Here are three displacements, each measured in meters: ${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{j}}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{k}}\mathbf{,}{\overrightarrow{\mathbf{d}}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{k}}$,and${\overrightarrow{\mathbf{d}}}_{\mathbf{3}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{j}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\stackrel{\mathbf{⏜}}{\mathbf{k}}$. (a) What is $\overrightarrow{\mathbf{r}}\mathbf{=}{\overrightarrow{\mathbf{d}}}_{\mathbf{1}}\mathbf{+}{\overrightarrow{\mathbf{d}}}_{\mathbf{2}}\mathbf{+}{\overrightarrow{\mathbf{d}}}_{\mathbf{3}}$? (b) What is the angle between$\overrightarrow{\mathbf{r}}$and the positive z axis? (c) What is the component of ${\mathit{d}}_{\mathbf{1}}$along the direction of ${\overrightarrow{\mathbf{d}}}_{\mathbf{2}}$(d) What is the component of ${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}$that is perpendicular to the direction of $\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}$and in the plane of role="math" localid="1658465314757" ${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}$and ${\overrightarrow{\mathbf{d}}}_{\mathbf{2}}$(Hint: For (c), consider Eq 3-20.and Fig, 3-18; for (d), consider Eq.3-24.)

The given vectors are,

${\overrightarrow{\mathrm{d}}}_1=4.0\stackrel{⏜}{\mathrm{i}}+5.0\stackrel{⏜}{\mathrm{j}}-6.0\stackrel{⏜}{\mathrm{k}}$

${\overrightarrow{\mathrm{d}}}_2=-1.0\stackrel{⏜}{\mathrm{i}}+2.0\stackrel{⏜}{\mathrm{j}}+3.0\stackrel{⏜}{\mathrm{k}}$

${\overrightarrow{\mathrm{d}}}_3=4.0\stackrel{⏜}{\mathrm{i}}+3.0\stackrel{⏜}{\mathrm{j}}+2.0\stackrel{⏜}{\mathrm{k}}$

Vector addition and subtraction can be used to find the displacement and magnitude can be found using the formula. To find the angle between $\overrightarrow{\mathbf{r}}$and positive z axis we can use a scalar product of two vectors. Unit vector along z axis is. The components of ${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}$can be found by using the property of the scalar product of two vectors.

Formula:

$\overrightarrow{\mathrm{r}}={\overrightarrow{\mathrm{d}}}_1+{\overrightarrow{\mathrm{d}}}_2+{\overrightarrow{\mathrm{d}}}_3............\left(\mathrm{i}\right)$

${\overrightarrow{\mathrm{d}}}_1.{\overrightarrow{\mathrm{d}}}_1={\mathrm{d}}_1{\mathrm{d}}_2\mathrm{cos\phi }.............\left(\mathrm{ii}\right)$

From equation (i), we can calculate the displacement vector $\overrightarrow{r}$as follows.

$$\begin{gathered} \overrightarrow{r}=\left(4.0\stackrel{⏜}{i}+5.0\stackrel{⏜}{j}-6.0\stackrel{⏜}{k}\right)-\left(-1.0\stackrel{⏜}{i}+2.0\stackrel{⏜}{j}+3.0\stackrel{⏜}{k}\right)+\left(4.0\stackrel{⏜}{i}+3.0\stackrel{⏜}{j}-2.0\stackrel{⏜}{k}\right) \\ =4.0\stackrel{⏜}{i}+5.0\stackrel{⏜}{j}-6.0\stackrel{⏜}{k}+1.0\stackrel{⏜}{i}-2.0\stackrel{⏜}{j}-3.0\stackrel{⏜}{k}+4.0\stackrel{⏜}{i}+3.0\stackrel{⏜}{j}+2.0\stackrel{⏜}{k} \\ =9.0\stackrel{⏜}{i}+6.0\stackrel{⏜}{j}-7.0\stackrel{⏜}{k} \end{gathered}$$

Therefore, the displacement$\overrightarrow{r}$is$9.0\stackrel{⏜}{i}+6.0\stackrel{⏜}{j}-7.0\stackrel{⏜}{k}$.

The magnitude of r is calculated as,

$$\begin{gathered} r=\sqrt{{\left(9.0\right)}^2+{\left(6.0\right)}^2+{\left(-7.0\right)}^2} \\ =12.9\mathrm{m} \end{gathered}$$

The angle between$\overrightarrow{r}$and positive z axis is θ

localid="1658466997778" $$\begin{gathered} \cos \theta =\frac{\overrightarrow{r}.\stackrel{⏜}{k}}{\left|\overrightarrow{r}\right|} \\ =\frac{-7.0}{12.0} \\ \theta ={\cos }^{-1}\left(-0.543\right) \\ =123° \end{gathered}$$

The angle betweenlocalid="1658467048657" $\overrightarrow{r}$and positive z axis is $123°$.

The component of ${\overrightarrow{d}}_1$along the direction of${\overrightarrow{d}}_1={\overrightarrow{d}}_{\left|\right|}$

$$\begin{gathered} {\overrightarrow{d}}_{▯}={\overrightarrow{d}}_1.\stackrel{⏜}{u} \\ =d_1\cos \varphi .............\left(\mathrm{iii}\right) \end{gathered}$$

$\stackrel{⏜}{u}$is a unit vector

$$\begin{gathered} {\overrightarrow{d}}_1.{\overrightarrow{d}}_2=d_1{d}_2\cos \varphi \\ \cos \varphi =\frac{{\overrightarrow{d}}_1\overrightarrow{d_2}}{d_1{d}_2}.............\left(iv\right) \end{gathered}$$

Equation (iii) becomes as

$$\begin{gathered} \overrightarrow{d_{▯}}=d_1\cos \varphi \\ =d_1\left(\frac{{\overrightarrow{d}}_1.{\overrightarrow{d}}_2}{d_1{d}_2}\right) \\ =\frac{{\overrightarrow{d}}_1.{\overrightarrow{d}}_2}{d_2} \end{gathered}$$

Now, calculate the dot product${\overrightarrow{d}}_1.{\overrightarrow{d}}_2$ as,

role="math" localid="1658467900456" $$\begin{gathered} {\overrightarrow{d}}_1.{\overrightarrow{d}}_2=d_{1x}{d}_{2x}+d_{1y}{d}_{2y}+d_{1z}+d_{2z} \\ =\left(4.0\right)\left(-1.0\right)+\left(5.0\right)\left(2.0\right)+\left(-6.0\right)\left(3.0\right) \end{gathered}$$

Substitute it in the above equation.

role="math" localid="1658468009605" $$\begin{gathered} \overrightarrow{d_{▯}}=\frac{\left(4.0\right)\left(-1.0\right)+\left(5.0\right)\left(2.0\right)+\left(-6.0\right)\left(3.0\right)}{\sqrt{{\left(-1.0\right)}^2\left(2.0\right)+{\left(3.0\right)}^2}} \\ =-3.2\mathrm{m} \end{gathered}$$

The component of ${\overrightarrow{d}}_1$in the perpendicular direction of$\overrightarrow{d_2}=\overrightarrow{d_{\perp }}$

$$\begin{gathered} {\overrightarrow{d}}_1={\overrightarrow{d}}_{\perp }+{\overrightarrow{d}}_{▯} \\ {\overrightarrow{d}}_{\perp }={\overrightarrow{d}}_1-{\overrightarrow{d}}_{▯} \\ {\overrightarrow{d}}_{\perp }^2={\overrightarrow{d}}_1^2-d_{▯} \\ {d}_{\perp }=\sqrt{\left[{\left(4.0\right)}^2+{\left(5.0\right)}^2+{\left(-6.0\right)}^2-{\left(-3.2\right)}^2\right]} \\ =8.2m \end{gathered}$$

Therefore, the component of ${\overrightarrow{d}}_1$in the perpendicular direction of ${\overrightarrow{d}}_2$is 8.2 m