left(A_x\frac{\partial B_y}{\partial x}+A_y\frac{\partial B_y}{\partial y}+A_z\frac{\partial B_y}{\partial z}\right)\hat{y} \\ +\left(A_x\frac{\partial B_z}{\partial x}+A_y\frac{\partial B_z}{\partial y}+A_z\frac{\partial B_z}{\partial z}\right)\hat{z}\right\} \end{gathered}$$.

Consider the expression for$\left(\hat{\mathrm{r}}·\nabla \right)\hat{\mathrm{r}}$is obtained as:

Solve for the x component as:

$\begin{array}{rcl}\hat{r}& =& \frac{\overrightarrow{r}}{\hat{r}}\\ & =& \frac{x\hat{x}+y\hat{y}+z\hat{z}}{\sqrt{x^2+y^2+z^2}}\\ & & \end{array}$

Solve for the x component of $\hat{r}·\nabla$as:

$$\begin{gathered} \left\{\frac{x}{\sqrt{x^2+y^2+z^2}}\frac{\partial }{dx}\left(\frac{x}{\sqrt{x^2+y^2+z^2}}\right)+\frac{x}{\sqrt{x^2+y^2+z^2}}\frac{\partial }{dy}\left(\frac{x}{\sqrt{x^2+y^2+z^2}}\right) \\ +\frac{x}{\sqrt{x^2+y^2+z^2}}\frac{\partial }{dz}\left(\frac{x}{\sqrt{x^2+y^2+z^2}}\right)\right\}=\frac{x{y}^2+x{z}^2-x{y}^2-x{z}^2}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}} \end{gathered}$$

Simliar values will be obtained fro y and z component.

Therefore, the values of$\left(\hat{\mathbf{r}}·\nabla \right)\hat{\mathbf{r}}=0$.

Consider the expressions for the vector as:

$$\begin{gathered} V_a=x^2\hat{x}+3x{z}^2\hat{y}-2xz\hat{z} \\ {V}_b=xy\hat{x}+2yz\hat{y}-3xz\hat{z} \end{gathered}$$

Solve for$\left({\mathbf{v}}_{\mathbf{a}}·\nabla \right){\mathbf{v}}_{\mathbf{b}}$as:

$$\begin{gathered} \begin{array}{rcl}\left(V_{\mathbf{a}}·\nabla \right)V_{\mathbf{b}}& =& \left\{\left(x^{2}y+3x^2{z}^2-2xz\left(0\right)\right)\hat{x}+\left(x^2\left(0\right)+\left(3z^2\right)\left(2xz\right)-\left(2xz\right)\left(2y\right)\right)\hat{y} \\ +\left(x^2\left(3z\right)+3x{z}^2\left(0\right)-2xz\left(3x\right)\right)\hat{z}\right\}\\ & =& \left(x^{2}y+3x^2{z}^2\right)\hat{x}+\left(6x{z}^2-4xyz\right)\hat{y}-3x^{2}z\hat{z}\\ & & \end{array} \end{gathered}$$

Therefore, the required expression is $\left(V_{\mathbf{a}}·\nabla \right)V_{\mathbf{b}}=\left(x^{2}y+3x^2{z}^2\right)\hat{x}+\left(6x{z}^2-4xyz\right)\hat{y}-3x^{2}z\hat{z}$.