91Ó°ÊÓ

Write the given vectors.

$$\begin{gathered} v_a=x^{2}i+3x{z}^{2}j+(-2xz)k \\ {v}_b=y^{2}i+2yzj+3zxk \\ {v}_c=y^{2}i+(2xy+z^2)j+2yzk \end{gathered}$$

The gradient of a scalar function $\mathit{F}$ is defined as$\mathbf{∇}\mathit{F}$and curl of a vector function$v$is defined as $\mathbf{∇}\mathbf{×}\mathit{v}$.

$$\begin{gathered} ∇.v_a=\left[\begin{array}{c}\frac{∂}{∂x}(x^{2}i+3x{z}^{2}j+(-2xz\left)k\right)+\frac{∂}{∂y}(x^{2}i+3x{z}^{2}j+(-2xz\left)k\right)\\ +\frac{∂}{∂z}(x^{2}i+3x{z}^{2}j+(-2xz\left)k\right)\end{array}\right] \\ =2x+0-2x \\ =0 \end{gathered}$$

The dot product of vector $v_b$is obtained as follows:

$$\begin{gathered} ∇.v_a=\left[\begin{array}{c}\frac{∂}{∂x}(y^{2}i+2yzj+3zxk)+\frac{∂}{∂y}(y^{2}i+2yzj+3zxk)\\ +\frac{∂}{∂z}(y^{2}i+2yzj+3zxk)\end{array}\right] \\ =y+2z+3x \end{gathered}$$

The dot product of vector $v_c$is obtained as follows:

$$\begin{gathered} ∇.v_a=\left[\begin{array}{c}\frac{∂}{∂x}(y^{2}i+(2xy+z^2)j+2yzk)+\frac{∂}{∂y}(y^{2}i+(2xy+z^2)j+2yzk)\\ +\frac{∂}{∂z}(y^{2}i+(2xy+z^2)j+2yzk)\end{array}\right] \\ =0+(2x+0)+2y \\ =2(x+y) \end{gathered}$$

The cross product of vector $v_a$is obtained as follows:

$$\begin{gathered} ∇×v_a=\left|\begin{array}{ccc}i& j& k\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ x^2& 3x{z}^2& -2xz\end{array}\right| \\ =\left[\begin{array}{c}i\left(\frac{∂}{∂y}(-2zx)-\frac{∂}{∂z}\left(3x{z}^2\right)\right)-j\left(\frac{∂}{∂x}(-2xz)-\frac{∂}{∂z}\left(x^2\right)\right)\\ +i\left(\frac{∂}{∂x}\left(3x{z}^2\right)-\frac{∂}{∂y}\left(x^2\right)\right)\end{array}\right]/ \\ =-6xzi+2zj+3z^{2}k \end{gathered}$$

The cross product of vector $v_b$is obtained as follows:

$$\begin{gathered} ∇×v_a=\left|\begin{array}{ccc}i& j& k\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ y^2& 2yz& 3zx\end{array}\right| \\ =\left[\begin{array}{c}i\left(\frac{∂}{∂y}\left(3zx\right)-\frac{∂}{∂z}\left(2yz\right)\right)-j\left(\frac{∂}{∂x}\left(3zx\right)-\frac{∂}{∂z}\left(y^2\right)\right)\\ +i\left(\frac{∂}{∂x}\left(2yz\right)-\frac{∂}{∂y}\left(y^2\right)\right)\end{array}\right] \\ =-2yi-3zj-xk \end{gathered}$$

The cross product of vector$v_c$ is obtained as follows:

$$\begin{gathered} ∇×v_a=\left|\begin{array}{ccc}i& j& k\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ y^2& 2xy+z^2& x\end{array}\right| \\ =\left[\begin{array}{c}i\left(\frac{∂}{∂y}\left(2yz\right)-\frac{∂}{∂z}(2xy+z^2)\right)-j\left(\frac{∂}{∂x}\left(2yz\right)-\frac{∂}{∂z}\left(y^2\right)\right)\\ +i\left(\frac{∂}{∂x}(2xy+z^2)-\frac{∂}{∂y}\left(y^2\right)\right)\end{array}\right] \\ =0 \end{gathered}$$

Out of the all the given functions, the curl of vector $v_c$is 0. So, it can be expressed as gradient of scalar function.

Let us assume $∇A-v_c$. Expand $∇A-v_c$ as,

$\frac{∂A}{∂x}i+\frac{∂A}{∂y}j+\frac{∂A}{∂z}k=y^{2}i+(2xy+z^2)j+2yzk$

On comparing the right and left side of the above equation, we obtain,

$$\begin{gathered} \frac{∂A}{∂x}=y^2 \\ A=y^{2}x+g(y,z) \end{gathered}$$

Where, $g(y,z)$ is a function of y , z.

Differentiate the above function with respect to y.\

$\frac{∂A}{∂y}=2yx+\frac{∂g(y,z)}{∂y}$

Where, role="math" localid="1655958836210" $g(y,z)$is a function of y , z.

comparing the right and left side of the above equation, as,

$2xy+z^2=2yx+\frac{∂g(y,z)}{∂y}$

We obtain the result, $\frac{∂g(y,z)}{∂y}=z^2$, On integrating this equation with respect to y on both sides, we get,

$g(y,z)=z^{2}y+f\left(z\right)$

Thus, the scalar function can be written as $A=y^{2}x+z^{2}y+f\left(z\right)$.

Differentiate the above function with respect to z.

$\frac{∂A}{∂z}-2yz+f^{\text{'}}\left(z\right)$

Comparing the right and left side of the above equation, as,

$2yz=2yz+f^{\text{'}}\left(z\right)$

Obtain the result, $f^{\text{'}}\left(z\right)=0$, On integrating this equation with respect to z on both sides, we get,

role="math" localid="1655958018234" $f^{\text{'}}\left(z\right)=C$

Here, C is some constant. Thus the scalar function is obtained as $A=y^{2}x+z^{2}y+C$ .

Out of the all the given functions, the dot product of vector $v_a$is 0. So, it can be expressed as curl of a vector, as divergence of curl is always 0.

Let us assume$∇×F=v_a$. Expand $∇×F=v_a$as,

$$\begin{gathered} ∇×F=v_a \\ \left|\begin{array}{ccc}i& j& k\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ F_x& F_y& F_z\end{array}\right|-x^{2}i+3z^{2}j+(-2xz)k \\ \left(\frac{∂F_z}{∂y}-\frac{∂F_y}{∂z}\right)i-\left(\frac{∂F_z}{∂x}-\frac{∂F_x}{∂z}\right)j+\left(\frac{∂F_y}{∂x}-\frac{∂F_x}{∂y}\right)k=x^{2}i+3z^{2}j+(-2xz)k \end{gathered}$$

On comparing the right and left side of the above equation, we obtain,

$$\begin{gathered} \left(\frac{∂F_z}{∂y}-\frac{∂F_y}{∂z}\right)=x^2......\left(1\right) \\ \left(\frac{∂F_x}{∂z}-\frac{∂F_z}{∂x}\right)=3x{z}^2.......\left(2\right) \\ \left(\frac{∂F_y}{∂x}-\frac{∂F_x}{∂y}\right)=2xz.........\left(3\right) \end{gathered}$$

To calculate $F_x$,$F_y$ , assume that $F_z=0$ . Substitute 0 for $F_z$int equation (2).

$$\begin{gathered} \left(\frac{∂F_z}{∂x}-\frac{∂\left(0\right)}{∂z}\right)=-3x{z}^2 \\ \frac{∂F_z}{∂x}=-3x{z}^2 \\ {F}_z=-\frac{3}{2}{x}^2{z}^2+f(x,y) \end{gathered}$$

Substitute 0 for $F_x$ int equation (3).

$$\begin{gathered} \left(\frac{∂F_y}{∂x}-\frac{∂\left(0\right)}{∂}\right)=-2xz \\ \frac{∂Fy}{∂x}=-2xz \\ {F}_y=-x^{2}z+g(y,z) \end{gathered}$$

Substitute $-x^{2}z+g(y,z)$ for $F_{y,}-\frac{3}{2}-x^2{z}^2+f(x,y)$for $F_z$ into equation (1).

$$\begin{gathered} \left(\frac{∂}{∂y}\left(-\frac{3}{2}-x^2{z}^2+f(x,y)\right)-\frac{∂}{∂z}(-x^{2}z+g(y,z\left)\right)\right)=x^2 \\ \frac{∂}{∂y}f(y,z)+x^2-\frac{∂}{∂z}g(y,z)=x^2 \\ \frac{∂}{∂y}f(y,z)+x^2-\frac{∂}{∂z}g(y,z)=0 \end{gathered}$$

The resulting expression $\frac{∂}{∂y}f(y,z)-\frac{∂}{∂z}g(y,z)=0$ is possible only when .

$f(y,z)=g(y,z)=0$

Thus, the vector $F=F_{x}i+F_{y}j+F_{z}k$ can be written as $F=x^{2}zj-\frac{3}{2}{x}^2{z}^{2}k$.