Question

The rectangular components of a vector are $(2,2)$. The corresponding rectangular components of another vector are $(1,\sqrt{3})$. Find the angle (in degree) between the two vectors.

Answer 1

Find the angle made by the $\overrightarrow{A}$ from the x axis.

Vectors can be written as,
$\overrightarrow{A}=2\hat{i}+2\hat{j}$
$\overrightarrow{B}=\hat{i}+\sqrt{3}\hat{j}$

As we know
$\tan \theta =\frac{y}{x}=\frac{2}{2}=1$
$\tan \theta =1$
$\theta ={45}^{\circ }$

Find the angle made by the $\overrightarrow{B}$ from the x axis.

As given, $\overrightarrow{B}=\hat{i}+\sqrt{3}\hat{j}$
Angle maade by the $\overrightarrow{B}$ from the $x$ axis is given by

$\tan \theta =\frac{y}{x}=\frac{\sqrt{3}}{1}=\sqrt{3}$
$\tan \theta =\sqrt{3}$
$\theta ={60}^{\circ }$

Find the angle made by the A ⃗.(B ) ⃗ from the x axis.

According to dot product of two vectors we get,
$\overrightarrow{A}\cdot \overrightarrow{B}=ABcos\theta \dots \left(i\right)$
By putting the values in equation $\left(i\right)$ we get,
$(2i+2j)(i+\sqrt{3}j)=2\sqrt{2}\times 2\times \cos \theta$
$2+2\sqrt{3}=4\sqrt{2}\cos \theta$
Therefore,
$\frac{1+\sqrt{3}}{2\sqrt{2}}=\cos \theta \dots \left(ii\right)$
Hence,
Angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $({60}^{\circ }-{45}^{\circ })={15}^{\circ }$.

Final Answer:${15}^{\circ }$