Question

$\overrightarrow{A}=2\hat{i}+3\hat{j}$ and $\overrightarrow{B}=3\hat{i}+2\hat{j}$ are two vectors. Find the angle between the two vectors.

Answer 1

Given that,

The vector is

$\overrightarrow{A}=2\hat{i}+3\hat{j}$

$\overrightarrow{B}=3\hat{i}+2\hat{j}$

We know that,$\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|}=\cos \theta$

Now, the dot product

$\overrightarrow{A}\cdot \overrightarrow{B}=(2\hat{i}+3\hat{j})\cdot (3\hat{i}+2\hat{j})$

$\overrightarrow{A}\cdot \overrightarrow{B}=6+6$

$\overrightarrow{A}\cdot \overrightarrow{B}=12$

Now, the magnitude

$|A|=\sqrt{4+9}$

$|A|=\sqrt{13}$

$|B|=\sqrt{9+4}$

$|B|=\sqrt{13}$

Now, the angle is

$\cos \theta =\frac{12}{13}$

$\theta ={\cos }^{-1}\frac{12}{13}$

Hence, the angle between the two vectors is ${\cos }^{-1}\frac{12}{13}$