Question

What is the angle $\theta$between vectors $\overrightarrow{A}$and $\overrightarrow{B}$in each part of $Exercise12?$

Short answer

a. $\theta ={125}^0$

b.$\theta ={68}^0$

Step-by-step solution

Given information (part a)

$\overrightarrow{A}=3\hat{i}+4\hat{j}$and$\overrightarrow{B}=2\hat{i}-6\hat{j}$

Explanation (part a)

The angle between $\overrightarrow{A}$and $\overrightarrow{B}$, we have

$\theta ={\cos }^{-1}\left(\frac{\overrightarrow{A}.\overrightarrow{B}}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}\right)$

we know that

role="math" localid="1649877276481" $$\begin{gathered} \overrightarrow{A}·\overrightarrow{B}=A_x{B}_x+A_y{B}_y \\ =\left(3\hat{i}\right)\left(2\hat{i}\right)+\left(4\hat{j}\right)(-6\hat{j}) \\ =6{\hat{i}}^2-24{\hat{j}}^2 \\ =6-24 \\ \overrightarrow{A}·\overrightarrow{B}=-18 \\ \left|\overrightarrow{A}\right|=\sqrt{9+16}=\sqrt{25}=5 \\ \left|\overrightarrow{B}\right|=\sqrt{4+36}=\sqrt{40}=2\sqrt{10} \end{gathered}$$

Substitute the values of $\overrightarrow{A}.\overrightarrow{B},\left|\overrightarrow{A}\right|and\left|\overrightarrow{B}\right|$in the above equation, we get

$$\begin{gathered} \theta ={\cos }^{-1}\left(\frac{-18}{5\times 2\sqrt{10}}\right) \\ ={\cos }^{-1}\left(\frac{-9}{5\sqrt{10}}\right)={\cos }^{-1}(-0.569)={125}^0 \end{gathered}$$

Given information (part b)

$\overrightarrow{A}=3\hat{i}-2\hat{j}$

$\overrightarrow{B}=6\hat{i}+4\hat{j}$

Explanation (part b)

The angle between$\overrightarrow{A}$ and$\overrightarrow{B}$ , we have

$\theta ={\cos }^{-1}\left(\frac{\overrightarrow{A}.\overrightarrow{B}}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}\right)$

we know that

$$\begin{gathered} \overrightarrow{A}·\overrightarrow{B}=A_x{B}_x+A_y{B}_y \\ =\left(3\hat{i}\right)\left(6\hat{i}\right)+(-2\hat{j})\left(4\hat{j}\right) \\ =18{\hat{i}}^2-8{\hat{j}}^2 \\ =18-8 \\ \overrightarrow{A}·\overrightarrow{B}=10 \\ \left|\overrightarrow{A}\right|=\sqrt{9+4}=\sqrt{13} \\ \left|\overrightarrow{B}\right|=\sqrt{36+16}=\sqrt{52}=2\sqrt{13} \end{gathered}$$

Substitute the values of $\overrightarrow{A}.\overrightarrow{B},\left|\overrightarrow{A}\right|and\left|\overrightarrow{B}\right|$in the above equation, we get

$$\begin{gathered} \theta ={\cos }^{-1}\left(\frac{10}{\sqrt{13}\times 2\sqrt{13}}\right) \\ ={\cos }^{-1}\left(\frac{10}{26}\right)={\cos }^{-1}(0.384)={68}^0 \end{gathered}$$