# How do you find the angle between the vectors u=2i-3j and v=4i+3j?

Jul 14, 2016

theta = arccos((-1)/(5*sqrt(13))) ≈ 93.18°

#### Explanation:

The angle between two vectors is found using the formula

$\cos \left(\theta\right) = \frac{\vec{u} \cdot \vec{v}}{| | \vec{u} | | | | \vec{v} | |}$, where $\vec{u} \cdot \vec{v}$ is known as the dot product, and $| | \vec{u} | |$ for instance is known as the magnitude of $\vec{u}$.

The dot product is calculated the following way:

$\vec{u} \cdot \vec{v} = {u}_{1} {v}_{1} + {u}_{2} {v}_{2} + \ldots {u}_{n} {v}_{n}$

The magnitude of the vector is found by squaring each term of vector vecu = ‹ u_1, u_2, ... u_n ›, add them all up, and take the square root of the result.

In our case we have

vecu = 2hati-3hatj = ‹2, -3› -> ||vecu|| = sqrt((2)^2+(-3)^(2))=sqrt(13)

vecv = 4hati +3hatj = ‹4, 3› -> ||vecv|| = sqrt((4)^2+(3)^(2))=sqrt(25) = 5

$\vec{u} \cdot \vec{v} = 2 \left(4\right) + - 3 \left(3\right) = 8 - 9 = - 1$

Using the formula

$\cos \left(\theta\right) = \frac{\vec{u} \cdot \vec{v}}{| | \vec{u} | | | | \vec{v} | |} = \frac{- 1}{5 \cdot \sqrt{13}}$

So theta = arccos((-1)/(5*sqrt(13))) ≈ 93.18°