Question

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

Answer 1

Please see the explanation for a description of the process.

Explanation:

Compute the dot-product by multiplying the `hati` coefficients and then adding the product of the `hatj` coefficients:

`baru*barv = (2)(1) + (-2)(-1) = 4`

A second way to compute the dot-product uses the magnitude of the two vectors and the cosine of the angle between the two vectors:

`baru*barv = |baru||barv|cos(theta)` [1]

We know that the dot-product is 4, therefore, we need to compute the magnitude of the two vectors and then solve for the angle between, `(theta)`;

`|baru| = sqrt(2^2 + (-1)^2) = sqrt(5)`

`|barv| = sqrt(1^2 + (-2)^2) = sqrt(5)`

Substitute into equation [1] and then solve for `theta`:

`4 = sqrt(5)sqrt(5)cos(theta)`

`4 = 5cos(theta)`

`cos(theta) = 4/5`

`theta = cos^-1(4/5)`

`theta ~~ 37°`

Answer 2

The angle is `=36.9º`

Explanation:

The angle between two vectors is given by the dot product definition.
`vecu.vecv=∥vecu∥*∥vecv∥costheta`
where `theta` is the angle between the 2 vectors

so, `costheta= (vecu.vecv) / (∥vecu∥*∥vecv∥)`
The dot product is `=〈2,-1〉〈1,-2〉=2+2=4`
The modulus of `vecu=∥vecu∥=sqrt(4+1)=sqrt5`
The modulus of `vecv=∥vecv∥=sqrt(1+4)=sqrt5`

So, `costheta=4/(sqrt5*sqrt5)=4/5`
`theta=arccostheta=36.9º`