Chapter 1, Problem 1.17P

(a)

To determine

Prove that scalar product is distributive that is $r\times \left(u+v\right)=\left(r\times u\right)+\left(r\times v\right)$.

Answer to Problem 1.17P

It is proved that scalar product is distributive that is $r\times \left(u+v\right)=\left(r\times u\right)+\left(r\times v\right)$.

Explanation of Solution

Let the vector $r$ as,

    $\begin{array}{l}r=r_x\text{x}+r_y\text{y}+r_z\text{z}\\ u=u_x\text{x}+u_y\text{y}+u_z\text{z}\\ v=v_x\text{x}+v_y\text{y}+v_z\text{z}\end{array}$

The LHs is given by,

    $r\times (u+v)=\left|\begin{array}{ccc}\text{x}& \text{y}& \text{z}\\ r_x& r_y& r_z\\ \left(u_x+v_x\right)& \left(u_y+v_y\right)& \left(u_z+v_z\right)\end{array}\right|$        (I)

Solve equation (I),

    $\begin{array}{c}r\times (u+v)=\left[r_y\left(u_z+v_z\right)-r_z\left(u_y+v_y\right)\right]\text{ x}-\left[r_x\left(u_z+v_z\right)-r_z\left(u_x+v_x\right)\right]\text{y}\\ +\left[r_x\left(u_y+v_y\right)-r_y\left(u_x+v_x\right)\right]\text{z }\\ \text{=}\left[\left(r_y{u}_z+r_y{v}_z\right)x-\left(r_z{u}_y+r_z{v}_y\right)x\right]-[\left(r_x{u}_z+r_x{v}_z\right)y-\left(r_z{u}_x+r_z{v}_x\right)y\\ +\left[\left(r_x{u}_y+r_x{v}_y\right)z-\left(r_y{u}_x+r_y{v}_x\right)z\right]\end{array}$

    $\begin{array}{c}r\times (u+v)=\left[\left(r_y{u}_z-r_z{u}_y\right)\text{x}-\left(r_x{u}_z-r_z{u}_x\right)\text{y}+\left(r_x{u}_y-r_y{u}_x\right)\text{z}\right]+\\ \left[\left(r_y{v}_z-r_z{v}_y\right)\text{x}-\left(r_x{v}_z-r_z{v}_x\right)\text{y}+\left(r_x{v}_y-r_y{v}_x\right)\text{z}\right]\\ =\left[\left(r_x\text{x}+r_y\text{y}+r_z\text{z}\right)\times \left(u_x\text{x}+u_y+u_z\text{z}\right)\right]+\left[\left(r_x\text{x}+r_y\text{y}+r_z\text{z}\right)\times \left(v_x\text{x}+v_y\text{y}+v_z\text{z}\right)\right]\\ =\left[\left(r\times u\right)+\left(r\times v\right)\right]\end{array}$        (II)

Conclusion:

Therefore, from equation (II) it is It is proved that scalar product is distributive that is $r\times \left(u+v\right)=\left(r\times u\right)+\left(r\times v\right)$.

(b)

To determine

Prove the product rule, $\frac{d}{dt}(r\times s)=r\times \frac{ds}{dt}+\frac{dr}{dt}\times s$.

Answer to Problem 1.17P

The product rule, $\frac{d}{dt}(r\times s)=r\times \frac{ds}{dt}+\frac{dr}{dt}\times s$ has been proved.

Explanation of Solution

In order to prove the product rule, consider the LHS of the equation.

    $(r\times s)=\left|\begin{array}{ccc}\text{x}& \text{y}& \text{z}\\ r_x& r_y& r_z\\ s_x& s_x& s_z\end{array}\right|$        (III)

The value of $\frac{d}{dt}\left(r\times s\right)$ is,

    $\begin{array}{c}\frac{d}{dt}\left(r\times s\right)=\frac{d}{dt}\left[\left(r_y{s}_z-r_z{s}_y\right)\text{x}-\left(r_x^{*}{s}_z-r_z{s}_x\right)\text{y}+\left(r_x{s}_y-r_y{s}_x\right)\text{z}\right]\\ =\left[r_y\frac{d{s}_z}{dt}+s_z\frac{d{r}_y}{dt}-r_z\frac{d{s}_y}{dt}-s_y\frac{d{r}_z}{dt}\right]\text{x}\\ +\left[r_x\frac{d{s}_z}{dt}+s_z\frac{d{r}_x}{dt}-r_z\frac{d{s}_x}{dt}-s_x\frac{d{r}_z}{dt}\right]\text{y}\\ +\left[r_x\frac{d{s}_y}{dt}+s_y\frac{d{r}_x}{dt}-r_y\frac{d{s}_x}{dt}-s_x\frac{d{r}_y}{dt}\right]z\end{array}$

    $\begin{array}{l}=\left(r_y\frac{d{s}_z}{dt}-r_z\frac{d{s}_y}{dt}\right)\text{x}+\left(s_z\frac{d{r}_y}{dt}-s_y\frac{d{r}_z}{dt}\right)\text{x}-\left(r_x\frac{d{s}_z}{dt}-r_z\frac{d{s}_x}{dt}\right)y-\left(s_z\frac{d{r}_x}{dt}-s_x\frac{d{r}_z}{dt}\right)y+\\ +\left(r_x\frac{d{s}_y}{dt}-r_y\frac{d{s}_x}{dt}\right)z+\left(s_y\frac{d{r}_x}{dt}-s_x\frac{d{r}_y}{dt}\right)z\\ \end{array}$

    $=$ $\begin{array}{c}=\left(r_x{\text{x+r}}_y\text{y}+r_z\text{z}\right)\times \left(\frac{d{s}_x}{dt}\text{x}+\frac{d{s}_y}{dt}\text{y}+\frac{d{s}_z}{dt}\text{z}\right)\left(\frac{d{r}_x}{dt}\text{x}+\frac{d{r}_y}{dt}\text{y}+\frac{d{r}_z}{dt}\text{z}\right)\times \left(s_x\text{x}+s_y\text{y}+s_z\text{z}\right)\\ =\left(r\times \frac{ds}{dt}\right)+\left(\frac{dr}{dt}\times s\right)\end{array}$

Conclusion:

Therefore, the product rule, $\frac{d}{dt}(r\times s)=r\times \frac{ds}{dt}+\frac{dr}{dt}\times s$ has been proved.