Chapter 1, Problem 15P

(a)

To determine

Simplify the given expression.

Answer to Problem 15P

The simplified form of the expression is $\underset{\_}{\left(B\cdot A\right)A-\left(A\cdot A\right)B}$.

Explanation of Solution

Calculation:

Consider the following two vectors.

$A=\left(A_x,A_y,A_z\right)$        (1)

$B=\left(B_x,B_y,B_z\right)$        (2)

Calculate the cross product $A\times B$ using Equations (1) and (2).

$A\times B=\left[\begin{array}{ccc}{a}_x& a_y& a_z\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right]$

$A\times B=\left(A_y{B}_z-A_z{B}_y\right)a_x-\left(A_x{B}_z-A_z{B}_x\right)a_y+\left(A_x{B}_y-A_y{B}_x\right)a_z$        (3)

Calculate the cross product $A\times \left(A\times B\right)$ using Equations (1) and (3).

$\begin{array}{c}A\times \left(A\times B\right)=\left[\begin{array}{ccc}{a}_x& a_y& a_z\\ A_x& A_y& A_z\\ \left(A_y{B}_z-A_z{B}_y\right)& -\left(A_x{B}_z-A_z{B}_x\right)& \left(A_x{B}_y-A_y{B}_x\right)\end{array}\right]\\ =\left\{\begin{array}{l}\left[\left(A_x{B}_y-A_y{B}_x\right)A_y+\left(A_x{B}_z-A_z{B}_x\right)A_z\right]a_x-\\ \left[\left(A_x{B}_y-A_y{B}_x\right)A_x-\left(A_y{B}_z-A_z{B}_y\right)A_z\right]a_y+\\ \left[-\left(A_x{B}_z-A_z{B}_x\right)A_x-\left(A_y{B}_z-A_z{B}_y\right)A_y\right]a_z\end{array}\right\}\\ =\left\{\begin{array}{l}\left[\left(A_x{A}_y{B}_y-A_y{A}_y{B}_x\right)+\left(A_x{A}_z{B}_z-A_z{A}_z{B}_x\right)\right]a_x-\\ \left[\left(A_x{A}_x{B}_y-A_x{A}_y{B}_x\right)-\left(A_y{A}_z{B}_z-A_z{A}_z{B}_y\right)\right]a_y+\\ \left[-\left(A_x{A}_x{B}_z-A_x{A}_z{B}_x\right)-\left(A_y{A}_y{B}_z-A_y{A}_z{B}_y\right)\right]a_z\end{array}\right\}\\ =\left\{\begin{array}{l}\left[A_x{A}_x{B}_x+A_x{A}_y{B}_y+A_x{B}_z{A}_z\right]a_x+\\ \left[A_y{A}_z{B}_z+A_y{A}_y{B}_y+A_z{A}_z{B}_z\right]a_y+\\ \left[A_x{A}_z{B}_x+A_y{A}_z{B}_y+A_z{A}_z{B}_z\right]a_z-\\ \left[A_x{A}_x{B}_x+A_y{A}_y{B}_x+A_z{A}_z{B}_x\right]a_x-\\ \left[A_z{A}_z{B}_z+A_y{A}_y{B}_y+A_x{A}_x{B}_y\right]a_y-\\ \left[A_x{A}_x{B}_z+A_y{A}_y{B}_z+A_z{A}_z{B}_z\right]a_z\end{array}\right\}\end{array}$

Simplify the equation as follows:

$\begin{array}{c}A\times \left(A\times B\right)=\left(A_x{B}_x+A_y{B}_y+A_z{B}_z\right)\left(A_x,A_y,A_z\right)-\left(A_x{A}_x+A_y{A}_y+A_z{A}_z\right)\left(B_x,B_y,B_z\right)\\ =\left(A\cdot B\right)A-\left(A\cdot A\right)B\\ =\left(B\cdot A\right)A-\left(A\cdot A\right)B\left\{\because A\cdot B=B\cdot A\right\}\end{array}$

Conclusion:

Thus, the simplified form of the expression is $\underset{\_}{\left(B\cdot A\right)A-\left(A\cdot A\right)B}$.

(b)

To determine

Simplify the given expression.

Answer to Problem 15P

The simplified form of the expression is $\underset{\_}{-A^2\left(A\times B\right)}$.

Explanation of Solution

Calculation:

Consider that $A\times B=C$.

Then, the expression for $A\times \left[A\times \left(A\times B\right)\right]$ is,

$A\times \left[A\times \left(A\times B\right)\right]=A\times \left[A\times C\right]$        (4)

Consider the expression for vector triple product.

$A\times \left(A\times C\right)=A\left(A\cdot C\right)-C\left(A\cdot A\right)$        (5)

Substitute Equation (5) in (4).

$\begin{array}{c}A\times \left[A\times \left(A\times B\right)\right]=A\left(A\cdot C\right)-C\left(A\cdot A\right)\\ =A\left(A\cdot \left(A\times B\right)\right)-\left(A\times B\right)\left(A\cdot A\right)\left\{\because C=A\times B\right\}\end{array}$

$A\times \left[A\times \left(A\times B\right)\right]=A\left(A\cdot \left(A\times B\right)\right)-\left(A\times B\right)\left(A\cdot A\right)$        (6)

Calculate the cross product $A\cdot \left(A\times B\right)$ using Equations (1) and (3).

$\begin{array}{c}A\cdot \left(A\times B\right)=\left(A_x,A_y,A_z\right)\cdot \left(\left(A_y{B}_z-A_z{B}_y\right),-\left(A_x{B}_z-A_z{B}_x\right),\left(A_x{B}_y-A_y{B}_x\right)\right)\\ =A_x\left(A_y{B}_z-A_z{B}_y\right)-\left(A_x{B}_z-A_z{B}_x\right)A_y+\left(A_x{B}_y-A_y{B}_x\right)A_z\\ =\left(A_x{A}_y{B}_z-A_x{A}_z{B}_y\right)-\left(A_x{A}_y{B}_z-A_y{A}_z{B}_x\right)+\left(A_x{A}_z{B}_y-A_y{A}_z{B}_x\right)\\ =0\end{array}$

Substitute 0 for $A\cdot \left(A\times B\right)$ in Equation (6).

$\begin{array}{c}A\times \left[A\times \left(A\times B\right)\right]=A\left(0\right)-\left(A\times B\right)\left(A\cdot A\right)\\ =-\left(A\times B\right)\left(A\cdot A\right)\\ =-A^2\left(A\times B\right)\end{array}$

Conclusion:

Thus, the simplified form of the expression is $\underset{\_}{-A^2\left(A\times B\right)}$.