You fly 32.0km in a straight line in still air in the direction 35.0° south of west. (a) Find the distances you would have to fly due south and then due west to arrive at the same point. (b) Find the distances you would have to fly first in a direction 45.0° south of west and then in a direction 45.0° west of north. Note these are the components of the displacement along a different set of axes—namely, the one rotated by 45° with respect to the axes in (a).

The diagram shows the flight direction as a vector labeled "a" in the third quadrant. With respect to the +x axis (east), the angle is 215° counterclockwise. The (E, N) components of the vector are ...

a = 32(cos(215°), sin(215°)) ≈ (-26.21, -18.35) . . . km

The minus signs indicate distance west and south, since the reference directions are east and north.

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Rotation of the axes to translate south to southwest, and west to northwest effectively adds 45° to the angle, making it 215° +45° = 260°. In the attachment, we have computed the new coordinates relative to axes rotated clockwise 45°. Then in (SE, NE) coordinates, the vector is ...

a' = 32(cos(260°), sin(260°)) = (-5.56, -31.51) . . . km

The minus signs indicate distance NW and SW, since the reference directions are SE and NE.