Question

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

Answer 1

So the acute angle between the vectors is `7.122°`(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 A` 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=2hati-3hatj` and `vec v=hati-2hatj`

The modulus is given by;

`||vec u|| = sqrt((2)^2+(-3)^2)=sqrt(4+9)=sqrt(13)`
`||vec v|| = sqrt((1)^2+ (-2)^2) =sqrt(1+4) =sqrt(5)`

And the scaler product is:

`vec u * vec v = (2)(1) + (-3)(-2)`
`\ \ \ \ \ \ \ \ \ \ = 2+6`
`\ \ \ \ \ \ \ \ \ \ = 8`

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

`8 = sqt(13)sqrt(5) cos theta`
`:. cos theta = 8/(sqrt(13)sqrt(5))`
`:. cos theta = 0.992277 ...`
`:. theta = 7.1250 ... °`
`:. theta = 7.13 °`(3sf)

So the acute angle between the vectors is `7.13°`(3sf)