# Homework Help: How to find the angle between an XYZ vector and the y-axis? Please help

1. Feb 4, 2010

### ian_spurlock

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

A vector is given as follows:

A = 2i - 2j - 3k

The angle between A and the y-axis is, in degrees, closest to?:

90
119
151
61
29

2. Relevant equations

c = ab sin$$\theta$$ ????

3. The attempt at a solution

I tried:

cos$$\theta$$ = (AdotB)/(A x B)

I tried to set a vector along the y-axis to use as B.

Isolating $$\theta$$ yielded 56.3 degrees, which is apparently incorrect.

2. Feb 4, 2010

### Staff: Mentor

Maybe try using the cross-product instead of the dot-product...

3. Feb 4, 2010

### ian_spurlock

To be honest, I don't know what formula to use and I don't even know if I can use the y-axis as a vector.

I don't want the answer... I just want to know where to start and what procedure I should use.

I'm pretty confused.

4. Feb 4, 2010

### vela

Staff Emeritus

You're just mixing up the dot product and the cross product. You should have

$$\vec{A}\cdot\vec{B} = |A||B|\cos \theta$$
$$|\vec{A}\times\vec{B}| = |A||B|\sin \theta$$

You can use either to find the angle $\theta$.

5. Feb 4, 2010

### Staff: Mentor

The unit vector in the direction of the y-axis is the j vector. Do you know the equation for the vector cross-product? How could you perhaps use it to find the angle?

6. Feb 4, 2010

### ian_spurlock

Well like vela posted the cross-product formula is LaTeX Code: $$|\vec{A} X \vec{B}| = |A||B|sin \theta$$

But I only have a vector A? Or can I use components of vector A in that formula?

Last edited by a moderator: Feb 4, 2010
7. Feb 4, 2010

### Staff: Mentor

I tried to clean up your latex -- hope I got it right.

Since you are asked to find the angle with the y axis, you have two vectors. you have A, and you have j. If you cross A into j as vectors, you will get an answer that has the inclusive angle in it. Do you know how to calculate the cross-product using a determinant?

8. Feb 4, 2010

### ian_spurlock

That part I do know how to do. But if I crossed vectors A and j, wouldn't I just get a vector in some form of i + j + k? How would I calculate an angle from that?

9. Feb 4, 2010

### Staff: Mentor

Use the magnitude equation that you posted a couple steps back. The magnitude of the vector resultant is related to the magnitude of A and j and ...

10. Feb 4, 2010

### ian_spurlock

Ahh, of sintheta

I'm getting it now.

Is j = (0i -2j + 0k)? Is it as simple as that?

11. Feb 4, 2010

### ian_spurlock

I tried to solve it as such... and I got 61 degrees which was wrong...

12. Feb 4, 2010

### Staff: Mentor

Post the details of your calculation. Also, make a good sketch for yourself in 3-space, to get an intuitive feel for what the answer should be.

13. Feb 4, 2010

### ian_spurlock

A = 2i - 2j - 3k ... A = 4.12
j = 0 + 1j + 0 ... j = 1
Code (Text):

|i   j   k |
|2  -2  -3 |
|0   1   0 |...... this matrix yielded [B]N[/B] = 3i + 2k ... N = 3.61

N = Aj sintheta
theta = sin^(-1) (N/Aj) = 61 degrees

Last edited by a moderator: Feb 4, 2010
14. Feb 4, 2010

### Staff: Mentor

Yeah, I get the same answer. It's not right? Do they want a couple figures to the right of the decimal point maybe?

15. Feb 4, 2010

### Staff: Mentor

Oops, I see it's multiple choice. What does it look like in your sketch?

16. Feb 4, 2010

### ian_spurlock

It's a multiple choice problem with the choices being:

90
119
151
61
29

I've already tried 2 answers, forfeiting 50% of the potential credit. Is it possible that it is 151 degrees? That is 61+90 degrees... Or 119? 180 - 61?

God I hate Mastering Physics. They gave us a free trial for it as LSU. What's the catch? We're basically beta testers. Infuriating.

17. Feb 4, 2010

### ian_spurlock

I'm bad at sketching 3d and I really don't know how to interpret my sketch.

18. Feb 4, 2010

### Staff: Mentor

I think the sketch is needed to see the answer. It's over 90 degrees to the positive y axis... Since the answer is multi-valued, our calculators only gave us the value for the first quadrant of the sine function...

19. Feb 4, 2010

### ian_spurlock

Eh, I am bad at sketching in 3d. What can I say about it really? It's below the XY plane?

Exactly what is theta in this case?

20. Feb 4, 2010