Direction of a 3D Vector

In summary, the given vectors \vec{A} and \vec{B} have magnitudes of \sqrt{44.625m/s} and \sqrt{14.625m/s} respectively. The expression for \vec{A} - \vec{B} is 4.25m/s\hat{_i} + 8.75m/s\hat{_j} - 2.00m/s\vec{_k}. The magnitude of \vec{A} \times \vec{B} is 16.3m/s and its direction can be determined using the right hand rule. Alternatively, the unit vector in the direction of \vec{A} \times \vec{B} can be
  • #1
odie5533
58
0

Homework Statement


Given two vectors [tex]\vec{A}=(4.25m/s)\hat{_i}+(5.00m/s)\hat{_j}-(1.25m/s)\hat{_k}[/tex] and [tex]\vec{B}=-(3.75m/s)\hat{_j} + (0.75m/s)\hat{_k}[/tex], a) find the magnitude of each vector; b) write an expression for [tex]\vec{A} - \vec{B}[/tex]. (c) Find the magnitude and direction of [tex]\vec{A} \times \vec{B}[/tex].

The Attempt at a Solution


a)[tex]|\vec{A}| = \sqrt{44.625m/s}[/tex]
[tex]|\vec{B}| = \sqrt{14.625m/s}[/tex]
b)[tex]\vec{A} - \vec{B} = 4.25m/s\hat{_i} + 8.75m/s\hat{_j} - 2.00m/s\vec{_k}[/tex]
c)[tex]\vec{A} \times \vec{B} = (4.25m/s\hat{_i} + 5.00m/s\hat{_j} - 1.25m/s\hat{_k}) \times (-3.75m/s\hat{_j} + 0.75m/s\hat{_k})[/tex]
[tex]\vec{C} = \vec{A} \times \vec{B}[/tex]
[tex]\vec{C}_x = (5.00m/s \times 0.75m/s) - (-1.25m/s \times -3.75m/s) = -0.9375m/s[/tex]
[tex]\vec{C}_y = (-1.25m/s \times 0) - (4.25m/s \times 0.75m/s) = -3.1875m/s[/tex]
[tex]\vec{C}_z = (4.25m/s \times -3.75m/s) - (5.00m/s \times 0) = -15.9m/s[/tex]
[tex]\vec{C} = -0.9375m/s\hat{_i} - 3.1875m/s\hat{_j} - 15.9m/s\hat{_k}[/tex]
[tex]|\vec{A} \times \vec{B}| = |\vec{C}| = \sqrt{(-0.9375m/s)^2 + (-3.1875m/s)^2 + (-15.9m/s)^2} = 16.3m/s[/tex]
Direction of [tex]\vec{A} \times \vec{B}[/tex] = ?

I'm not sure how to define the direction of a 3D vector is what it really comes down to.
 
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  • #2
I'm guessing they want the unit vector in the direction of [tex]\vec{A} \times \vec{B}[/tex] for direction...
 
  • #3
The direction of A x B can be determined by using right hand rule. Use all 4 fingers as vector A, then make a 90 degree for the direction of vector B. The thumb will show you the vector C direction.
 
  • #4
As mattakir said, the cross product of two vectors is perpendicular to both in the direction of the "right hand rule".

As learningphysics said, you can also show direction by taking the unit vector in the direction of the cross product vector. You can also use the "direction angles", [itex]\theta[/itex], [itex]\phi[/itex], and [itex]\psi[/itex], the angles the vector makes with the x-axis, y-axis, and z-axis respectively. If ai+ bj+ ck is a unit vector, the "direction cosines" are given by [itex]cos(\theta)= a[/itex], [itex]cos(\phi)= b[/itex], and [itex]cos(\psi)= c[/itex].
 
  • #5
odie5533 said:

Homework Statement


Given two vectors [tex]\vec{A}=(4.25m/s)\hat{_i}+(5.00m/s)\hat{_j}-(1.25m/s)\hat{_k}[/tex] and [tex]\vec{B}=-(3.75m/s)\hat{_j} + (0.75m/s)\hat{_k}[/tex], a) find the magnitude of each vector; b) write an expression for [tex]\vec{A} - \vec{B}[/tex]. (c) Find the magnitude and direction of [tex]\vec{A} \times \vec{B}[/tex].

The Attempt at a Solution


a)[tex]|\vec{A}| = \sqrt{44.625m/s}[/tex]
[tex]|\vec{B}| = \sqrt{14.625m/s}[/tex]
b)[tex]\vec{A} - \vec{B} = 4.25m/s\hat{_i} + 8.75m/s\hat{_j} - 2.00m/s\vec{_k}[/tex]
c)[tex]\vec{A} \times \vec{B} = (4.25m/s\hat{_i} + 5.00m/s\hat{_j} - 1.25m/s\hat{_k}) \times (-3.75m/s\hat{_j} + 0.75m/s\hat{_k})[/tex]
[tex]\vec{C} = \vec{A} \times \vec{B}[/tex]
[tex]\vec{C}_x = (5.00m/s \times 0.75m/s) - (-1.25m/s \times -3.75m/s) = -0.9375m/s[/tex]
[tex]\vec{C}_y = (-1.25m/s \times 0) - (4.25m/s \times 0.75m/s) = -3.1875m/s[/tex]
[tex]\vec{C}_z = (4.25m/s \times -3.75m/s) - (5.00m/s \times 0) = -15.9m/s[/tex]
[tex]\vec{C} = -0.9375m/s\hat{_i} - 3.1875m/s\hat{_j} - 15.9m/s\hat{_k}[/tex]
[tex]|\vec{A} \times \vec{B}| = |\vec{C}| = \sqrt{(-0.9375m/s)^2 + (-3.1875m/s)^2 + (-15.9m/s)^2} = 16.3m/s[/tex]
Direction of [tex]\vec{A} \times \vec{B}[/tex] = ?

I'm not sure how to define the direction of a 3D vector is what it really comes down to.

If it is about the direction of A X B. Then you can find it by the right hand rule.
But if they are asking the unit vector in the direction of A X B. Then first of all find A(vector) X B(vector) which you have already found out easily. Then find the magnitude of A X B which you have found with equal ease. Now unit vector in the direction of A X B is A vector divided by the magnitude of A vector.
 

1. What is the direction of a 3D vector?

A 3D vector is a mathematical object that has both magnitude and direction in three-dimensional space. The direction of a 3D vector is determined by the coordinates of its endpoint, which represents the vector's direction relative to the origin.

2. How is the direction of a 3D vector represented?

The direction of a 3D vector can be represented using a variety of conventions, such as spherical coordinates, cylindrical coordinates, or unit vectors. The most common representation is the use of unit vectors, which are vectors with a magnitude of 1 that point in a specific direction.

3. How do you find the direction of a 3D vector?

The direction of a 3D vector can be found by taking the vector's coordinates and normalizing them to create a unit vector. This unit vector will have the same direction as the original vector, but with a magnitude of 1. Alternatively, the direction of a 3D vector can also be described using angles, such as azimuth and inclination, or by using the dot product with another vector to find the angle between them.

4. Can a 3D vector have multiple directions?

No, a 3D vector can only have one direction at a time. However, it is possible for a 3D vector to change direction if its coordinates are altered or if it is transformed by a rotation matrix. Additionally, a 3D vector can have different directions relative to different reference frames.

5. How does the direction of a 3D vector affect its properties?

The direction of a 3D vector affects its properties, such as its dot product and cross product with other vectors, as well as its angle with other vectors. Additionally, the direction of a 3D vector can determine the direction of its resultant vector when added to another vector. In certain applications, the direction of a 3D vector may also be used to represent physical quantities, such as force or velocity.

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