Chapter 6, Problem 12P

(a)

To determine

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

Answer to Problem 12P

The angle between the vectors $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ is $\underset{\_}{11.3°}$.

Explanation of Solution

Given that $\overrightarrow{\text{A}}=3.00\widehat{\text{i}}-\text{2}\text{.00}\widehat{\text{j}}$ and $\overrightarrow{\text{B}}=4.00\widehat{\text{i}}-\text{4}\text{.00}\widehat{\text{j}}$.

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

    $\overrightarrow{\text{A}}\cdot \overrightarrow{\text{B}}=AB\cos \theta$        (I)

Here, $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ are the vectors, and $\theta$ is the angle between $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$.

Rearrange equation (I), to find $\theta$.

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

Conclusion:

Substitute $3.00\widehat{\text{i}}-\text{2}\text{.00}\widehat{\text{j}}$ for $\overrightarrow{\text{A}}$, and $4.00\widehat{\text{i}}-\text{4}\text{.00}\widehat{\text{j}}$ for $\overrightarrow{\text{B}}$ in equation (II), to find $\theta$.

    $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{\left(3.00\widehat{\text{i}}-\text{2}\text{.00}\widehat{\text{j}}\right)\cdot \left(4.00\widehat{\text{i}}-\text{4}\text{.00}\widehat{\text{j}}\right)}{\left(\sqrt{9+4}\right).\left(\sqrt{16+16}\right)}\right)\\ ={\cos }^{-1}\left(\frac{\left(12.0+8.00\right)}{\left(\sqrt{9+4}\right).\left(\sqrt{16+16}\right)}\right)\\ =11.3°\end{array}$

Therefore, The angle between the vectors $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ is $\underset{\_}{11.3°}$.

(b)

To determine

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

Answer to Problem 12P

The angle between the vectors $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ is $\underset{\_}{156°}$.

Explanation of Solution

Given that $\overrightarrow{\text{A}}=-2.00\widehat{\text{i}}+4.\text{00}\widehat{\text{j}}$ and $\overrightarrow{\text{B}}=3.00\widehat{\text{i}}-\text{4}\text{.00}\widehat{\text{j}}+\text{2}\text{.00}\widehat{\text{k}}$.

Conclusion:

Substitute $-2.00\widehat{\text{i}}+4.\text{00}\widehat{\text{j}}$ for $\overrightarrow{\text{A}}$, and $3.00\widehat{\text{i}}-\text{4}\text{.00}\widehat{\text{j}}+\text{2}\text{.00}\widehat{\text{k}}$ for $\overrightarrow{\text{B}}$ in equation (II), to find $\theta$.

    $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{\left(-2.00\widehat{\text{i}}+4.\text{00}\widehat{\text{j}}\right)\cdot \left(3.00\widehat{\text{i}}-\text{4}\text{.00}\widehat{\text{j}}+\text{2}\text{.00}\widehat{\text{k}}\right)}{\left(\sqrt{4+16}\right).\left(\sqrt{9+16+4}\right)}\right)\\ ={\cos }^{-1}\left(\frac{\left(-6.0-16.00\right)}{\left(\sqrt{20.0}\right).\left(\sqrt{29.0}\right)}\right)\\ =156°\end{array}$

Therefore, The angle between the vectors $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ is $\underset{\_}{156°}$.

(c)

To determine

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

Answer to Problem 12P

The angle between the vectors $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ is $\underset{\_}{82.3°}$.

Explanation of Solution

Given that $\overrightarrow{\text{A}}=\widehat{\text{i}}-2.00\widehat{\text{j}}+2.\text{00}\widehat{\text{k}}$ and $\overrightarrow{\text{B}}=3.\text{00}\widehat{\text{j}}+4.\text{00}\widehat{\text{k}}$.

Conclusion:

Substitute $\widehat{\text{i}}-2.00\widehat{\text{j}}+2.\text{00}\widehat{\text{k}}$ for $\overrightarrow{\text{A}}$, and $3.\text{00}\widehat{\text{j}}+4.\text{00}\widehat{\text{k}}$ for $\overrightarrow{\text{B}}$ in equation (II), to find $\theta$.

    $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{\left(\widehat{\text{i}}-2.00\widehat{\text{j}}+2.\text{00}\widehat{\text{k}}\right)\cdot \left(3.\text{00}\widehat{\text{j}}+4.\text{00}\widehat{\text{k}}\right)}{\left(\sqrt{1+4+4}\right).\left(\sqrt{9+16}\right)}\right)\\ ={\cos }^{-1}\left(\frac{\left(-6.0+8.00\right)}{\left(\sqrt{9}\right).\left(\sqrt{25.0}\right)}\right)\\ =82.3°\end{array}$

Therefore, The angle between the vectors $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ is $\underset{\_}{82.3°}$.