Chapter 3, Problem 3.1P

Find $A\times B$ for the following:

1. $A=2a_x-3a_y+4a_z,B=5a_y-1a_z$
2. $A=a_{\rho }+2a_{\varphi }+4a_z,B=2a_{\rho }+6a_z$
3. $A=2a_r+5a_{\theta }+1a_{\varphi },B=a_r+3a_{\varphi }$

To determine

The cross product of the two fields.

Answer to Problem 3.1P

The cross product of the two vectors is $-17a_x+2a_y+10a_z$ .

Explanation of Solution

Given:

The field $A$ is $2a_x-3a_y+4a_z$ .

The field $B$ is $5a_y-1a_z$ .

Concept used:

Write the expression for the cross-product of the two fields.

  $A\times B=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right|$.........(1)

Here, $A_x,A_y,A_z$ are the components of the field $A$ in $x,y,z$ directions and $B_x,B_y,B_z$ are the components of the field $B$ in $x,y,z$ directions.

Calculation:

Substitute $2$ for $A_x$ , $-3$ for $A_y$ , $4$ for $A_z$ , $0$ for $B_x$ , $5$ for $B_y$ and $-1$ for $B_z$ in equation (1).

  $\begin{array}{c}A\times B=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ 2& -3& 4\\ 0& 5& -1\end{array}\right|\\ =a_x\left(3-20\right)-a_y\left(-2-0\right)+a_z\left(10-0\right)\\ =-17a_x+2a_y+10a_z\end{array}$

Conclusion:

Thus, the cross product of the two vectors is $-17a_x+2a_y+10a_z$ .

To determine

The cross product of the two fields.

Answer to Problem 3.1P

The cross product of the two vectors is $12a_{\rho }+2a_{\varphi }-4a_z$ .

Explanation of Solution

Given:

The field $A$ is $a_{\rho }+2a_{\varphi }+4a_z$ .

The field $B$ is $2a_{\rho }+6a_z$ .

Concept used:

Write the expression for the cross-product of the two fields.

  $A\times B=\left|\begin{array}{ccc}{a}_{\rho }& a_{\varphi }& a_z\\ A_{\rho }& A_{\varphi }& A_z\\ B_{\rho }& B_{\varphi }& B_z\end{array}\right|$ ........(2)

Here, $A_{\rho },A_{\varphi },A_z$ are the components of the field $A$ in $\rho ,\varphi ,z$ directions and $B_{\rho },B_{\varphi },B_z$ are the components of the field $B$ in $\rho ,\varphi ,z$ directions.

Calculation:

Substitute $1$ for $A_{\rho }$ , $2$ for $A_{\varphi }$ , $4$ for $A_z$ , $2$ for $B_{\rho }$ , $0$ for $B_{\varphi }$ and $6$ for $B_z$ in equation (2).

  $\begin{array}{c}A\times B=\left|\begin{array}{ccc}{a}_{\rho }& a_{\varphi }& a_z\\ 1& 2& 4\\ 2& 0& 6\end{array}\right|\\ =a_{\rho }\left(12-0\right)-a_{\varphi }\left(6-8\right)+a_z\left(0-4\right)\\ =12a_{\rho }+2a_{\varphi }-4a_z\end{array}$

Conclusion:

Thus, the cross product of the two vectors is $12a_{\rho }+2a_{\varphi }-4a_z$ .

To determine

The cross product of the two fields.

Answer to Problem 3.1P

The cross product of the two vectors is $15a_r-5a_{\theta }-5a_{\varphi }$ .

Explanation of Solution

Given:

The field $A$ is $2a_r+5a_{\theta }+1a_{\varphi }$ .

The field $B$ is $a_r+3a_{\varphi }$ .

Concept used:

Write the expression for the cross-product of the two fields.

  $A\times B=\left|\begin{array}{ccc}{a}_r& a_{\theta }& a_{\varphi }\\ A_r& A_{\theta }& A_{\varphi }\\ B_r& B_{\theta }& B_{\varphi }\end{array}\right|$ ........(3)

Here, $A_r,A_{\theta },A_{\varphi }$ are the components of the field $A$ in $r,\theta ,\varphi$ directions and $B_r,B_{\theta },B_{\varphi }$ are the components of the field $B$ in $r,\theta ,\varphi$ directions.

Calculation:

Substitute $2$ for $A_r$ , $5$ for $A_{\theta }$ , $1$ for $A_{\varphi }$ , $1$ for $B_r$ , $0$ for $B_{\theta }$ and $3$ for $B_{\varphi }$ in equation (3).

  $\begin{array}{c}A\times B=\left|\begin{array}{ccc}{a}_r& a_{\theta }& a_{\varphi }\\ 2& 5& 1\\ 1& 0& 3\end{array}\right|\\ =a_r\left(15-0\right)-a_{\theta }\left(6-1\right)+a_{\varphi }\left(0-5\right)\\ =15a_r-5a_{\theta }-5a_{\varphi }\end{array}$

Conclusion:

Thus, the cross product of the two vectors is $15a_r-5a_{\theta }-5a_{\varphi }$ .