91影视

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

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

Formulae:

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

Then,

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

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

$\stackrel{鈫�}{a}路\stackrel{鈫�}{b}=\left|\stackrel{鈫�}{a}\right|\left|\stackrel{鈫�}{b}\right|\cos \left(胃\right)$ (iii)

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

$a_x=0$

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

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

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

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

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

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

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

Use equation (i) to find the dot product of $\stackrel{鈫�}{a}路\stackrel{鈫�}{b}$.

$$\begin{gathered} \stackrel{鈫�}{a}路\stackrel{鈫�}{b}=\left(0脳0.93678\right)+\left(1.4527脳0\right)+\left(2.851脳1.0404\right) \\ =2.966 \\ 鈮�2.97m \end{gathered}$$

Therefore, the dot product$\stackrel{鈫�}{a}路\stackrel{鈫�}{b}$ is 2.97 m.

Use equation (ii) to find the cross product$\stackrel{鈫�}{a}脳\stackrel{鈫�}{b}$

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

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

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

$$\begin{gathered} \stackrel{鈫�}{a}脳\stackrel{鈫�}{b}=\left|\stackrel{鈫�}{a}\right|\left|\stackrel{鈫�}{b}\right|\cos \left(胃\right) \\ 2.966=3.20脳1.40脳\cos \left(胃\right) \\ 胃={\cos }^{-1}\left(0.66205\right) \\ =48.5掳 \end{gathered}$$

Therefore, the angle between$\stackrel{鈫�}{a}$ and$\stackrel{鈫�}{b}$ is $48.5掳$.