91Ó°ÊÓ

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

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

Formulae:

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

Then,

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

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

$\stackrel{→}{a}·\stackrel{→}{b}=\left|\stackrel{→}{a}\right|\left|\stackrel{→}{b}\right|\cos \left(\mathrm{θ}\right)$ (iii)

Vector $\stackrel{→}{a}$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" $$\begin{gathered} a_y=3.20×\cos \left(63°\right) \\ =1.4527m \end{gathered}$$

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

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

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

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

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

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

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

$$\begin{gathered} \stackrel{→}{a}·\stackrel{→}{b}=\left(0×0.93678\right)+\left(1.4527×0\right)+\left(2.851×1.0404\right) \\ =2.966 \\ ≈2.97m \end{gathered}$$

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

Use equation (ii) to find the cross product$\stackrel{→}{a}×\stackrel{→}{b}$

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

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

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

$$\begin{gathered} \stackrel{→}{a}×\stackrel{→}{b}=\left|\stackrel{→}{a}\right|\left|\stackrel{→}{b}\right|\cos \left(\mathrm{θ}\right) \\ 2.966=3.20×1.40×\cos \left(\mathrm{θ}\right) \\ \mathrm{θ}={\cos }^{-1}\left(0.66205\right) \\ =48.5° \end{gathered}$$

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