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

1 Answer
May 6, 2018

Answer:

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#?

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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^@#.