SOLUTION: What is the direction and magnitude of v = < -3.5, -12>

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Question 1062725: What is the direction and magnitude of v = < -3.5, -12>
Answer by math_helper(2461) About Me  (Show Source):
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A vector of the form is taken to be "from the origin" to the point (x,y), and in that direction.
Magnitude = +sqrt%28%28-3.5%29%5E2+%2B+%28-12%29%5E2%29+ = +sqrt%28156.25%29+ = +highlight%2812.5%29+

—
Direction can be found from ++tan%28theta%29+=+y%2Fx+ —> +theta+=+arctan%28-12%2F-3.5%29+
= +73.740%5Eo+. There is another step: we must keep in mind that the point (-3.5,-12) is in quadrant 3
while +73.740%5Eo+ is in quadrant 1 (because positive angles are measured counter clockwise from +x-axis), so we must add 180%5Eo (the period of the tan() function) to get the proper angle: +73.740%5Eo+%2B+180%5Eo+=+highlight%28253.740%5Eo%29+ as measured CCW from the +x-axis.