Question

How do you use the definition of the scalar product, find the angles between the following pairs of vectors: A = - 3 i + j and B = 21 j - 3k?

Answer 1

`71.8 °`(3sf)

Explanation:

The angle `theta` between two vectors `vec A` and `vec B` is related to the modulus (or magnitude) and scaler (or dot) product of `vec A` and `vec B` by the relationship:

`vec A * vec B = ||A|| * ||B|| * cos theta`

By convention when we refer to the angle between vectors we choose the acute angle.

So for this problem, let the angle betwen `vecu` and `vecv` be `theta` then:

`vec u=<<3, 1, 0>>` and `vec v=<<0, 21, -3>>`

The modulus is given by;

`||vec u|| = ||<<3, 1, 0>>|| = sqrt(3^2+1^2+0^2)=sqrt(9+1)=sqrt(10)`
`||vec v|| = ||<<0, 21,-3>>|| = sqrt(0^2+21^2+(-3)^2)=sqrt(441+9)=sqrt(450)`

And the scaler product is:

`vec u * vec v = <<3, 1,0>> * <<0,21,-3>>`
`\ \ \ \ \ \ \ \ \ \ = (3)(0) + (1)(21) + (0)(-3)`
`\ \ \ \ \ \ \ \ \ \ = 21`

And so using `vec A * vec B = ||A|| * ||B|| * cos theta` we have:

`21 = sqrt(10) * sqrt(450) * cos theta`
`:. cos theta = (21)/sqrt(4500)`
`:. cos theta = (7sqrt(5))/50`
`:. cos theta = 0.31304 ...`
`:. theta = 71.7568 ... °`
`:. theta = 71.8 °`(3sf)