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windsylph

  • 3 years ago

: Linear Algebra: Describe the intersection of the two hyperplanes x1 + 2x2 - x3 + x4 = 8 and -x1 + 3x2 + x3 - x4 = 2.

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  1. victorarana
    • 3 years ago
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    They intersecate on a plane

  2. windsylph
    • 3 years ago
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    Is there a way to get an equation for that plane?

  3. tdabboud
    • 3 years ago
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    You could set up the augmented matrix, row reduce to get the values and that should be the point of intersection

  4. windsylph
    • 3 years ago
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    yep I did that.. and haha seemingly im stuck on a dumb question: let a and b be free variables. my solution then is (a-b+4, 2, a, b)^T. How do I express that as a span, then as an equation?

  5. tdabboud
    • 3 years ago
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    i get x2 = 2; x3 = x1 + x4 - 4; So like you said x1 and x4 are free so the span is: =(2 , -4) + x1(1, 0, 1, 0) + x4(0, 0, 1, 1) (these are vertical vectors I just cant write it on here) =(2, -4) + x1u + x4v (u and v are vectors above)

  6. tdabboud
    • 3 years ago
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    understand?

  7. tdabboud
    • 3 years ago
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    I made a small mistake: It is x3 and x4 that are free so it becomes x1=4 +x3 - x4 x2 =2 which translates to... =(2 , -4) + x3(1, 0, 1, 0) + x4(-1, 0, 0, 1)

  8. tdabboud
    • 3 years ago
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    (4, 2, 0, 0) sorry

  9. windsylph
    • 3 years ago
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    thanks, i was looking for an answer of the form span(u,v), but if that's that then that's okay. how about an equation for the plane?

  10. tdabboud
    • 3 years ago
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    well span(u,v) is just the span of those vectors: span{ (1, 0, 1, 0) , (-1, 0, 0, 1)}

  11. windsylph
    • 3 years ago
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    oh, so (4,2,0,0)^T doesn't matter when expressing it in that form?

  12. tdabboud
    • 3 years ago
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    nope because it only depends on the free variables

  13. windsylph
    • 3 years ago
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    okay thanks. is there a way to express that as an equation?

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