Find all vectors u that satisfy the equation
<1,1,1> x u = <-3,12>

- anonymous

Find all vectors u that satisfy the equation
<1,1,1> x u = <-3,12>

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- anonymous

Shouldn´t be (-3,1,2)?

- anonymous

So the cross product of \(\langle 1,1,1\rangle\) and some vector \(\vec{u}\) is \(\langle -3,1,2\rangle\) ?

- anonymous

yes

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## More answers

- anonymous

|dw:1378085626125:dw|

- anonymous

then equal that to <-3, 1, 2 >

- anonymous

But i can't figure it out

- anonymous

Given \(a\times b=c\), you know that \(c\) is orthogonal to \(a\) and \(b\), right?
Two vectors are orthogonal if their dot product is 0. This is obviously the case for \(\langle1,1,1\rangle\) and \(\langle-3,1,2\rangle\).
If you can find some combo of \(u_1,u_2,u_3\) that makes \(\vec{u}\cdot\langle-3,1,2\rangle=0\), then you have another way of figuring this problem out.

- anonymous

Other than \(\vec{u}=\vec{0}\), of course.

- anonymous

so i don't do the cross product?

- anonymous

You could, I'm just considering an alternative method.

- anonymous

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
If you can find some combo of \(u_1,u_2,u_3\) that makes \(\vec{u}\cdot\langle-3,1,2\rangle=0\), then you have another way of figuring this problem out.
\(\color{blue}{\text{End of Quote}}\)
Also consider \(\vec{u}\cdot\langle1,1,1\rangle=0\)

- anonymous

would that be no solution?

- anonymous

there is a solution

- anonymous

That gives 2 equations with three variables. You'd need a third equation.
For now, I'd just try using the cross product.

- anonymous

Yes, there's at least one.

- anonymous

i can't solve this: i (C-B) + j(A-C) +k(B-A) = -3i + 1j + 2k

- anonymous

so i set up three equations: C-B= -3 A-C= 1 and B-A= 2

- anonymous

The third equation is probably utilizing something like \[
\|\vec a \times \vec b\| = \|\vec a\| \|\vec b\| \sin(\theta)
\]

- anonymous

A B and C just represent the x, y, z of vector u

- anonymous

now it's just systems of equation. it shouldn't be hard i'm just struggling to isolate the variables. i keep running into something like 2= B +2 - B

- anonymous

@kimmy0394 Each vector component makes an equation.

- anonymous

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
If you can find some combo of \(u_1,u_2,u_3\) that makes \(\vec{u}\cdot\langle-3,1,2\rangle=0\), then you have another way of figuring this problem out.
\(\color{blue}{\text{End of Quote}}\)
This means \(-3u_1+u_2+2u_3=0\). You just have to figure out some \(u_1,u_2,u_3\) that make this equation work.
For example, if you let \(\vec{u}=\langle1,3,0\rangle\), then indeed you have \(\langle1,1,1\rangle\times\vec{u}=\langle-3,1,2\rangle\).

- anonymous

so what am i doing wrong?

- anonymous

In my opinion, although your method might be right, I would avoid using it. Too much work, but it'd be good to practice finding cross products.

- anonymous

\[\begin{align*}\langle{1,1,1}\rangle\times\langle{a,b,c}\rangle&=\langle-3,1,2\rangle\\
\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&1&1\\a&b&c\end{vmatrix}&=\\
(c-b)\vec{i} -(c-a)\vec{j}+(b-a)\vec{k}&=\end{align*}\]
giving you
\[\begin{cases}c-b=-3\\a-c=1\\b-a=2\end{cases}\]
The system has multiple solutions. Like my method suggests, it might be easier to fix values of one of the variables and solve for the other two.

- anonymous

@SithsAndGiggles You have one equation and 3 variables.
You can get another equation by using the fact that: \[
\|\vec a \times \vec b \| = \|\vec a\|\|\vec b\|\sin(\theta)
\]Which would give you one free variable.

- anonymous

apparently the answer is "there are infinitely many solutions of the form u= <1+z, 3+z, z>

- anonymous

and that was with the cross product method
so i'm not sure how they got that

- anonymous

i don't understand the 1 and 3

- anonymous

@wio I don't think that would be necessary.
@kimmy0394, exactly, and what happens when \(z=0\)? You have \(\vec{u}=\langle1,3,0\rangle!\).

- anonymous

how are you getting <1,3,0>???

- anonymous

@kimmy0394, that's just one possible solution. The problem wants you to find every possible solution.

- anonymous

Without factoring in length, \(\langle 0,0,0\rangle\) is a valid solution. That is why your method is incomplete.

- anonymous

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
\[\begin{cases}c-b=-3\\a-c=1\\b-a=2\end{cases}\]
\(\color{blue}{\text{End of Quote}}\)
Let \(c=z\), then (from the second equation) \(a=1+z\) and (from the first) \(b=z+3\), so \(\vec{u}=\langle a,b,c\rangle=\langle1+z,z+3,z\rangle\).

- anonymous

This is a generalization of what I did to find (1,3,0). Here you "fix" one of the variables, then express the remaining ones in terms of this "fixed" variable.

- anonymous

gotcha! thanks

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