anonymous
  • anonymous
Find all vectors u that satisfy the equation <1,1,1> x u = <-3,12>
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
Shouldn´t be (-3,1,2)?
anonymous
  • anonymous
So the cross product of \(\langle 1,1,1\rangle\) and some vector \(\vec{u}\) is \(\langle -3,1,2\rangle\) ?
anonymous
  • anonymous
yes

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anonymous
  • anonymous
|dw:1378085626125:dw|
anonymous
  • anonymous
then equal that to <-3, 1, 2 >
anonymous
  • anonymous
But i can't figure it out
anonymous
  • 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
  • anonymous
Other than \(\vec{u}=\vec{0}\), of course.
anonymous
  • anonymous
so i don't do the cross product?
anonymous
  • anonymous
You could, I'm just considering an alternative method.
anonymous
  • 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
  • anonymous
would that be no solution?
anonymous
  • anonymous
there is a solution
anonymous
  • 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
  • anonymous
Yes, there's at least one.
anonymous
  • anonymous
i can't solve this: i (C-B) + j(A-C) +k(B-A) = -3i + 1j + 2k
anonymous
  • anonymous
so i set up three equations: C-B= -3 A-C= 1 and B-A= 2
anonymous
  • anonymous
The third equation is probably utilizing something like \[ \|\vec a \times \vec b\| = \|\vec a\| \|\vec b\| \sin(\theta) \]
anonymous
  • anonymous
A B and C just represent the x, y, z of vector u
anonymous
  • 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
  • anonymous
@kimmy0394 Each vector component makes an equation.
anonymous
  • 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
  • anonymous
so what am i doing wrong?
anonymous
  • 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
  • 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
  • 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
  • anonymous
apparently the answer is "there are infinitely many solutions of the form u= <1+z, 3+z, z>
anonymous
  • anonymous
and that was with the cross product method so i'm not sure how they got that
anonymous
  • anonymous
i don't understand the 1 and 3
anonymous
  • 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
  • anonymous
how are you getting <1,3,0>???
anonymous
  • anonymous
@kimmy0394, that's just one possible solution. The problem wants you to find every possible solution.
anonymous
  • anonymous
Without factoring in length, \(\langle 0,0,0\rangle\) is a valid solution. That is why your method is incomplete.
anonymous
  • 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
  • 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
  • anonymous
gotcha! thanks

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