91Ó°ÊÓ

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

$\stackrel{→}{r}=xi+yj+zk$

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

a)

Consider the expression is $\left(A·∇\right)B$.

Write the expression as:

$\begin{array}{rcl}\left(A·∇\right)B& =& \left(A_x\hat{x}+A_y\hat{y}+A_z\hat{z}\right)·\left(\frac{∂}{∂x}\hat{x}+\frac{∂}{∂y}\hat{y}+\frac{∂}{∂z}\hat{z}\right)\\ & =& \left(A_x\frac{∂}{∂x}+A_y\frac{∂}{∂y}+A_z\frac{∂}{∂z}\right)\stackrel{→}{B}\\ & & \end{array}$

Solve further as:

$$\begin{gathered} \begin{array}{rcl}\left(A·∇\right)B& =& \left(A_x\frac{∂}{∂x}+A_y\frac{∂}{∂y}+A_z\frac{∂}{∂z}\right)\stackrel{→}{B}\\ & =& \left(A_x\frac{∂}{∂x}+A_y\frac{∂}{∂y}+A_z\frac{∂}{∂z}\right)\left(B_x\hat{x}+B_y\hat{y}+B_z\hat{z}\right)\\ & =& \left\{\backslash \mathrm{begingathered}\left(A_x\frac{∂B_x}{∂x}+A_y\frac{∂B_x}{∂y}+A_z\frac{∂B_x}{∂z}\right)\hat{x}+\left(A_x\frac{∂B_y}{∂x}+A_y\frac{∂B_y}{∂y}+A_z\frac{∂B_y}{∂z}\right)\hat{y}\backslash \\ +\left(A_x\frac{∂B_z}{∂x}+A_y\frac{∂B_z}{∂y}+A_z\frac{∂B_z}{∂z}\right)\hat{z} \\ \right\}\\ & & \end{array} \end{gathered}$$

Therefore, the reuired expression is .

$$\begin{gathered} \left(A·∇\right)B=\left\{\left(A_x\frac{∂B_x}{∂x}+A_y\frac{∂B_x}{∂y}+A_z\frac{∂B_x}{∂z}\right)\hat{x}+\left(A_x\frac{∂B_y}{∂x}+A_y\frac{∂B_y}{∂y}+A_z\frac{∂B_y}{∂z}\right)\hat{y}\backslash \\ +\left(A_x\frac{∂B_z}{∂x}+A_y\frac{∂B_z}{∂y}+A_z\frac{∂B_z}{∂z}\right)\hat{z} \\ \right\} \end{gathered}$$

(b)

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

Solve for the x component as:

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

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

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

Simliar values will be obtained fro y and z component.

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

(c)

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

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

Therefore, the required expression is .

$\left(V_{\mathbf{a}}·∇\right)V_{\mathbf{b}}=\left(x^{2}y+3x^2{z}^2\right)\hat{x}+\left(6x{z}^2-4xyz\right)\hat{y}-3x^{2}z\hat{z}$