(a) In unit-vector notation, what is ${}^{\overrightarrow{\mathbf{r}}\mathbf{=}\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}\mathbf{+}\overrightarrow{\mathbf{c}}}$if ${}^{\overrightarrow{\mathbf{a}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}}$, ,and ${}^{\overrightarrow{\mathbf{b}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}}$ ? (b) Calculate the angle between ${}^{\overrightarrow{\mathbf{r}}}$and the positive z axis. (c) What is the component of${}^{\overrightarrow{\mathbf{a}}}$ along the direction of ${}^{\overrightarrow{\mathbf{b}}}$? (d) What is the component of ${}^{\overrightarrow{\mathbf{b}}}$perpendicular to the direction of ${}^{\overrightarrow{\mathbf{b}}}$but in the plane of ${}^{\overrightarrow{\mathbf{a}}}$and ${}^{\overrightarrow{\mathbf{b}}}$ ? (Hint: For (c), see Eq.${}^{\mathbf{3}\mathbf{-}\mathbf{20}}$ and Fig. ${}^{\mathbf{3}\mathbf{-}\mathbf{18}}$)

The vectors ${}^{\overrightarrow{\mathbf{a}}\mathbf{,}\mathbf{}\overrightarrow{\mathbf{b}}}$and ${}^{\overrightarrow{\mathbf{c}}}$are,

$\overrightarrow{\mathbf{a}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$

role="math" localid="1654745877724" $\overrightarrow{\mathbf{b}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$

$\overrightarrow{c}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$

This problem is based on dot product which is nothing but the scalar product. The scalar product of two vectors gives a scalar quantity.

The expression for the scalar product is as follows:

$$\begin{gathered} \overrightarrow{\mathbf{a}}\mathbf{.}\overrightarrow{\mathbf{b}}\mathbf{=}{\mathbf{a}}_{\mathbf{x}}{\mathbf{b}}_{\mathbf{x}}\mathbf{+}{\mathbf{a}}_{\mathbf{y}}{\mathbf{b}}_{\mathbf{y}}\mathbf{+}{\mathbf{a}}_{\mathbf{z}}{\mathbf{b}}_{\mathbf{z}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{ab}\mathbf{}\mathbf{cos\theta } \end{gathered}$$ … (i)

The expression for the magnitude of a vector is as follows:

$\mathit{A}\mathit{=}\sqrt{{\mathit{A}}_{\mathit{x}}^{\mathit{2}}\mathit{+}{\mathit{A}}_{\mathit{y}}^{\mathit{2}}\mathit{+}{\mathit{A}}_{\mathit{z}}^{\mathit{2}}}$ … (ii)

The value of vector ${}^{\overrightarrow{\mathbf{r}}}$is calculated as:

role="math" localid="1654746771591" $$\begin{gathered} \overrightarrow{\mathbf{r}}\mathbf{=}\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}\mathbf{+}\overrightarrow{\mathbf{c}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)}\mathbf{-}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{11}\hat{\mathbf{i}}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{0}\hat{\mathbf{k}} \end{gathered}$$

Thus, the value of vector role="math" localid="1654746992113" ${}^{\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{11}\hat{\mathbf{i}}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}}$.

Equation (i) can be used to calculate the angle between ${}^{\overrightarrow{\mathbf{r}}}$and positive z axis.

$\overrightarrow{\mathbf{r}}=\mathbf{11}\hat{\mathbf{i}}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{7}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$

The unit vector along positive z-axis is,

$\hat{\mathbf{k}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}$

Therefore, the dot product,

$$\begin{gathered} \overrightarrow{\mathbf{r}}\mathbf{.}\hat{\mathbf{k}}\mathbf{}\mathbf{=}\left(\overrightarrow{r}=11\hat{i}+5.0\hat{j}-7.0\hat{k}\right)\mathbf{.}\left(1.0\hat{k}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{0} \end{gathered}$$

The magnitude of ${}^{\overrightarrow{\mathbf{r}}}$and ${}^{\hat{\mathbf{k}}}$is given by,

$\left|\overrightarrow{r}\right|\mathbf{=}\sqrt{{\mathbf{11}}^{\mathbf{2}}\mathbf{+}{\mathbf{5}}^{\mathbf{2}}\mathbf{+}{\mathbf{7}}^{\mathbf{2}}\mathbf{}}\mathbf{=}\mathbf{14}$

$\left|\hat{k}\right|\mathbf{}\mathbf{=}\mathbf{1}$

The angle between $\overrightarrow{{}^{\mathbf{r}}}$and positive z axis is calculated as follows:

$$\begin{gathered} \overrightarrow{\mathbf{r}}\mathbf{.}\hat{\mathbf{k}}\mathbf{}\mathbf{cos}\mathit{\theta } \\ \left(-7.0\right)\mathbf{=}\left(14\right)\left(1\right)\mathbf{cos}\mathit{\theta } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{cos}\mathit{\theta }\mathbf{=}\frac{\mathbf{-}\mathbf{7}}{\mathbf{14}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{\theta }\mathbf{=}{\mathbf{120}}^{\mathbf{\theta }} \end{gathered}$$

Thus, the angle between ${}^{\overrightarrow{\mathbf{r}}}$ and the positive z axis is ${}^{{\mathbf{120}}^{\mathbf{°}}}$.

Component of ${}^{\overrightarrow{\mathbf{a}}}$along the direction of ${}^{\overrightarrow{\mathbf{b}}}$is given by ${}^{\mathbf{a}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}$where ${}^{\mathbf{\theta }}$is the angle between ${}^{\overrightarrow{\mathbf{a}}}$and${}^{\overrightarrow{\mathbf{b}}}$.

The magnitude of vector ${}^{\overrightarrow{\mathbf{a}}}$and ${}^{\overrightarrow{\mathbf{b}}}$is,

$$\begin{gathered} \left|\overrightarrow{a}\right|\mathbf{=}\sqrt{{\mathbf{5}}^{\mathbf{2}}\mathbf{+}{\mathbf{4}}^{\mathbf{2}}\mathbf{+}{\mathbf{6}}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{77} \end{gathered}$$

$$\begin{gathered} \mathbf{\left|}\overrightarrow{b}\mathbf{\right|}\mathbf{=}\sqrt{{\mathbf{2}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}{\mathbf{3}}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{12} \end{gathered}$$

The dot product of vector ${}^{\overrightarrow{\mathbf{a}}}$and ${}^{\overrightarrow{\mathbf{b}}}$is,

$$\begin{gathered} \overrightarrow{\mathbf{a}}\mathbf{.}\overrightarrow{\mathbf{b}}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)}\mathbf{.}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{20} \end{gathered}$$

The angle between vector ${}^{\overrightarrow{\mathbf{a}}}$and ${}^{\overrightarrow{\mathbf{b}}}$is,

$$\begin{gathered} \left(-20\right)\mathbf{=}\left(8.77\right)\left(4.12\right)\mathit{c}\mathit{o}\mathit{s}\mathit{\theta } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{=}\frac{\mathbf{-}\mathbf{20}}{\mathbf{36}\mathbf{.}\mathbf{13}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{\theta }\mathbf{=}\mathbf{123}\mathbf{.}{\mathbf{6}}^{\mathbf{°}} \end{gathered}$$

Now, the component of ${}^{\overrightarrow{\mathbf{a}}}$along the direction of ${}^{\overrightarrow{\mathbf{b}}}$is given by,

$$\begin{gathered} \mathbf{acos\theta }\mathbf{=}\left(8.77\right)\mathbf{cos}\left(123.6\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{9} \end{gathered}$$

Thus, the component of ${}^{\overrightarrow{\mathbf{a}}}$along the direction of ${}^{\overrightarrow{\mathbf{b}}}$is ${}^{\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{9}}$.

Component of ${}^{\overrightarrow{\mathbf{a}}}$along the direction perpendicular to $\overrightarrow{b}$and in the plane of ${}^{\overrightarrow{\mathbf{a}}}$and ${}^{\overrightarrow{\mathbf{b}}}$is given by ${}^{\mathbf{a}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}$, where ${}^{\mathbf{\theta }}$is the angle between ${}^{\overrightarrow{\mathbf{a}}}$and ${}^{\overrightarrow{\mathbf{b}}}$.

Therefeore,

localid="1654749201788" $$\begin{gathered} \mathbf{a}\mathbf{}\mathbf{sin\theta }\mathbf{}\mathbf{=}\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{77}\mathbf{)}\mathbf{sin}\mathbf{(}\mathbf{123}\mathbf{.}\mathbf{6}\mathbf{)}\mathbf{} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{3} \end{gathered}$$

Thus, the component of ${}^{\overrightarrow{\mathbf{a}}}$along the direction perpendicular to ${}^{\overrightarrow{\mathbf{b}}}$and in the plane of ${}^{\overrightarrow{\mathbf{a}}}$and ${}^{\overrightarrow{\mathbf{b}}}$is ${}^{\mathbf{7}\mathbf{.}\mathbf{3}}$.