Chapter 11, Problem 3P

(a)

To determine

The product of vectors $\overrightarrow{A}$ and $\overrightarrow{B}$.

Answer to Problem 3P

The product of vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is $\underset{\_}{7.00\widehat{\text{k}}}$.

Explanation of Solution

Write the expression for cross product of two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ as.

  $\begin{array}{c}\overrightarrow{A}\times \overrightarrow{B}=\left|\begin{array}{ccc}\widehat{\text{i}}& \widehat{\text{j}}& \widehat{\text{k}}\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right|\\ =\left|\begin{array}{cc}{A}_y& A_z\\ B_y& B_z\end{array}\right|\widehat{\text{i}}+\left|\begin{array}{cc}{A}_z& A_x\\ B_z& B_x\end{array}\right|\widehat{\text{j}}+\left|\begin{array}{cc}{A}_x& A_y\\ B_x& B_y\end{array}\right|\widehat{\text{k}}\end{array}$

Simplify the above obtained expression as.

  $\overrightarrow{A}\times \overrightarrow{B}=\left(A_y{B}_z-A_z{B}_y\right)\widehat{i}+\left(A_z{B}_x-A_x{B}_z\right)\widehat{j}+\left(A_x{B}_y-A_y{B}_x\right)\widehat{k}$                        (I)

Here, $A_x$ is $x$-component of $\overrightarrow{A}$, $A_y$ is $y$-component of $\overrightarrow{A}$, $A_z$ is $z$-component of $\overrightarrow{A}$, $B_x$ is $x$-component of $\overrightarrow{B}$, $B_y$ is $y$-component of $\overrightarrow{B}$, $B_z$ is $z$-component of $\overrightarrow{B}$.

Conclusion:

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

  $\begin{array}{c}\overrightarrow{A}\times \overrightarrow{B}=\left(\left(2\right)\left(0\right)-\left(0\right)\left(3\right)\right)\widehat{i}+\left(\left(1\right)\left(0\right)-\left(-2\right)\left(0\right)\right)\widehat{j}+\left(\left(1\right)\left(3\right)-\left(-2\right)\left(3\right)\right)\widehat{k}\\ =7.00\widehat{\text{k}}\end{array}$

Thus, the product of vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is $\underset{\_}{7.00\widehat{\text{k}}}$.

(b)

To determine

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

Answer to Problem 3P

The angle between vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is $\underset{\_}{60.3°}$.

Explanation of Solution

Write the expression for angle between two vectors as.

  $\theta ={\sin }^{-1}\left(\frac{\left|\overrightarrow{A}\times \overrightarrow{B}\right|}{AB}\right)$                                                                                        (II)

Here, $\left|\overrightarrow{A}\times \overrightarrow{B}\right|$ is magnitude of $\overrightarrow{A}\times \overrightarrow{B}$ vector, $A$ is magnitude of vector $\overrightarrow{A}$ and $B$ is magnitude of vector $\overrightarrow{B}$.

Write the expression for magnitude of $\overrightarrow{A}\times \overrightarrow{B}$ vector as.

  $\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\sqrt{{\left({\left(AB\right)}_x\right)}^2+{\left({\left(AB\right)}_y\right)}^2+{\left({\left(AB\right)}_z\right)}^2}$                                              (III)

Here, ${\left(AB\right)}_x$ is $x$-component of $\left(\overrightarrow{A}\times \overrightarrow{B}\right)$, ${\left(AB\right)}_y$ is $y$-component of $\left(\overrightarrow{A}\times \overrightarrow{B}\right)$, ${\left(AB\right)}_z$ is $z$-component of $\left(\overrightarrow{A}\times \overrightarrow{B}\right)$.

Write the expression for magnitude of $\overrightarrow{A}$ vector as.

  $A=\sqrt{{\left(A_x\right)}^2+{\left(A_y\right)}^2+{\left(A_z\right)}^2}$                                                                        (IV)

Here, $A_x$ is $x$-component of $\overrightarrow{A}$, $A_y$ is $y$-component of $\overrightarrow{A}$, $A_z$ is $z$-component of $\overrightarrow{A}$ and $A$ is magnitude of $\overrightarrow{A}$.

Write the expression for magnitude of $\overrightarrow{B}$ vector as.

  $B=\sqrt{{\left(B_x\right)}^2+{\left(B_y\right)}^2+{\left(B_z\right)}^2}$                                                                         (V)

Here, $B_x$ is $x$-component of $\overrightarrow{B}$, $B_y$ is $y$-component of $\overrightarrow{B}$, $B_z$ is $z$-component of $\overrightarrow{B}$ and $B$ is magnitude of $\overrightarrow{B}$.

Substitute the value of $A$, $B$ and $\left|\overrightarrow{A}\times \overrightarrow{B}\right|$ from equation (IV),(V) and (III) in equation (II) as.

  $\theta ={\sin }^{-1}\left(\frac{\sqrt{{\left({\left(AB\right)}_x\right)}^2+{\left({\left(AB\right)}_y\right)}^2+{\left({\left(AB\right)}_z\right)}^2}}{\left(\sqrt{{\left(A_x\right)}^2+{\left(A_y\right)}^2+{\left(A_z\right)}^2}\right)\left(\sqrt{{\left(A_x\right)}^2+{\left(A_y\right)}^2+{\left(A_z\right)}^2}\right)}\right)$               (VI)

Conclusion:

Substitute $0$ for ${\left(AB\right)}_x$, $0$ for ${\left(AB\right)}_y$, $7$ for ${\left(AB\right)}_z$, $1$ for $A_x$, $2$ for $A_y$, $0$ for $A_z$, $-2$ for $B_x$, $3$ for $B_y$, $0$ for $B_z$ in equation (VI).

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\sqrt{{\left(0\right)}^2+{\left(0\right)}^2+{\left(7\right)}^2}}{\left(\sqrt{{\left(1\right)}^2+{\left(2\right)}^2+{\left(0\right)}^2}\right)\left(\sqrt{{\left(-2\right)}^2+{\left(3\right)}^2+{\left(0\right)}^2}\right)}\right)\\ =60.3°\end{array}$

Thus, the angle between vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is $\underset{\_}{60.3°}$.