# Three dimensional vector help

1. Oct 3, 2007

### physstudent1

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

vector A=(3i+2j+k)m. The angle vector makes with the y axis is nearest
a 52 degrees
b 54 degrees
c 56 degrees
d 58 degrees
e 60 degrees

2. Relevant equations

3. The attempt at a solution

I did arctan of (2/3) to get 34 then subtracted it from 90 and got C but I'm not sure about this one the third component is kind of screwing with me a little can anyone clear up how to do this with a third dimension.

2. Oct 3, 2007

### bob1182006

for this you need to use the geometric definition of the dot product:

$$\vec{a}\cdot\vec{b}=\parallel\vec{a}\parallel\parallel\vec{b}\parallel \cos \theta$$

where ||a|| is the distance of a to the origin (0,0,0).
$$\vec{a}$$=[x,y,z]
$$\parallel\vec{a}\parallel=\sqrt{x^2+y^2+z^2}$$

and theta is the angle between the 2 vectors.

solving for theta gives you:

$$\theta=arccos\{\frac{\vec{a}\cdot\vec{b}}{\parallel\vec{a}\parallel\parallel\vec{b}\parallel}\}$$

you know A in vector form and you just have to find the vector form of the y-axis and do the dot product. find the distace's and the arccos

3. Oct 3, 2007

### physstudent1

is the vector component form of the y axis 0,1,0

Last edited: Oct 3, 2007
4. Oct 3, 2007

### physstudent1

I got arccos ( 2/ sqrt(14)) = 57.7 degrees this is with assuming the vector form of the y axis is 0,1,0

5. Oct 3, 2007

### bob1182006

yes for the y-axis you would use the unit vector j which has coordinates of (0,1,0).

6. Oct 3, 2007

thanks alot