# How do you use the definition of the scalar product, find the angles between the following pairs of vectors: - 4i + 5 j- k and 3i + 4j - k?

Jul 14, 2018

The angle is $= {74.2}^{\circ}$

#### Explanation:

The vectors are

$\vec{A} = < - 4 , 5 , - 1 >$

and

$\vec{B} = < 3 , 4 , - 1 >$

The angle between $\vec{A}$ and $\vec{B}$ is given by the dot product definition.

vecA.vecB=∥vecA∥*∥vecB∥costheta

Where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$

The dot product is

vecA.vecB=〈-4,5,-1〉.〈3,4,-1〉=-12+20+1=9

The modulus of $\vec{A}$= ∥〈-4,5,-1〉∥=sqrt(16+25+1)=sqrt42

The modulus of $\vec{B}$= ∥〈3,4,-1〉∥=sqrt(9+16+1)=sqrt26

So,

costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=9/(sqrt42*sqrt26)=0.27

$\theta = \arccos \left(0.27\right) = {74.2}^{\circ}$