Cross Product and Vector Multiplication

If i have the following 4 vectors:

D=Ax(B*C)

How would I solve for C?

Try and make this a tab bit more clear. I have A is a 1x3 matrix, B is a 3x3 matrix C is a 3x1 matrix and D is a 1x3 matrix. I am trying to solve for C. The problem is stated as A cross the product B*C equals D.

[Azzi Abdelmalek]

What are your known parameters ?

[the cyclist]

Can you please verify that in MATLAB syntax,

D = cross(A,B*C)

?

[Melissa]

Its a stiffness and force equation for displacement. So D is reaction force of [Fx,Fy,Fz]. A is the position vector [x;y;d]. B is the transformation to global stiffness [3x3 DCM] and C is the displacement [x.y.z]

[Melissa]

>> A=[1 2 3]; B=[1 2 3; 4 5 6; 7 8 9]; C=[4; 5; 6];
>> D=cross(A,B*C)

D =

      13   -26    13

Answer 1

[Roger Stafford]

 C = [ cross(A',B(:,1)) , cross(A',B(:,2)) , cross(A',B(:,3)) ]\(D');

[Melissa]

When attempting to use your code I receive:

C =

   NaN
   NaN
   NaN

[Roger Stafford]

I was overly hasty. It is true that if

 D=cross(A,B*C)

then

 [cross(A',B(:,1)),cross(A',B(:,2)),cross(A',B(:,3))] * C = D'

However this matrix

 [cross(A',B(:,1)),cross(A',B(:,2)),cross(A',B(:,3))]

is inherently singular, that is, its determinant is zero. In other words the equations you can obtain from

 D=cross(A,B*C)

have infinitely many solutions for C. You cannot determine C uniquely from A, B, and D.

To visualize this better, consider this equation

 Z = cross(X,Y)

where X and Y are vectors. If you are then given Z and X, there will obviously be infinitely many Y vectors that will satisfy the above equation. Any Y which is orthogonal to Z and has the appropriate component orthogonal to the X direction will be a solution. The same applies to your situation. You cannot determine C from A, B, and D.

[Melissa]

That makes sense. I went through the properties of cross product and multiplication and arrived at your equation but since it is singular then I cannot obtain C. Dang. Thank you for your time and explanation.

[Roger Stafford]

There is a way to obtain a unique value for C that satisfies your equation. Let C be given by:

 C = B\(cross(D',A')/dot(A,A));

This is the unique C such that

 D' = cross(A',B*C)

and

 dot(B*C,A') = 0

That is, such that B*C is orthogonal to A'. It assumes that D is orthogonal to A and that B is non-singular.

Answer 2

[joseph agno]

• Find the Matlab command to carry out the cross product of two vectors b and c and try it out on two vectors.

Answer 3

[muhammad asif]

if n corss n = ny

how to write cross in symbol notation form in matlab

[Nitin Phadkule]

cross(n,n)

Answer 4

[muhammad asif]

circle statement

[Steven Lord]

Do you mean n times n? If so use the * operator.

But please do not put your own files in any subdirectory under the matlabroot directory!