Q42E - University Physics with Modern Physics

Question

(I) (a) Find the scalar product of the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ given in Exercise 1.38. (b) Find the angle between these two vectors.

Step-by-Step Solution

Answer

The angle between these two vector is $1.43°$

Given data

$\overrightarrow{A}=4.00\hat{i}+7.00\hat{j}$

$\overrightarrow{B}=5.00\hat{i}-2.00\hat{j}$

The scalar product of vector

 a.b = $\left(4.00\hat{i}+7.00\hat{j}\right).\left(5.00\hat{i}-2.00\hat{j}\right)$

       $$\begin{gathered} =\left(4\times 5\right)+\left(7\times \left(-2\right)\right) \\ =20-14 \\ =6 \end{gathered}$$

The angle between two vector

The magnitude of vector is,

$$\begin{gathered} |A⃗|=\surd (4^2+7^2)=\surd 65=8.06 \\ |B⃗|=\surd (5^2+2^2)=\surd 29=5.3 \end{gathered}$$

The angle between the two vector is,

$\theta ={\cos }^{-1}\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}$

${\cos }^{-1}$= $\frac{6}{\left(8.06\right)\left(5.3\right)}$

            =${\cos }^{-1}\left(0.1399\right)$

$\theta =1.43°$