# I need to figure this out: (A×B)⋅C

## Homework Statement

Calculate (A×B)⋅C for the three vectors A with magnitude A = 5.00 and angle θA = 25.1∘ measured in the sense from the +x - axis toward the +y - axis, B with B = 4.18 and θB = 62.0∘, and C with magnitude C = 5.82 and in the +z - direction. Vectors A and B are in the xy-plane.

## Homework Equations

Magnitude of (AXB)=A*B*sin(theta)
Magnitude of (A dot B)=A*B*cos(theta)

## The Attempt at a Solution

To be quite honest, I wasn't sure where to start. I know that A and B are 36.9 degrees apart and C is perpendicular to both of them, but I don't know where to go after this...Please help me!

Quantum Defect
Homework Helper
Gold Member

## Homework Statement

Calculate (A×B)⋅C for the three vectors A with magnitude A = 5.00 and angle θA = 25.1∘ measured in the sense from the +x - axis toward the +y - axis, B with B = 4.18 and θB = 62.0∘, and C with magnitude C = 5.82 and in the +z - direction. Vectors A and B are in the xy-plane.

## Homework Equations

Magnitude of (AXB)=A*B*sin(theta)
Magnitude of (A dot B)=A*B*cos(theta)

## The Attempt at a Solution

To be quite honest, I wasn't sure where to start. I know that A and B are 36.9 degrees apart and C is perpendicular to both of them, but I don't know where to go after this...Please help me!

Remember that the cross product gives you a vector, and remember to keep your left hand in your pocket!

I think I am supposed to use trig to figure out the vector components, but I never got the hang of trigonometry...Could I get help with solving?

Quantum Defect
Homework Helper
Gold Member
I think I am supposed to use trig to figure out the vector components, but I never got the hang of trigonometry...Could I get help with solving?

The equation that you give for the cross product is correct for the magnitude. (theta = angle between vectors A and B). The result of a cross product is a vector, however, which has magnitude and direction.

I'll try to figure out the vectors for each...it might take a while.

berkeman
Mentor
I'll try to figure out the vectors for each...it might take a while.

Do you know how to convert from the "polar" representation of vectors (which is what you were given) to the "rectangular" representation? You use the rectangular representation to do the cross product more easily...

haruspex
Homework Helper
Gold Member
convert from the "polar" representation
No, that's quite unnecessary here. Much easier to stay in polar for this one.
Magnitude of (AXB)=A*B*sin(theta)
I know that A and B are 36.9 degrees apart
Right, so what is theta in that cross product equation?
The next challenge is to figure out the direction of AxB. There are two parts to this.
In general, if I take the cross product of two vectors to get a third, a x b = c, what is the angle between a and c?
The second part is to get the sign right. This is tricky because a x b is equal and opposite to b x a. As Quantum Defect indicated, you need to use the 'right hand rule'. (This is because the answer is governed by a convention and is part of the definition of cross product. It could have been defined the other way and all of the mathematics would still work, but it's necessary to have an agreed definition and the right hand rule is it.)

berkeman