# Homework Help: Vector Cross Product Problem

1. Aug 26, 2014

### Fetch

1. The problem statement, all variables and given/known data
Vectors A & B lie in an xy plane. A has a magnitude 7.4 and an angle 142(deg) with respect to the +x direction. B has components (-6.84i, -7.37j)
B) What is the angle between the -y axis and the direction of the Cross product between A and B?

2. Relevant equations
Cross Product Formula

3. The attempt at a solution

So I layed out my grid.

i----------j-------k
(-5.83)----(4.55)----(0) (A->)
(-6.84)---(-7.37)----(0) (B->)

I got the -5.83 & 4.55 from the formula (7.4cos142, 7.4sin142)
Welp, I added everything up and got 74.1
Wiley plus isn't accepting it. I have everything else right on this problem save for this part. The answer is expressed in degrees. I very much gave this question an honest amount of attempts. Running out./:

Last edited: Aug 26, 2014
2. Aug 26, 2014

### ehild

You miss a minus in front of 7.37. But the direction of the cross-product vector is needed. Express it with i, j, k. Does it have any components in the plane?

ehild

3. Aug 26, 2014

### Fetch

When I work it out I have [(0)-(0)]i - [(0)-(0)]j + [(-5.83*-7.37)-(4.55*-6.84)]k

Which comes out to (0)i + (0)j + (74.0891)k

4. Aug 26, 2014

### ehild

No, you need to give an angle.
The resultant vector has only k component. What do you get if you make the dot product of it with any vectors of the plane?
If the dot product of two vectors is zero, what is the angle between them?

ehild

5. Aug 27, 2014

### Fetch

I'm stupid. Its 90.
Lol.

Hey, while you're at it can you explain part C?

It asks for the angle between the -y axis and (A-> x (b-> +3k)

The answer is 105, but it looks like I lucked into it, would love to understand what's actually going on. I know that the K vector is going upwards, but since there's and x and y involved then it goes out at an angle. Could I use tangent to discover the angle it makes with the floor, then take that number away from 180 to get the 105?

6. Aug 27, 2014

### ehild

I do not understand your notation. Do you need the cross product $\vec A \times (\vec B + 3 \vec k)$?

Remember, the components of a vector are magnitude times cosine of the angle they enclose with a positive axis.

ehild

7. Aug 27, 2014

### Fetch

Yes that is what I meant.

I understand that for an xy plane, but when I do it with respect to the 3d plane I get a little lost.
I'm having trouble understanding how to find this angle?

8. Aug 27, 2014

### ehild

Do you know how to determine the angle between two vectors, using the dot product?

ehild

9. Aug 27, 2014

### Fetch

Its something like Vector A dot Vector B = abcosθ right?
But wouldn't that only give me an angle with respect to the xy plane and not with respect to the variation due to traveling along the z axis?

10. Aug 27, 2014

### ehild

You need the angle of the cross product vector with the unit vector -j .

What is the cross product $\vec A \times (\vec B + 3 \vec k)$?

ehild

11. Aug 27, 2014

### Fetch

I got 13.65i - (-17.49)j + 74.08k?

12. Aug 27, 2014

### ehild

Well, what is the magnitude of this? And what angle does it enclose with a unit vector along the -y axis?

ehild

13. Aug 27, 2014

### Fetch

The magnitude is 73.33.
I used the j component (17.49j) as my magnitude for the -y vector (represented as a positive y vector so I could find the reference angle).
Then used the dot product to find the angle that it enclosed.
It encloses an angle of like 76 or 75 degrees with the positive y-axis, so the angle with the negative -y axis was like 104 or 105 degrees, which is the answer.

Seem legit?

14. Aug 27, 2014

### ehild

Yes, it is correct.
If you multiply 17.49 j with (-j) and divide it by the magnitude 73.33, you get a negative number, cos(104°)

ehild