How Do You Calculate the Direction of a Cross Product Vector?

[odie5533]

Homework Statement

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

The Attempt at a Solution

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

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

 

[learningphysics]

I'm guessing they want the unit vector in the direction of \vec{A} \times \vec{B} for direction...

 

[mattakir]

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.

 

[HallsofIvy]

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", \theta, \phi, and \psi, 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 cos(\theta)= a, cos(\phi)= b, and cos(\psi)= c.

 

[FedEx]

Homework Statement

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

The Attempt at a Solution

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

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.