Chapter 9, Problem 67PQ

(a)

To determine

The angle between the pair of vectors.

Answer to Problem 67PQ

The angle between the pair of vectors is $\underset{\_}{90°}$.

Explanation of Solution

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

    $A=\sqrt{A_x^2+A_y^2+A_z^2}$ (I)

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

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

    $B=\sqrt{B_x^2+B_y^2+B_z^2}$ (II)

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

Write the expression for dot product between $\overrightarrow{A}$ and $\overrightarrow{B}$.

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

Write the expression for angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.

    $\theta ={\cos }^{-1}\left(\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{AB}\right)$ (IV)

Conclusion:

Substitute $-3.00$ for $A_x$, $-1.00$ for $A_y$ and $+4.00$ for $A_z$ in equation (I).

    $\begin{array}{c}A=\sqrt{{\left(-3.00\right)}^2+{\left(-1.00\right)}^2+{\left(+4.00\right)}^2}\\ =5.10\end{array}$

Substitute $+2.00$ for $B_x$, $+2.00$ for $B_y$ and $+2.00$ for $B_z$ in equation (II).

    $\begin{array}{c}B=\sqrt{{\left(+2.00\right)}^2+{\left(+2.00\right)}^2+{\left(+2.00\right)}^2}\\ =3.46\end{array}$

Substitute $-3.00$ for $A_x$, $-1.00$ for $A_y$, $+4.00$ for $A_z$, $+2.00$ for $B_x$, $+2.00$ for $B_y$ and $+2.00$ for $B_z$ in equation (III).

    $\begin{array}{c}\overrightarrow{A}\cdot \overrightarrow{B}=\left(-3.00\right)\left(+2.00\right)+\left(-1.00\right)\left(+2.00\right)+\left(+4.00\right)\left(+2.00\right)\\ =0\end{array}$

Substitute $0$ for $\overrightarrow{A}\cdot \overrightarrow{B}$, $5.10$ for $A$ and $3.46$ for $B$ in equation (IV).

    $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{0}{\left(5.10\right)\left(3.46\right)}\right)\\ =90°\end{array}$

Therefore, the angle between the pair of vectors is $\underset{\_}{90°}$.

(b)

To determine

The angle between the pair of vectors.

Answer to Problem 67PQ

The angle between the pair of vectors is $\underset{\_}{120°}$.

Explanation of Solution

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

    $A=\sqrt{A_x^2+A_y^2+A_z^2}$ (I)

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

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

    $B=\sqrt{B_x^2+B_y^2+B_z^2}$ (II)

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

Write the expression for dot product between $\overrightarrow{A}$ and $\overrightarrow{B}$.

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

Write the expression for angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.

    $\theta ={\cos }^{-1}\left(\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{AB}\right)$ (IV)

Conclusion:

Substitute $+1.00$ for $A_x$, $+2.00$ for $A_y$ and $0$ for $A_z$ in equation (I).

    $\begin{array}{c}A=\sqrt{{\left(+1.00\right)}^2+{\left(+2.00\right)}^2+{\left(0\right)}^2}\\ =2.24\end{array}$

Substitute $0$ for $B_x$, $-2.00$ for $B_y$ and $-3.00$ for $B_z$ in equation (II).

    $\begin{array}{c}B=\sqrt{{\left(0\right)}^2+{\left(-2.00\right)}^2+{\left(-3.00\right)}^2}\\ =3.61\end{array}$

Substitute  $+1.00$ for $A_x$, $+2.00$ for $A_y$, $0$ for $A_z$, $0$ for $B_x$, $-2.00$ for $B_y$ and $-3.00$ for $B_z$ in equation (III).

    $\begin{array}{c}\overrightarrow{A}\cdot \overrightarrow{B}=\left(+1.00\right)\left(0\right)+\left(+2.00\right)\left(-2.00\right)+\left(0\right)\left(-3.00\right)\\ =-4.00\end{array}$

Substitute $-4.00$ for $\overrightarrow{A}\cdot \overrightarrow{B}$, $2.24$ for $A$ and $3.61$ for $B$ in equation (IV).

    $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{-4.00}{\left(2.24\right)\left(3.61\right)}\right)\\ =120°\end{array}$

Therefore, the angle between the pair of vectors is $\underset{\_}{120°}$.

(c)

To determine

The angle between the pair of vectors.

Answer to Problem 67PQ

The angle between the pair of vectors is $\underset{\_}{86°}$.

Explanation of Solution

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

    $A=\sqrt{A_x^2+A_y^2+A_z^2}$ (I)

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

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

    $B=\sqrt{B_x^2+B_y^2+B_z^2}$ (II)

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

Write the expression for dot product between $\overrightarrow{A}$ and $\overrightarrow{B}$.

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

Write the expression for angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.

    $\theta ={\cos }^{-1}\left(\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{AB}\right)$ (IV)

Conclusion:

Substitute $+4.00$ for $A_x$, $0$ for $A_y$ and $+2.00$ for $A_z$ in equation (I).

    $\begin{array}{c}A=\sqrt{{\left(+4.00\right)}^2+{\left(0\right)}^2+{\left(+2.00\right)}^2}\\ =4.47\end{array}$

Substitute $-1.00$ for $B_x$, $+5.00$ for $B_y$ and $+3.00$ for $B_z$ in equation (II).

    $\begin{array}{c}B=\sqrt{{\left(-1.00\right)}^2+{\left(+5.00\right)}^2+{\left(+3.00\right)}^2}\\ =5.92\end{array}$

Substitute $+4.00$ for $A_x$, $0$ for $A_y$, $+2.00$ for $A_z$, $-1.00$ for $B_x$, $+5.00$ for $B_y$ and $+3.00$ for $B_z$ in equation (III).

    $\begin{array}{c}\overrightarrow{A}\cdot \overrightarrow{B}=\left(+4.00\right)\left(-1.00\right)+\left(0\right)\left(+5.00\right)+\left(+2.00\right)\left(+3.00\right)\\ =2.00\end{array}$

Substitute $0$ for $\overrightarrow{A}\cdot \overrightarrow{B}$, $4.47$ for $A$ and $5.92$ for $B$ in equation (IV).

    $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{2.00}{\left(4.47\right)\left(5.92\right)}\right)\\ =86°\end{array}$

Therefore, the angle between the pair of vectors is $\underset{\_}{86°}$.