Vector (cross) product: What is the magnitude of the cross product of a vector of magnitude 2.00 m pointing east and a vector of magnitude 4.00 m pointing 30.0° west of north?

Question

Quizplus · Accepted Answer

The magnitude of the cross product of two vectors is given by: 
|A x B| = |A| |B| sin(theta)

where A and B are the vectors and theta is the angle between them.

First, we need to find the vector product of the two given vectors.

Let vector A be the vector of magnitude 2.00 m pointing east. 
Let vector B be the vector of magnitude 4.00 m pointing 30.0° west of north.
Vector A can be written as:
A = 2.00 m (cos 0° i + sin 0° j) = 2.00 m i

Vector B can be written as:
B = 4.00 m (cos 30° (-j) + sin 30° i) = 4.00 m (cos 30° (-j) + sin 30° i) = 4.00 m (-√3 / 2 j + 1 / 2 i)

The cross product of A and B is given by:
A x B = |A| |B| sin(theta) n

where n is the unit vector perpendicular to both A and B. In this case, since A is pointing in the positive x-direction and B is pointing in the second quadrant, the cross product will be in the negative z-direction.

The magnitude of the cross product is then given by:
|A x B| = |2.00 m| |4.00 m| sin(30°) = 4.00 m^2 (1 / 2) = 2.00 m^2

Therefore, the answer is: 
A) 6.93

Vector (Cross) Product: What Is the Magnitude of the Cross