Q 42E - University Physics with Modern Physics

Question

(I) (a) Find the scalar product of the vectors $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$ 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 $1.43°$.

Step 1: Given data

$\begin{array}{l}\overrightarrow{A}=4.00\hat{i}+7.00\hat{j}\\ \overrightarrow{B}=5.00\hat{i}-2.00\hat{j}\end{array}$

Step 2: The scalar product of vector

$$\begin{gathered} a.b=\left(4.00\hat{i}+7.00\hat{j}\right).\left(5.00\hat{i}-7200\hat{j}\right) \\ =\left(4\times 5\right)+\left(7\times \left(-2\right)\right) \\ =20-14 \\ =6 \end{gathered}$$

Step 3: The angle between two vector

The magnitude of vector is,

               

$\begin{array}{l}\left|\overrightarrow{A}\right|=\sqrt{4^2+7^2}=\sqrt{65}=8.06\\ \left|\overrightarrow{B}\right|=\sqrt{5^2+2^2}=\sqrt{29}=5.3\end{array}$               

The angle between the two vector is,

 

$$\begin{gathered} \theta ={\cos }^{-1}\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|} \\ ={\cos }^{-1}\frac{6}{\left(8.06\right)\left(5.3\right)} \\ ={\cos }^{-1}\frac{6}{\left(42.87\right)} \\ ={\cos }^{-1}\left(0.1399\right) \\ \theta =1.43° \end{gathered}$$