Chapter 1 Vector Analysis: Problem 1.1

Problem 1.1: Using the definitions in Eqs. 1.1 and 1.4, and appropriate diagrams,
show that the dot product and cross product are distributive,
a) when the three vectors are co-planar.

b) in the general case.

Solution: 

Eq${}^n$ 1.1

$$\overrightarrow{A}\cdot \overrightarrow{B}=|A||B|\cos \theta$$where $\theta$ is the angle between the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$

Eq${}^n$ 1.4

$$\overrightarrow{A}\times \overrightarrow{B}=|A||B|\sin \theta \hat{n}$$where $\theta$ is the angle between the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, and $\hat{n}$ is a unit vector normal to both $\overrightarrow{A}$ and $\overrightarrow{B}$.

We have to show that:

$$\overrightarrow{A}\cdot (\overrightarrow{B}+\overrightarrow{C})=\overrightarrow{A}\cdot \overrightarrow{B}+\overrightarrow{A}\cdot \overrightarrow{C}$$

$$\overrightarrow{A}\times (\overrightarrow{B}+\overrightarrow{C})=\overrightarrow{A}\times \overrightarrow{B}+\overrightarrow{A}\times \overrightarrow{C}$$

$a)$ When $\overrightarrow{A}$, $\overrightarrow{B}$ and $\overrightarrow{C}$ are co-planar. 

${\theta }_1$ and ${\theta }_2$ are the angles the vector $\overrightarrow{B}$ and $\overrightarrow{C}$ makes with the $\overrightarrow{A}$ respectively.

To get $\overrightarrow{B}+\overrightarrow{C}$ put the tail of $\overrightarrow{C}$ on the head of $\overrightarrow{B}$.

 

Let ${\theta }_3$ be the angle the resultant $\overrightarrow{B}+\overrightarrow{C}$ makes with the $\overrightarrow{A}$.

Draw the components of $\overrightarrow{B},\overrightarrow{C}$ and $\overrightarrow{B}+\overrightarrow{C}$ along $\overrightarrow{A}$.

$$\therefore |\overrightarrow{B}+\overrightarrow{C}|\cos {\theta }_3=|\overrightarrow{B}|\cos {\theta }_1+|\overrightarrow{C}|\cos {\theta }_2$$

 

Multiplying both sides by $|A|$

 

$$\therefore |A||\overrightarrow{B}+\overrightarrow{C}|\cos {\theta }_3=|A||\overrightarrow{B}|\cos {\theta }_1+|A||\overrightarrow{C}|\cos {\theta }_2$$

 

From eq${}^n$ 1.1 this is nothing but,

 

$$\overrightarrow{A}\cdot (\overrightarrow{B}+\overrightarrow{C})=\overrightarrow{A}\cdot \overrightarrow{B}+\overrightarrow{A}\cdot \overrightarrow{C}$$

In the same way, take components along the normal to the vector $\overrightarrow{A}$.

 

$$\therefore |\overrightarrow{B}+\overrightarrow{C}|\sin {\theta }_3=|\overrightarrow{B}|\sin {\theta }_1+|\overrightarrow{C}|\sin {\theta }_2$$

 

Multiplying both sides by $|A|\hat{n}$

 

where $\hat{n}$  is the vector normal to the plane containing the vectors $\overrightarrow{A},\overrightarrow{B}$ and $\overrightarrow{C}$.

 

$$\therefore |A||\overrightarrow{B}+\overrightarrow{C}|\sin {\theta }_3\hat{n}=|A||\overrightarrow{B}|\sin {\theta }_1\hat{n}+|A||\overrightarrow{C}|\sin {\theta }_2\hat{n}$$

 

From eq${}^n$ 1.4 this is nothing but,

$$\overrightarrow{A}\times (\overrightarrow{B}+\overrightarrow{C})=\overrightarrow{A}\times \overrightarrow{B}+\overrightarrow{A}\times \overrightarrow{C}$$

$b)$ For the general case.

As suggested by the author of the book, you should read the corresponding proof in the book G.E Hay' Vector and Tensor Analysis. Link to the book: http://bayanbox.ir/view/7717647651178235179/Hay-Vector-and-Tensor-Analysis.pdf

Consider

$\overrightarrow{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$, $\overrightarrow{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$, and $\overrightarrow{C}=C_x\hat{i}+C_y\hat{j}+C_z\hat{k}$

 Scalar product of two vectors is given by

$$\overrightarrow{A}\cdot \overrightarrow{B}=A_x{B}_x+A_y{B}_y+A_z{B}_z$$

 

$$\overrightarrow{A}\cdot (\overrightarrow{B}+\overrightarrow{C})=(A_x\hat{i}+A_y\hat{j}+A_z\hat{k})\cdot ((B_x\hat{i}+B_y\hat{j}+B_z\hat{k})+(C_x\hat{i}+C_y\hat{j}+C_z\hat{k}))$$

$$\overrightarrow{A}\cdot (\overrightarrow{B}+\overrightarrow{C})=(A_x\hat{i}+A_y\hat{j}+A_z\hat{k})\cdot ((B_x+C_x)\hat{i}+(B_y+C_y)\hat{j}+(B_z+C_z)\hat{k})$$

$$\overrightarrow{A}\cdot (\overrightarrow{B}+\overrightarrow{C})=A_x(B_x+C_x)+A_y(B_y+C_y)+A_z(B_z+C_z)$$

$$\overrightarrow{A}\cdot (\overrightarrow{B}+\overrightarrow{C})=A_x{B}_x+A_y{B}_y+A_z{B}_z+A_x{C}_x+A_y{C}_y+A_z{C}_z$$

$$\overrightarrow{A}\cdot (\overrightarrow{B}+\overrightarrow{C})=\overrightarrow{A}\cdot \overrightarrow{B}+\overrightarrow{A}\cdot \overrightarrow{C}$$

Vector Product of two vectors is given by:

$$\overrightarrow{A}\times \overrightarrow{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right|$$

Consider

$$\overrightarrow{A}\times (\overrightarrow{B}+\overrightarrow{C})=(A_x\hat{i}+A_y\hat{j}+A_z\hat{k})\times ((B_x+C_x)\hat{i}+(B_y+C_y)\hat{j}+(B_z+C_z)\hat{k})$$

$$\overrightarrow{A}\times (\overrightarrow{B}+\overrightarrow{C})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ A_x& A_y& A_z\\ (B_x+C_x)& (B_y+C_y)& (B_z+C_z)\end{array}\right|$$

Using the property of addition of determinants:

$$\overrightarrow{A}\times (\overrightarrow{B}+\overrightarrow{C})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right|+\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ A_x& A_y& A_z\\ C_x& C_y& C_z\end{array}\right|$$

$$\overrightarrow{A}\times (\overrightarrow{B}+\overrightarrow{C})=\overrightarrow{A}\times \overrightarrow{B}+\overrightarrow{A}\times \overrightarrow{C}$$

 Hence, proved.

If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.

 

If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.