1. ## Cross product problem

Hello,
I'm facing problems while trying to solve this problem

we have u and v and we need to find b that satisfies the following equation:

u x b = v

I've tried to solve it and i got only one coordinates for B and it we should get infinity number of vectors satisfies this equation up there

2. ## Re: Cross product problem

Originally Posted by ahmedzoro10
Hello,
I'm facing problems while trying to solve this problem

we have u and v and we need to find b that satisfies the following equation:

u x b = v

I've tried to solve it and i got only one coordinates for B and it we should get infinity number of vectors satisfies this equation up there
it would be helpful to know what u and v are ...

3. ## Re: Cross product problem

Originally Posted by skeeter
it would be helpful to know what u and v are ...
u= [-1 3 2] and v= [1 1 -1]

4. ## Re: Cross product problem

if $\displaystyle b = (b_1,b_2,b_3)$ then $\displaystyle u \times b = v$ means that $\displaystyle (3b_3 - 2b_2,2b_1+b_3,-b_2-3b_1) = (1,1,-1)$.

this is a system of 3 linear equations in 3 unknowns:

$\displaystyle \begin{bmatrix}0&-2&3\\2&0&1\\-3&-1&0\end{bmatrix} \begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix} = \begin{bmatrix}1\\1\\-1\end{bmatrix}$

can you continue?

5. ## Re: Cross product problem

Originally Posted by Deveno
if $\displaystyle b = (b_1,b_2,b_3)$ then $\displaystyle u \times b = v$ means that $\displaystyle (3b_3 - 2b_2,2b_1+b_3,-b_2-3b_1) = (1,1,-1)$.

this is a system of 3 linear equations in 3 unknowns:

$\displaystyle {\color{red}\begin{bmatrix}0&-2&3\\2&0&1\\-3&-1&0\end{bmatrix}} \begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix} = \begin{bmatrix}1\\1\\-1\end{bmatrix}$
Be careful:
That matrix is singular.

You might try rref.

6. ## Re: Cross product problem

Originally Posted by Plato
Be careful:
That matrix is singular.

You might try rref.
I got it . Thanks alot

7. ## Re: Cross product problem

I got A^(-1) = 0! is that right!
I've used adjoint method getting the inverse of A

8. ## Re: Cross product problem

Originally Posted by ahmedzoro10
I got A^(-1) = 0! is that right!
I've used adjoint method getting the inverse of A
No, that is not right. Plato told you that A does not have an inverse.

(And no invertible matrix has 0 as its inverse.)

9. ## Re: Cross product problem

Originally Posted by HallsofIvy
No, that is not right. Plato told you that A does not have an inverse.

(And no invertible matrix has 0 as its inverse.)
oh sorry I haven't noticed his post saying "this matrices is singular"

10. ## Re: Cross product problem

b will equal [ t , 1-3t , 1-2t ] assuming that b1 = t ,right?

11. ## Re: Cross product problem

You certainly can do that. In particular, If you multiply the last equation by 2 and subtract it from the first equation, you get $\displaystyle 6b_1+ 3b_3= 3$. From the second equation, $\displaystyle b_3= 1- 2b_1$ so $\displaystyle 6b_1+ 3b_3= 6b_1+3- 6b_1= 3$ which is true for all x. Taking $\displaystyle b_1= t$ gives $\displaystyle b2= 1- 3t$ and $\displaystyle b_3= 1- 2t$, exactly what you give above.