document.write( "Question 1062725: What is the direction and magnitude of v = < -3.5, -12> \n" ); document.write( "
Algebra.Com's Answer #677748 by math_helper(2461)\"\" \"About 
You can put this solution on YOUR website!
A vector of the form is taken to be \"from the origin\" to the point (x,y), and in that direction.\r
\n" ); document.write( "\n" ); document.write( "Magnitude = \"+sqrt%28%28-3.5%29%5E2+%2B+%28-12%29%5E2%29+\" = \"+sqrt%28156.25%29+\" = \"+highlight%2812.5%29+\" \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "—
\n" ); document.write( "Direction can be found from \"++tan%28theta%29+=+y%2Fx+\" —> \"+theta+=+arctan%28-12%2F-3.5%29+\"
\n" ); document.write( "= \"+73.740%5Eo+\". There is another step: we must keep in mind that the point (-3.5,-12) is in quadrant 3
\n" ); document.write( "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.
\n" ); document.write( "
\n" );