Question

Vector $\overrightarrow{\mathbf{a}}$lies in the yz plane $\mathbf{63}\mathbf{.}\mathbf{0}\mathbf{°}$from the positive direction of the y axis, has a positive z component, and has magnitude 3.20units. Vector $\overrightarrow{\mathit{b}}$lies in xz the plane 48.0from the positive direction of the x axis, has a positive z component, and has magnitude$\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{°}$units. Find (a)role="math" localid="1661144136421" $\overrightarrow{\mathbf{a}}·\overrightarrow{\mathbf{b}}$, (b)$\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{b}}$, and (c) the angle between$\overrightarrow{\mathbf{a}}$and$\overrightarrow{\mathbf{b}}$.

Short answer

1. The dot product $\overrightarrow{a}·\overrightarrow{b}$is 2.97.
2. The cross product$\overrightarrow{a}\times \overrightarrow{b}$ is $1.51\hat{\mathrm{i}}+2.67\stackrel{⏜}{\mathrm{j}}-1.36\hat{\mathrm{k}}$.
3. The angle between$\overrightarrow{a}$ and$\overrightarrow{b}$ is $48.5°$.

Step-by-step solution

Given data

$$\begin{gathered} \left|\overrightarrow{a}\right|=3.20\mathrm{units} \\ \left|\overrightarrow{\mathrm{b}}\right|=1.40\mathrm{units} \end{gathered}$$

To understand the concept

To solve for given identities we must use the concept of dot product and cross product of two vectors.

Formulae:

If $\overrightarrow{a}=a_x\hat{i}+a_y\stackrel{⏜}{j}+a_z\hat{k}$; $\overrightarrow{b}=b_x\hat{i}+b_y\stackrel{⏜}{j}+b_z\hat{k}$

Then,

$\overrightarrow{a}.\overrightarrow{b}=a_X{b}_X+a_y{b}_y+a_z{b}_z$ (i)

$\overrightarrow{a}\times \overrightarrow{b}=\hat{i}\left(a_y{b}_z-a_z{b}_y\right)-\hat{j}\left(a_x{b}_z-a_z{b}_x\right)+\hat{k}\left(a_x{b}_y-a_y{b}_x\right)$ (ii)

$\overrightarrow{a}·\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \left(\theta \right)$ (iii)

Find a→ and b→

Vector $\overrightarrow{a}$lies in the yz plane $63.0°$from the positive direction of y axis. So, components will be,

$a_x=0$

role="math" localid="1661145307164" $$\begin{gathered} a_y=3.20\times \cos \left(63°\right) \\ =1.4527m \end{gathered}$$

$$\begin{gathered} a_z=3.20\times \sin \left(63°\right) \\ =2.851m \end{gathered}$$

Therefore, the vector$\overrightarrow{A}$ in unit notation is,

$\begin{array}{l}\overrightarrow{a}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}\\ =0\hat{i}+1.4527\stackrel{⏜}{j}+2.851\hat{k}\end{array}$

Vector$\overrightarrow{b}$ lies in xz plane$48°$ from the positive direction of x axis, so components will be,

$$\begin{gathered} b_x=1.4\times \cos \left(48°\right) \\ =0.93678 \\ {b}_y=0m \\ {b}_z=1.4\times \sin \left(48°\right) \\ =1.0404m \end{gathered}$$

Therefore, the vector $\overrightarrow{b}$in unit notation is, $\overrightarrow{b}=0.93678\hat{i}+0\hat{j}+1.0404\hat{k}$.

(a) Find the dot product a→·b→

Use equation (i) to find the dot product of $\overrightarrow{a}·\overrightarrow{b}$.

$$\begin{gathered} \overrightarrow{a}·\overrightarrow{b}=\left(0\times 0.93678\right)+\left(1.4527\times 0\right)+\left(2.851\times 1.0404\right) \\ =2.966 \\ \approx 2.97m \end{gathered}$$

Therefore, the dot product$\overrightarrow{a}·\overrightarrow{b}$ is 2.97 m.

(b) Find the cross product a→×b→

Use equation (ii) to find the cross product$\overrightarrow{a}\times \overrightarrow{b}$

$$\begin{gathered} \overrightarrow{a}\times \overrightarrow{b}=\hat{i}\left(1.4527\times 1.0404-2.9456\times 0\right)-\hat{j}\left(0\times 1.0404-2.851\times 0.93678\right)+\hat{k}\left(0\times 0-1.4527\times 0.93678\right) \\ =1.51\hat{i}+2.67\hat{j}-1.36\hat{k} \end{gathered}$$

The cross product$\overrightarrow{a}\times \overrightarrow{b}$ is .$1.51\hat{i}+2.67\hat{j}-1.36\hat{k}$

(c) Find the angle between a→ and b→

Now, use the equation (iii) to find the angle between$\overrightarrow{a}$and $\overrightarrow{b}$.

$$\begin{gathered} \overrightarrow{a}\times \overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \left(\theta \right) \\ 2.966=3.20\times 1.40\times \cos \left(\theta \right) \\ \theta ={\cos }^{-1}\left(0.66205\right) \\ =48.5° \end{gathered}$$

Therefore, the angle between$\overrightarrow{a}$ and$\overrightarrow{b}$ is $48.5°$.