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

[ian_spurlock]

How to find the angle between an XYZ vector and the y-axis? Please help!

Homework Statement

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

Homework Equations

c = ab sin$$\theta$$ ?

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.

 

[berkeman]

Homework Statement

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

Homework Equations

c = ab sin$$\theta$$ ?

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.

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

 

[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.

 

[vela]

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 $\theta$.

 

[berkeman]

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.

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?

 

[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?

 

[berkeman]

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?

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?

 

[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?

 

[berkeman]

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?

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 ...

 

[ian_spurlock]

Ahh, of sintheta

I'm getting it now.

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

 

[ian_spurlock]

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

 

[berkeman]

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

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.

 

[ian_spurlock]

A = 2i - 2j - 3k ... A = 4.12
j = 0 + 1j + 0 ... j = 1

|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

 

[berkeman]

A = 2i - 2j - 3k ... A = 4.12
j = 0 + 1j + 0 ... j = 1

|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

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

 

[berkeman]

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

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

 

[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.

 

[ian_spurlock]

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

 

[berkeman]

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

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...

 

[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?

 

[ian_spurlock]

The answer was 119 apparently. I don't even know why. Maybe my sketch is just extremely difficult to interpret? Blah.

Anyway, thanks for the time and input. At least I have the raw calculation part down.

Thanks a bunch.

 

[berkeman]

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.

I think I know which one it is based on my sketch. I've requested help from the other Homework Helpers, to be sure that I'm not leading you astray...

 

[ian_spurlock]

OK great. I'd like to know how to solve this in case I encounter a problem like this on an exam.

 

[berkeman]

OK great. I'd like to know how to solve this in case I encounter a problem like this on an exam.

I'm still hopeful that another HH'er will check all of this. From my sketch, it looks like the 61 degree angle is to the negetive y axis, not the positive y axis. That's why the answer was 180-61, to get the answer to the + y axis. Without the sketch or some other intuition, it would be hard to know that the angle calculated with the cross-product technique is in the first quadrant of the sine function, and may not match the physical situation.

 

[Subdot]

Well, using the dot product to find theta, isn't it simply because cos(theta) < 0 which only occurs in the second and third quadrants? A sketch, or in this case the multiple choice answers, shows that the angle is in the second quadrant and so the angle is 180 - 61.

 

[ian_spurlock]

OK that makes crystal clear sense.

I feel kinda dumb for not thinking of that haha. Looooooooong day.

:epic facepalm:

Thanks a lot man. Much obliged.

 

[ian_spurlock]

And thanks @ Subdot

I'll remember that trick from now on