Question

Vector $\overrightarrow{\mathbf{A}}$has a magnitude of 5 units, and vector $\overrightarrow{\mathbf{B}}$ has a magnitude of 9 units. The two vectors make an angle of ${\mathbf{50}}^{\mathbf{\circ }}$ with each other. Find $\overrightarrow{\mathbf{A}}\mathbf{·}\overrightarrow{\mathbf{B}}$.

Short answer

The dot product of vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is $28.93{\mathrm{unit}}^2$.

Step-by-step solution

Given data

Magnitude of vector $\overrightarrow{A}$

$A=5\mathrm{units}$

Magnitude of vector $\overrightarrow{B}$

$B=9\mathrm{units}$

Angel between vectors $\overrightarrow{A}$ and $\overrightarrow{B}$

$\theta ={50}^{\circ }$

Dot product of two vectors

The dot product of two vectors $\overrightarrow{\mathbf{A}}$and role="math" localid="1668408307321" $\overrightarrow{\mathit{B}}$is

$\overrightarrow{\mathbf{A}}\mathbf{·}\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{ABcos\theta }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(I\right)$

Determining the dot product of the given vectors

From equation (I), the dot product of vectors $\overrightarrow{A}$and $\overrightarrow{B}$ is

$$\begin{gathered} \overrightarrow{A}·\overrightarrow{B}=5\mathrm{units}\times 9\mathrm{units}\times \cos {50}^{\circ } \\ =28.93{\mathrm{unit}}^2 \end{gathered}$$

Thus the required dot product is 28.93