Question

Find the direction and magnitude of v = <13,-11>

Answer 1

magnitude will be the distance from (0,0) to (13,-11). The distance formula is
\[d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\] but in this case it simplifies due to the use of (0,0) as the other endpoint to \[d = \sqrt{v_x^2+v_y^2}\] where your vector is \(v = <v_x, v_y>\)

The direction can be found by \[\theta = \tan^{-1}(\frac{v_y}{v_x})\] You may have to adjust the result to put the direction in the proper quadrant. Sketching to check the result is a good idea.

Answer 2

im confused

Answer 3

If v is a vector v = <13,-11>
then the magnitude of v is |v| = sqrt(13^2+11^2)= 17.03
the direction is toward south east or from the origin in the forth quadrant

Answer 4

Thank you!!!

Answer 5

Of course, that only gives you the general direction, not the actual answer :-)

Answer 6

okay thanks

Answer 7

The direction is from the origin (0,0) to the point (13,-11), an arrow is usually placed at end of line at point (13,-11)

Answer 8

thank you for taking the time to show me

Answer 9

A calculation using Mathematica 9 is attached.

Answer 10

Thank you for the medal. Going back to V8 Home Edition of Mathematica. V9 has too many problems.