Chapter 10, Problem 30P

(a)

To determine

To Calculate:The value of $\left|\overrightarrow{A}\times \overrightarrow{B}\right|$ and compared with $\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|$ .

The angle between the vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ .

Explanation of Solution

Given data:

  $\overrightarrow{A}=4\overrightarrow{i},\overrightarrow{B}=6\overrightarrow{i}+6\overrightarrow{j}$

Formula Used:

Consider vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ are perpendicular the vector product of two vectors, that is,

  $\begin{array}{c}\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \varphi \\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin {90}^{\text{o}}\\ \frac{\left|\overrightarrow{A}\times \overrightarrow{B}\right|}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}=1\mathrm{.....}\left(1\right)\end{array}$

Using scalar product of two vectors we can find angle between the vectors.

  $\begin{array}{c}\left|\overrightarrow{A}\cdot \overrightarrow{B}\right|=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \varphi \\ \varphi ={\cos }^{-1}\left(\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}\right)\end{array}$$$

Calculation:

The vector product of vector $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is

  $\begin{array}{c}\overrightarrow{A}=4\overrightarrow{i},\overrightarrow{B}=6\overrightarrow{i}+6\overrightarrow{j}\\ \overrightarrow{A}\times \overrightarrow{B}=\left(4\overrightarrow{i}\right)\times \left(6\overrightarrow{i}+6\overrightarrow{j}\right)\\ =\left(4\overrightarrow{i}\times 6\overrightarrow{i}\right)+\left(4\overrightarrow{i}\times 6\overrightarrow{j}\right)\\ =0+24\widehat{k}\\ =24\widehat{k}\end{array}$

The magnitude of vector $\overrightarrow{A}\text{ and }\overrightarrow{B}$ are

  $\begin{array}{c}\left|\overrightarrow{A}\right|=\sqrt{{\left(4\widehat{i}\right)}^2+0^2+0^2}\\ =4\\ \left|\overrightarrow{B}\right|=\sqrt{{\left(6\widehat{i}\right)}^2+{\left(6\widehat{j}\right)}^2+0^2}\\ =6\sqrt{2}\\ \left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|=4\left(6\sqrt{2}\right)\\ =24\sqrt{2}\\ \frac{\left|\overrightarrow{A}\times \overrightarrow{B}\right|}{\left(\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\right)}=\frac{\left|24\widehat{k}\right|}{24\sqrt{2}}\\ =0.707\end{array}$

Comparing with equation $\left(1\right)$ , the vector $\overrightarrow{A}\text{ and }\overrightarrow{B}$ are not perpendicular.

The scalar product of vector $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is

  $\begin{array}{c}\overrightarrow{A}\cdot \overrightarrow{B}=4\widehat{i}\cdot \left(6\widehat{i}+6\widehat{j}\right)\\ =24\left(\widehat{i}\cdot \widehat{i}\right)+24\left(\widehat{i}\cdot \widehat{j}\right)\\ =24\end{array}$

The angle between the vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is

  $\begin{array}{c}\varphi ={\cos }^{-1}\left(\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}\right)\\ ={\cos }^{-1}\left(\frac{24}{24\sqrt{2}}\right)\\ ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)\\ ={45}^{\text{o}}\end{array}$

Conclusion:

The angle between the vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is ${45}^{\text{o}}$ .

(b)

To determine

To Calculate: The value of $\left|\overrightarrow{A}\times \overrightarrow{B}\right|$ and compared with $\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|$ .

The angle between the vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ .

Explanation of Solution

Given data:

  $\overrightarrow{A}=4\widehat{i},\overrightarrow{B}=6\widehat{i}+6\widehat{k}$

Formula used:

We consider vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ are perpendicular the vector product of two vectors, that is

  $\begin{array}{c}\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \varphi \\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin {90}^{\text{o}}\\ \frac{\left|\overrightarrow{A}\times \overrightarrow{B}\right|}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}=1\mathrm{.....}\left(1\right)\end{array}$

Using scalar product of two vectors we can find angle between the vectors.

  $\begin{array}{c}\left|\overrightarrow{A}\cdot \overrightarrow{B}\right|=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \varphi \\ \varphi ={\cos }^{-1}\left(\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}\right)\end{array}$

Calculation:

The vector product of vector $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is

  $\begin{array}{c}\overrightarrow{A}=4\widehat{i},\overrightarrow{B}=6\widehat{i}+6\widehat{k}\\ \overrightarrow{A}\times \overrightarrow{B}=\left(4\widehat{i}\right)\times \left(6\widehat{i}+6\widehat{k}\right)\\ =\left(4\widehat{i}\times 6\overrightarrow{i}\right)+\left(4\widehat{i}\times 6\widehat{k}\right)\\ =0+24\left(-\widehat{j}\right)\\ =-24\widehat{j}\end{array}$

The magnitude of vector $\overrightarrow{A}\text{ and }\overrightarrow{B}$ are

  $\begin{array}{c}\left|\overrightarrow{A}\right|=\sqrt{{\left(4\widehat{i}\right)}^2+0^2+0^2}\\ =4\\ \left|\overrightarrow{B}\right|=\sqrt{{\left(6\widehat{i}\right)}^2+{\left(6\widehat{k}\right)}^2+0^2}\\ =6\sqrt{2}\\ \left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|=4\left(6\sqrt{2}\right)\\ =24\sqrt{2}\\ \frac{\left|\overrightarrow{A}\times \overrightarrow{B}\right|}{\left(\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\right)}=\frac{\left|-24\widehat{j}\right|}{24\sqrt{2}}\\ =0.707\end{array}$

Comparing with equation $\left(1\right)$ , the vector $\overrightarrow{A}\text{ and }\overrightarrow{B}$ are not perpendicular.

The scalar product of vector $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is

  $\begin{array}{c}\overrightarrow{A}\cdot \overrightarrow{B}=4\widehat{i}\cdot \left(6\widehat{i}+6\widehat{k}\right)\\ =24\left(\widehat{i}\cdot \widehat{i}\right)+24\left(\widehat{i}\cdot \widehat{k}\right)\\ =24\end{array}$

The angle between the vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is

  $\begin{array}{c}\varphi ={\cos }^{-1}\left(\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}\right)\\ ={\cos }^{-1}\left(\frac{24}{24\sqrt{2}}\right)\\ ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)\\ ={45}^{\text{o}}\end{array}$

Conclusion:

The angle between the vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is $45°$ .

(c)

To determine

To Calculate: The value of $\left|\overrightarrow{A}\times \overrightarrow{B}\right|$ and compared with $\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|$ .

The angle between the vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ .

Explanation of Solution

Given data:

  $\overrightarrow{A}=2\widehat{i}+3\widehat{j},\overrightarrow{B}=3\widehat{i}+2\widehat{j}$

Formula used:

We consider vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ are perpendicular the vector product of two vectors, that is,

  $\begin{array}{c}\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \varphi \\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin {90}^{\text{o}}\\ \frac{\left|\overrightarrow{A}\times \overrightarrow{B}\right|}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}=1\mathrm{.....}\left(1\right)\end{array}$

Using scalar product of two vectors we can find angle between the vectors.

  $\begin{array}{c}\left|\overrightarrow{A}\cdot \overrightarrow{B}\right|=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \varphi \\ \varphi ={\cos }^{-1}\left(\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}\right)\end{array}$

Calculation:

The vector product of vector $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is

  $\begin{array}{c}\overrightarrow{A}=2\widehat{i}+3\widehat{j},\overrightarrow{B}=3\widehat{i}+2\widehat{j}\\ \overrightarrow{A}\times \overrightarrow{B}=6\left(\widehat{i}\times \widehat{i}\right)+4\left(\widehat{i}\times \widehat{j}\right)+9\left(\widehat{j}\times \widehat{i}\right)+6\left(\widehat{j}\times \widehat{j}\right)\\ =6\left(0\right)+4\widehat{k}-9\widehat{k}+6\left(0\right)\\ =-5\widehat{k}\end{array}$

The magnitude of vector $\overrightarrow{A}\text{ and }\overrightarrow{B}$ are

  $\begin{array}{c}\left|\overrightarrow{A}\right|=\sqrt{{\left(2\widehat{i}\right)}^2+{\left(3\widehat{j}\right)}^2+0^2}\\ =\sqrt{13}\\ \left|\overrightarrow{B}\right|=\sqrt{{\left(3\widehat{i}\right)}^2+{\left(2\widehat{j}\right)}^2+0^2}\\ =\sqrt{13}\\ \left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|=\sqrt{13}\sqrt{13}\\ =13\\ \frac{\left|\overrightarrow{A}\times \overrightarrow{B}\right|}{\left(\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\right)}=\frac{\left|-5\widehat{k}\right|}{13}\\ =0.385\end{array}$

Comparing with equation $\left(1\right)$ , the vector $\overrightarrow{A}\text{ and }\overrightarrow{B}$ are not perpendicular.

The scalar product of vector $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is

  $\begin{array}{c}\overrightarrow{A}\cdot \overrightarrow{B}=\left(2\widehat{i}+3\widehat{j}\right)\cdot \left(3\widehat{i}+2\widehat{j}\right)\\ =6\left(\widehat{i}\cdot \widehat{i}\right)+6\left(\widehat{i}\cdot \widehat{k}\right)\\ =6\left(1\right)+6\left(1\right)\\ =12\end{array}$

The angle between the vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is

  $\begin{array}{c}\varphi ={\cos }^{-1}\left(\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}\right)\\ ={\cos }^{-1}\left(\frac{12}{13}\right)\\ ={23}^{\text{o}}\end{array}$

Conclusion:

The angle between the vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is $23°$ .