3 Vectors Issuing from a Common Point

[Saladsamurai]

Homework Statement

This one is a good one. This is in the section on cross products:

If A, B, and C are distinct vectors issuing from a common point, show that the vector $(\mathbf{A}\times\mathbf{B})+(\mathbf{B}\times\mathbf{C})+(\mathbf{C}\times\mathbf{A})$ is perpendicular to the plane containing their heads.

Homework Equations

Definition of cross product

The Attempt at a Solution

I am jammed at the start. Can I get just a little hint here? Also, what exactly does it mean for vectors to be distinct? Does it just mean they are not equal to each other? Or that each is not zero?

 

[Mark44]

Homework Statement

This one is a good one. This is in the section on cross products:

If A, B, and C are distinct vectors issuing from a common point, show that the vector $(\mathbf{A}\times\mathbf{B})+(\mathbf{B}\times\mathbf{C})+(\mathbf{C}\times\mathbf{A})$ is perpendicular to the plane containing their heads.

Homework Equations

Definition of cross product

The Attempt at a Solution

I am jammed at the start. Can I get just a little hint here? Also, what exactly does it mean for vectors to be distinct? Does it just mean they are not equal to each other? Or that each is not zero?

It means that none of them is equal to any of the others. It doesn't say that any of them is not zero.

 

[Saladsamurai]

It means that none of them is equal to any of the others. It doesn't say that any of them is not zero.

Ok great. It wasn't that clear to me. Any thoughts on how to start this? I need to establish a way of orienting things. First I need to determine how the plane containig the heads is related to the vectors themselves. Perhaps I should construct 2 more vectors, one that connect the heads of A and B and one that connect the heads of A and C? I'm not sure why yet Just trying to get the gears moving.

 

[Mark44]

You can think of the three vectors as being the supports of the plane that their heads lie in. Yes, construct two more vectors, AB and AC (or BC). The cross product of these vectors will give you the normal to the plane. That vector should be parallel to (a multiple of) the vector (A x B) + (B x C) + (C X A).

 

[Saladsamurai]

Ok. I am just now having time to get back to this one. So we have two new vectors: one from head of A to head of B called AB and one from head of A to head of C called AC. By crossing AB with AC I get

ABxAC = {(AB)_y(AC)_z - (AB)_z(AC)_y}i - {(AB)_x(AC)_z - (AB)_z(AC)_x}j + {(AB)_x(AC)_y - (AB)_y(AC)_x}k.

I know that ABxAC is normal to the plane in which the heads of A, B, and C rest.

Now this should be parallel to $(\mathbf{A}\times\mathbf{B})+(\mathbf{B}\times\mathbf{C})+(\mathbf{C}\times\mathbf{A})$.

If that is the cae, then I can use either cross product or dot product to show that is true. My problem is that I don't think that my expression for ABxAC is very useful since I don't see how it can cancel any terms that would arise of I "dot it" with the vector $(\mathbf{A}\times\mathbf{B})+(\mathbf{B}\times\mathbf{C})+(\mathbf{C}\times\mathbf{A})$ since they don't share any common terms.

Any thoughts on a better approach/naming convention for these vectors?

 

[Mark44]

Ok. I am just now having time to get back to this one. So we have two new vectors: one from head of A to head of B called AB and one from head of A to head of C called AC. By crossing AB with AC I get

ABxAC = {(AB)_y(AC)_z - (AB)_z(AC)_y}i - {(AB)_x(AC)_z - (AB)_z(AC)_x}j + {(AB)_x(AC)_y - (AB)_y(AC)_x}k.

I know that ABxAC is normal to the plane in which the heads of A, B, and C rest.

Now this should be parallel to $(\mathbf{A}\times\mathbf{B})+(\mathbf{B}\times\mathbf{C})+(\mathbf{C}\times\mathbf{A})$.

Yes, that's what you're trying to show.

If that is the cae, then I can use either cross product or dot product to show that is true. My problem is that I don't think that my expression for ABxAC is very useful since I don't see how it can cancel any terms that would arise of I "dot it" with the vector $(\mathbf{A}\times\mathbf{B})+(\mathbf{B}\times\mathbf{C})+(\mathbf{C}\times\mathbf{A})$ since they don't share any common terms.

You don't need either the cross product or the dot product to show that two vectors are parallel. If u and v are parallel, then each one is a scalar multiple of the other. (If they're pointing in opposite directions, the scalar will be negative.)

Any thoughts on a better approach/naming convention for these vectors?