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

In summary, Wiley Plus is not accepting the homework because they are missing an angle in one of the equations.
  • #1
Fetch
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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./:
 
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  • #2
Fetch said:

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
 
  • #3
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? o_O
 
  • #4
Fetch said:
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? o_O
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
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
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
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
Do you know how to determine the angle between two vectors, using the dot product?

ehild
 
  • #9
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
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
I got 13.65i - (-17.49)j + 74.08k?
 
  • #12
Well, what is the magnitude of this? And what angle does it enclose with a unit vector along the -y axis?

ehild
 
  • #13
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
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
 
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1. What is a vector cross product?

A vector cross product, also known as a vector product or cross product, is an operation that takes two vectors as input and produces a third vector that is perpendicular to both of the input vectors. It is denoted by the symbol "×" or "x".

2. How is the vector cross product calculated?

The vector cross product is calculated by taking the determinant of a 3x3 matrix composed of the unit vectors i, j, and k and the components of the two input vectors. The resulting vector is in the direction determined by the right-hand rule.

3. What is the physical significance of the vector cross product?

The vector cross product has several physical interpretations, including calculating torque, determining the direction of a magnetic field, and calculating the area of a parallelogram or triangle formed by the two input vectors.

4. What are the properties of the vector cross product?

The vector cross product has several properties, including being non-commutative, distributive, and having a magnitude equal to the product of the magnitudes of the input vectors multiplied by the sine of the angle between them.

5. In what situations is the vector cross product used?

The vector cross product is used in many fields of science and engineering, including physics, mathematics, and computer graphics. It is particularly useful in calculating forces and motion in three-dimensional systems and in determining the direction of electromagnetic fields.

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