Calculating resultant torque using cross product

[steffercakes]

1."In this exercise, you will be finding the resultant torque from the cross product of a lever arm with a force vector. The lever arm vector is A=2.0i+3.0j. The force vector is B=3.0i-4.0j.
Find A x B
B x A
and 2A x 3B



2.My teacher has been sick the past few days so hasnt taught us anything about torque yet, but the homework is still due. I'm not sure where to start because every formula I've read about how to do these says to include sinθ, but an angle was not given.



3. is it just: 6i-12j+1k
-6i+12j-1k
and 36i-72j+6k? Someone please help me understand how these are done.

 

[Hardik Batra]

http://en.wikipedia.org/wiki/Cross_product

This link will help you.
In this page see ...Matrix notation part.

 

[steffercakes]

http://en.wikipedia.org/wiki/Cross_product

This link will help you.
In this page see ...Matrix notation part.

Nevermind! Thanks. :)

 

[CKOMETTER]

1."In this exercise, you will be finding the resultant torque from the cross product of a lever arm with a force vector. The lever arm vector is A=2.0i+3.0j. The force vector is B=3.0i-4.0j.
Find A x B
B x A
and 2A x 3B



2.My teacher has been sick the past few days so hasnt taught us anything about torque yet, but the homework is still due. I'm not sure where to start because every formula I've read about how to do these says to include sinθ, but an angle was not given.



3. is it just: 6i-12j+1k
-6i+12j-1k
and 36i-72j+6k? Someone please help me understand how these are done.

There are two interpretations of the cross-product:

Geometrical Interpretation:

\vec{A}x\vec{B} = |A||B|\sin \theta \hat{n}

Algebraic Interpretation:

\vec{A} = A_x i + A_y j + A_z k
\vec{B} = B_x i + B_y j + B_z k
\vec{A}x\vec{B} = (A_y B_z-A_z B_y)i + (A_z B_x - A_x B_z)j + (A_x B_y - A_y B_x)k

 

[Hardik Batra]

Or you can learn from this video also.

 

[steffercakes]

There are two interpretations of the cross-product:

Geometrical Interpretation:

\vec{A}x\vec{B} = |A||B|\sin \theta \hat{n}

Algebraic Interpretation:

\vec{A} = A_x i + A_y j + A_z k
\vec{B} = B_x i + B_y j + B_z k
\vec{A}x\vec{B} = (A_y B_z-A_z B_y)i + (A_z B_x - A_x B_z)j + (A_x B_y - A_y B_x)k

ohhh okay, I see now. and if there's no Z (z=0), that makes that direction 0? So in this case the i and j are 0?

 

[CKOMETTER]

ohhh okay, I see now. and if there's no Z (z=0), that makes that direction 0? So in this case the i and j are 0?

Yes, the cross-product of two vectors results in a vector that it's perpendicular to the plane that contains both vectors: in this case the Z direction.