# Vector with angles with axis (book wrong?)

1. Mar 30, 2014

### kkinsky

1. The problem statement, all variables and given/known data

got a vector F (105,300,140)

magnitude: 347.3

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

2. Relevant equations

why is that?

3. 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

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

2. Mar 30, 2014

### Staff: Mentor

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?

3. Mar 30, 2014

### D H

Staff Emeritus
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.

4. Mar 31, 2014

### HallsofIvy

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

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