Vector with angles with axis (book wrong?)

[kkinsky]

Homework Statement

got a vector F (105,300,140)

magnitude: 347.3

it wants the angles, 3 of them with the axis x y z

Homework Equations

all my calculations has lead to the wrong answer, but the correct answer is (180-myanswer)

why is that?

The Attempt at a Solution

using AB = |A||b| cos α formula

so with an x-axis the coordinates are (1,0,0)

FX/|F||X|=cos(x-axis angle)

arcos(105/347.3) = x-axis angle

getting the answer: 72.4degrees

the "correct" answer is 107.6

meaning 180-72.4, why am i getting this 180degree descrepency?

 

[jedishrfu]

try drawing the vector on the three axes and you'll see that your answer reasonable.

Check to see if one of the coordinates have the correct signs in your book like is x=-105?

 

[D H]

Your F=(105,300,140) represents a point in the first octant. The angles between the +x, +y, and +z axes must all be less than 90 degrees.However, perhaps you misread the question and missed seeing a minus sign. If your vector F was -(105,300,140), or (-105,300,140), then the correct answer would be 107.6.

 

[HallsofIvy]

By the way, a nice way to do this problem is to use the "direction cosines". If v is a unit vector, [itex]<v_x, v_y, v_z>[/itex], then its components are the cosines of the angles the vector makes with the three axes. That is, if [itex]\theta_x[/itex], [itex]\theta_y[/itex] and [itex]\theta_z[/itex] are the angles the vector makes with the respective axes, then [itex]cos(\theta_x)= v_x[/itex], [itex]cos(\theta_y)= v_y[/itex], and [itex]cos(\theta_z)= v_z[/itex].

Here, F has length [itex]\sqrt{(105)^2+ (300)^2+ (140)^2}= \sqrt{120625}= 347.3[/itex] (approximately). So a unit vector in the direction of F is [itex]<104/347.3, 300/347.3, 140/347.3>= <0.3023, 0.8638, 0.4034>[/itex] so that the angles are [itex]cos^{-1}(0.3023)= 72.4[/itex] degrees, [itex]cos^{-1}(0.8638)= 32.2[/itex] degrees, [itex]cos^{-1}(.4034)= 66.2[/itex] degrees.