Question

If `A = <5 ,2 ,5 >`, `B = <6 ,5 ,3 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=92.1º`

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈5,2,5〉-〈6,5,3〉=〈-1,-3,2〉`

The angle between `vecA` and `vecC` is given by the dot product definition.

`vecA.vecC=∥vecA∥*∥vecC∥costheta`

Where `theta` is the angle between `vecA` and `vecC`

The dot product is

`vecA.vecC=〈5,2,5〉.〈-1,-3,2〉=-5-6+10=-1`

The modulus of `vecA`= `∥〈5,2,5〉∥=sqrt(25+4+25)=sqrt54`

The modulus of `vecC`= `∥(-1,-3,2〉∥=sqrt(1+9+4)=sqrt14`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-1/(sqrt54*sqrt14)=-0.036`

`theta=92.1`º

Answer 2

We can start by finding `vecC`. To subtract vectors, we subtract the corresponding components. For `vecA-vecB`, we have:

`vecC = <5,2,5> - <6,5,3>`

`= <(5-6),(2-5),(5-3)>`

`=<-1,-3,2>`

The angle between two vectors can be found using this general formula:

`cos(theta)=(veca*vecb)/(|veca|*|vecb|)`

We can start by finding the dot product of A and C.

`vecA*vecC=<5,2,5> *<-1,-3,2>`

`=(5*-1)+(2*-3)+(5*2)`

`=-5-6+10`

`=-1`

Now we can find the product of the magnitude of each vector.

`|vecA|=|<5,2,5>`

`=sqrt(5^2+2^2+5^2)`

`=sqrt(54)`

`=3sqrt(6)`

`|vecB|=|<-1,-3,2>`

`=sqrt((-1)^2+(-3)^2+2^2)`

`=sqrt(14)`

So the product of the magnitudes is `3sqrt6*sqrt14=3sqrt(84)=6sqrt(21)`

We now have:

`cos(theta)=-1/(6sqrt21)=-sqrt21/126`

Solving for `theta`:

`theta=arccos(-sqrt21/126)`

`=92 ^o`