Question

How do you find the angle between the vectors `u=<1,0>` and `v=<0,-2>`?

Answer 1

Please see the explanation.

Explanation:

Two methods to dot-product to compute the dot product are:

`"[1] "baru*barv = (u_x)(v_x) + (u_y)(v_y) + ...`

Where `u_x` and `v_x` are the x components, `u_y` and `v_y` are the y components, and so on.

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

where the magnitudes are `|baru| = sqrt(u_x^2 + u_y^2 + ...)` and `|barv| = sqrt(v_x^2 + v_y^2 + ...)` and `theta` is the angle between the two vectors.

To do this problem, you compute the dot-product using method [1]:

`baru*barv = (1)(0) + (0)(-2)`

`baru*barv = 0`

This is a special case where the dot-product is zero and we instantly know that `theta = pi/2` but let's proceed as, if it were not this special case.

Next, compute the magnitudes:

`|baru| = sqrt(1^2 + 0^2)`

`|baru| = 1`

`|barv| = sqrt(0^2 + 2^2)`

`|barv| = 2`

Into the equation for method [2], substitute 0 for `baru*barv`, 1 for `|baru|`, and 2 for `|barv|`:

`0 = (1)(2)cos(theta)`

Solve the equation for `theta`

`cos(theta) = 0`

`theta = pi/2`

This is the answer.

Answer 2

`pi/2 and` (, for the opposite sense of measurement, ) `3/2pi`

Explanation:

The dot ( scalar ) product

`u.v= <1, 0>.<0, -2>=(1)(0)+(0)(-2)=0`..

So, the cosine of the angle, `(u.v)/(|u||v|)` is 0.

In `theta in [0, 2pi]`, the angle is `pi/2`, in the clockwise sense,

from `v to u`, and `3/2pi`, for measurement in the opposite sense.