Question

Vector $\overrightarrow{\mathbf{A}}$has a magnitude of$\mathbf{6}\mathbf{.}\mathbf{00}$units, vector$\overrightarrow{\mathbf{B}}$has magnitude of$\mathbf{7}\mathbf{.}\mathbf{00}$units, and$\overrightarrow{\mathbf{A}}\mathbf{.}\overrightarrow{\mathbf{B}}$has a value of$\mathbf{14}\mathbf{.}\mathbf{0}$What is the angle between the directions of $\overrightarrow{\mathbf{A}}$and $\overrightarrow{\mathbf{B}}$?

Short answer

The angle $\left(\theta \right)$between $\overrightarrow{A}$and $\overrightarrow{B}$is $70.5°$.

Step-by-step solution

Vector operations

The dot product of the vectors is equal to the product of the magnitude of the vectors and the cosine of the angle between them. Using this definition, find the value of the angle between the two vectors when the magnitude and dot product are given. The formula for the dot product is,

$\overrightarrow{a}.\overrightarrow{b}=a.b.c0s\theta$ (i)

The magnitude of the vector $\overrightarrow{A}$is $6.00\mathrm{units}$. The magnitude of vector $\overrightarrow{B}$is$7.00\mathrm{units}$. The dot product of two vector is equal to$14.0\mathrm{units}$

Finding the direction between two vectors

Substitute the given values in equation (i)

$$\begin{gathered} \overrightarrow{a}.\overrightarrow{b}=a.b.\cos \theta \\ 14.0=6.00\times 7.00\times \cos \theta \\ 14.0=42\times \cos \theta \\ \cos \theta =\frac{14.0}{42} \\ \theta ={\cos }^{-1}\left(\frac{14.0}{42}\right) \\ =70.5° \end{gathered}$$

Therefore, the angle between the two vectors is $70.5°$.