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quasar987

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I have this problem that reads

a) Prove that the three-product (

With some effort, I did that.

b) Use this result to prove the following identity geometrically:

I proved the identity geometrically and then I argued that both members were the same in magnitude because geometrically they both represented by parallelepiped of equal edged and all parallelepiped of equal edges have equal volume, hence...etc.

But the sign part, I can't find the trick.

The identity gives |ABCsin(x)cos(y)| = |ABCsin(w)cos(z)| where x is the angle between

All of these angles are between 0 and 180°, which means sin(x) and sin(w) are always positive and the negative sign comes from y and z being between 90° and 180°.

If I could show that when one angle on the cos is between 90° and 180°, so is the other it would be done, but I can't find a relationship between any of the 4 above angles except that sin(x)|cos(y)| = sin(w)|cos(z)|.

If you have any idea, let me know. Big thanks!

a) Prove that the three-product (

**A.**(**B x C**)) of the vectors**A**,**B**and**C**where**A**,**B**and**C**are not lying in a single plane, is the volume of the parallelepiped whose edges are**A**,**B**and**C**with positive or negative sign according to wheter a right-hand screw rotated from**A**toward**B**would advance along**C**in the positive or negative direction.With some effort, I did that.

b) Use this result to prove the following identity geometrically:

**A.**(**B x C**) = (**A x B**)**.C**. Verify that the right and left members of the identity are equal in sign as well as in magnitude.I proved the identity geometrically and then I argued that both members were the same in magnitude because geometrically they both represented by parallelepiped of equal edged and all parallelepiped of equal edges have equal volume, hence...etc.

But the sign part, I can't find the trick.

The identity gives |ABCsin(x)cos(y)| = |ABCsin(w)cos(z)| where x is the angle between

**B**and**C**, y the angle between**A**and**B x C**, w the angle between**A**and**B**and z the angle between**C**and**A x B**.All of these angles are between 0 and 180°, which means sin(x) and sin(w) are always positive and the negative sign comes from y and z being between 90° and 180°.

If I could show that when one angle on the cos is between 90° and 180°, so is the other it would be done, but I can't find a relationship between any of the 4 above angles except that sin(x)|cos(y)| = sin(w)|cos(z)|.

If you have any idea, let me know. Big thanks!

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