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

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Answer 1 · 2018-05-06T21:04:27

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

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 $-45^@$ .

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