Work/Angle?

Question

Given the vector V = (3.80i) + (5.40j) + (-7.00k), calculate the angle between V and the positive x, y, and z axes. Your answers should be in degrees and may be greater than 90.

Wow...how can physics do this to me...In the book it explains how to do it with only the x and y coordinates but it also gave a force and distance vector and to find the angle between the two forces = inverse cos (W/delta-distance).

There are no examples like this in the book nor in the questions...I have no clue how to do this if it has 3 axis and no force or distance vector stated. please help....

Answer 1

Calculate the magnitude of the vector M = √[3.8^2 + 5.4^2 + 7^2]

Then the angle of the vector to each axis is

angle to x = arccos(3.8/M)
angle to y = arccos(5.4/M)
angle to z = arccos(-7.0/M)

Answer 2

1st answer's method is correct for this problem which is limited to angles between a vector and the basis vectors. A more general solution is usable for angles between any two 3D vectors. First you convert the two vectors to unit vectors. (I.e., the vector [3.8,5.4,-7.0] and the axis [1,0,0] for x axis, etc.). The unit vector is the vector divided by its magnitude (the axes [1,0,0] etc. are already unit vectors). Then take the dot product of the two unit vectors. Finally, the angle = arccos(dot product). The ref. is a tutorial that illustrates the concept.