Question

How do I find the magnitude and direction angle of the vector `v =6i-6j`?

Answer 1

The magnitude of `v` is `|vec v| = 6sqrt2` and the direction angle `theta` is `theta= -45^@`.

Explanation:

In this answer, we'll define the direction angle to be `theta` and the magnitude of the vector to be `|vec v|`.

Now, for a more general case, how do we find the magnitude and direction angle of the vector `vec nu = aveci-bvecj`?

As you can see, the sides with lenghts `a`, `b` and `|vec nu|` form a right triangle. Thus,

`|vec nu|^2 = a^2+b^2`

`sin theta = a/|vec nu|"`

`cos theta = b/|vec nu|`

Using these identities, we can solve for our particular case with `vec v = 6veci-6vecj`.

`:. |vec v|^2 = 6^2+(-6)^2 = 72 => |vec v| = 6sqrt2`

`sin theta = b/|vec v| = -6/(6sqrt2) = -1/sqrt2`

`cos theta = a/|vec v| = 6/(6sqrt2) = 1/sqrt2`

Now, the sine function is negative in the fourth quadrant while the cosine is positive, so we must have `theta = -45^@`.

Our vector has magnitude `6sqrt2` and direction angle `-45^@`.