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

Dec 20, 2016

The answer is =143.1º

Explanation:

The angle between 2 vectors is given by the dot product

vecu.vecv=∥vecu∥*∥vecv∥ costheta

Here,

vecu=〈3,4〉

vecv=〈0,-2〉

The dot product is vecu.vecv=〈3,4〉.〈0,-2〉=0-8=-8

The modulus of $\vec{u}$ is =∥〈3,4〉∥=sqrt(9+16)=sqrt25=5

The modulus of $\vec{v}$ is =∥〈0,-2〉∥=sqrt(4)=2

The angle is

$\cos \theta = \frac{\vec{u} . \vec{v}}{| | \vec{u} | | \cdot | | \vec{v} | |}$

$= - \frac{8}{5 \cdot 2} = - 0.8$

theta=143.1º