Finding the Angle Between Vectors A and B in the Cross Product

[Fetch]

Homework Statement

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?

Homework Equations

Cross Product Formula

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./:

 

[ehild]

Homework Statement

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

So I layed out my grid.

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

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

 

[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
Is the answer not 74.1?

 

[ehild]

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
Is the answer not 74.1?

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

 

[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?

 

[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

 

[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?

 

[ehild]

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

ehild

 

[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?

 

[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

 

[Fetch]

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

 

[ehild]

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

ehild

 

[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?

 

[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