Question

What is the angle between `<7 , -2 , 7 >` and `< 2 , 5 , 1 >`?

Answer 1

The angle is `=78.5`º

Explanation:

We use the dot product definition,

`veca.vecb=∥veca∥*∥vecb∥*cos(veca,vecb)`

Here, `veca=〈7,-2,7〉` and `vecb=〈2,5,1〉`

The dot product is

`veca.vecb=〈7,-2,7〉.〈2,5,1〉=14-10+7=11`

The modulus of `veca=∥〈7,-2,7〉∥=sqrt(49+4+49)=sqrt102`

The modulus of `vecb=∥〈2,5,1〉∥=sqrt(4+25+1)=sqrt30`

We can now calculate the angle,

`cos(veca,vecb)=(veca.vecb)/(∥veca∥*∥vecb∥)=11/(sqrt102*sqrt30)=0.2`

The angle is `(veca,vecb)=78.5`º