Vector/cross product problem

[Syringer]

"a = <1,2,3> and b = <1,-1,-1> Sketch the collection of all position vectors c satisfying a x b = a x c"

One obvious choice for c would be b. Are there any other choices though? I've tried to find the cross product of a x b, and set the x, y, and z coords equal to a x c..using subscripts for c, but there aren't any solutions for that system of equation. So is b = c the only solution for the problem?

 

[zzzz]

2c-3b=1
c-3a=-4
b-2a=-3

solve for a,b and c.
One solution is (1, -1, -1)
see if there are others.

 

[Syringer]

Originally posted by: zzzz
2c-3b=1
c-3a=-4
b-2a=-3

solve for a,b and c.
One solution is (1, -1, -1)
see if there are others.

Yeah I tried using the same system of equations..but there's no solution. I'm guessing <1,-1,-1> is the only one then right?

 

[speg]

Wouldn't the middle equation be: 3a+c=4?

 

[Syringer]

Originally posted by: speg
Wouldn't the middle equation be: 3a+c=4?

He multiplied both sides by -1, thereby reversing the left side..

 

[speg]

Edit: I fixed myself. Don't mind me 😛

Okay so all vectors k[1,-1,-1] {k|keR} Would satisfy that would they not?

 

[Syringer]

Originally posted by: speg
Edit: I fixed myself. Don't mind me 😛

Okay so all vectors k[1,-1,-1] {k|keR} Would satisfy that would they not?

Well then that would satisfy

a x k*b = a x k*c

but it wouldn't really work for

a x b = a x k*c..

 

[Armitage]

Originally posted by: Syringer
"a = <1,2,3> and b = <1,-1,-1> Sketch the collection of all position vectors c satisfying a x b = a x c"

One obvious choice for c would be b. Are there any other choices though? I've tried to find the cross product of a x b, and set the x, y, and z coords equal to a x c..using subscripts for c, but there aren't any solutions for that system of equation. So is b = c the only solution for the problem?

The solution to a x b is perpendicular to a and b.
As long as a and b are not colinear, vector a and b define a plane.
Any vector that lies in that plane that is not colinear with a satisfies the equation for c.
Any vector in the plane defined by a and b can be expressed as a linear combination of a and b,
so we can say that c = D a + E b where D and E are constants and c != +-a

 

[Syringer]

Originally posted by: ergeorge

Originally posted by: Syringer
"a = <1,2,3> and b = <1,-1,-1> Sketch the collection of all position vectors c satisfying a x b = a x c"

One obvious choice for c would be b. Are there any other choices though? I've tried to find the cross product of a x b, and set the x, y, and z coords equal to a x c..using subscripts for c, but there aren't any solutions for that system of equation. So is b = c the only solution for the problem?

The solution to a x b is perpendicular to a and b.
As long as a and b are not colinear, vector a and b define a plane.
Any vector that lies in that plane that is not colinear with a satisfies the equation for c.
Any vector in the plane defined by a and b can be expressed as a linear combination of a and b,
so we can say that c = D a + E b where D and E are constants and c != +-a

Hmm, I tried multiplying a and b by a constant and then adding them..but those don't seem to work. The only one that does work though is simply a + b..by multiplying by the constants and then taking the cross product of "a x c", the result is a vector that is a multiple of a x b..even if the constants are the same.

 

[Syringer]

Actually, if you just multiply "a" by a constant, and add it to just "b" to get "c", then it actually works. Cool.

 

[Armitage]

Originally posted by: Syringer
Actually, if you just multiply "a" by a constant, and add it to just "b" to get "c", then it actually works. Cool.

Ok, I did leave out one step as an exercise for the reader 😀
c = D a + E b will give you the right direction (try normalizing the results), but you will need to solve for D and E such that the magnitudes match.