Question

How do I find the direction angle of vector `<1, -sqrt3>`?

Answer 1

The direction angle of a vector is given by the formula

`arctan(b/a) = theta`

where `b = -sqrt(3)` and `a = 1`

because `tan(theta) = (opposite)/(adjacent)`

The opposite is `b` and adjacent is `a` draw it on a paper if you don't visualize

here we have `arctan(-sqrt(3)) = theta`

So `theta = -pi/3`

Here a graph of the form `y = mx` where `m = tan(theta)` you can see that when `x = 1 => y = -sqrt(3)`

graph{y=-tan(22/21)x [-20, 20, -10.42, 10.42]}