Question

How do you find the angel between u=27i+14j, v=i-7j?

Answer 1

`71°`(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=27 ulhati+14 ulhatj=((27),(14))` and `vec v=1 ulhati -7 ulhatj=((1),(-7))`

The modulus is given by;

`|vec u| = |((27),(14))| \ \ = sqrt(27^2+14^2)=sqrt(729+196)=sqrt(925)`
`|vec v| = |((1),(-7))| = sqrt(1^2+ -7^2) =sqrt(1+49) =sqrt(50)`

And the scaler product is:

`vec u * vec v = ((27),(14)) * ((1),(-7))`
`\ \ \ \ \ \ \ \ \ \ = (27)(1) + (14)(-7)`
`\ \ \ \ \ \ \ \ \ \ = 27-98`
`\ \ \ \ \ \ \ \ \ \ = -71`

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

`-71 = sqrt(925) * sqrt(50) * cos theta`
`:. cos theta = (-71)/(sqrt(46250))`
`:. cos theta = -0.33014 ...`
`:. theta = 109.277 °`
`:. theta = 109 °`(3sf)

So the acute angle between the vectors is `71°`(3sf)