# Cross product problem

• December 14th 2011, 12:54 PM
ahmedzoro10
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
• December 14th 2011, 02:34 PM
skeeter
Re: Cross product problem
Quote:

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 ...
• December 14th 2011, 11:51 PM
ahmedzoro10
Re: Cross product problem
Quote:

Originally Posted by skeeter
it would be helpful to know what u and v are ...

u= [-1 3 2] and v= [1 1 -1]
• December 15th 2011, 08:13 AM
Deveno
Re: Cross product problem
if $b = (b_1,b_2,b_3)$ then $u \times b = v$ means that $(3b_3 - 2b_2,2b_1+b_3,-b_2-3b_1) = (1,1,-1)$.

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

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

can you continue?
• December 15th 2011, 08:22 AM
Plato
Re: Cross product problem
Quote:

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

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

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

Be careful:
That matrix is singular.

You might try rref.
• December 15th 2011, 09:10 AM
ahmedzoro10
Re: Cross product problem
Quote:

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

You might try rref.

I got it . Thanks alot :)
• December 15th 2011, 10:18 AM
ahmedzoro10
Re: Cross product problem
I got A^(-1) = 0! is that right!
I've used adjoint method getting the inverse of A
• December 15th 2011, 10:41 AM
HallsofIvy
Re: Cross product problem
Quote:

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.)
• December 15th 2011, 10:55 AM
ahmedzoro10
Re: Cross product problem
Quote:

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" :)
• December 15th 2011, 11:11 AM
ahmedzoro10
Re: Cross product problem
b will equal [ t , 1-3t , 1-2t ] assuming that b1 = t ,right?
• December 15th 2011, 11:38 AM
ahmedzoro10
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 $6b_1+ 3b_3= 3$. From the second equation, $b_3= 1- 2b_1$ so $6b_1+ 3b_3= 6b_1+3- 6b_1= 3$ which is true for all x. Taking $b_1= t$ gives $b2= 1- 3t$ and $b_3= 1- 2t$, exactly what you give above.