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Find k if (4,1,k) & (5,1,-3) are perpendicular?

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Vectors
    find k if (4,1,k)τ & (5,1,-3)τ are perpendicular?

    From the answer sheet I know the answer is k = 7

    2. Relevant equations
    I believe I need these two but I'm not certain:

    Dot-product: v*w = v1w1 + v2w2 + ... vdwd
    cos θ = v*w / ||v|| * ||w||


    3. The attempt at a solution

    [STRIKE]Because θ = arccos(v*w / ||v|| * ||w||) = the angle between two vectors. θ should be 90 = perpendicular.

    90 = arccos ((4 * 5 + 1 * 1 + k * -3) / √(4²+1²+k²) * √(5²+1²+(-3)²)) =
    90 = arccos (21-3k / √(17+k²) * √35)

    90 = arccos(21-3k / √(17+k²) * √35)[/STRIKE]

    cos(90) = 0 = perpendicular

    [itex]\frac{21-3k}{\sqrt{17+k²} * \sqrt{35}} = 0[/itex]

    what would be the easiest way to get to k = 7 ?
     
    Last edited: Oct 2, 2011
  2. jcsd
  3. Oct 2, 2011 #2

    HallsofIvy

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    What you need is that cos(90)= 0! So that two vectors are perpendicular if and only if their dot product is 0.
     
  4. Oct 2, 2011 #3
    ok. so let me reform:

    (21-3k) / √(17+k²) * √35 = 0

    [itex]\frac{21-3k}{\sqrt{17+k²} * \sqrt{35}} = 0[/itex]

    what would be the easiest way to get to k = 7? (when I put 7 as k I get zero, so the equation is correct)
     
    Last edited: Oct 2, 2011
  5. Oct 2, 2011 #4

    HallsofIvy

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    If you already knew that the dot product is 21- 3k, why even use that denominator?

    The two vectors are perpendicular if and only if their dot product, 21- 3k= 0. Solve that for k?
     
  6. Oct 2, 2011 #5
    [STRIKE]Ow yeah. I'm being stupid.[/STRIKE]

    And if two vectors are parallel? [STRIKE]then the angle is, I suppose 180 degrees? What equation would I need to solve in that case?

    (i'm sorry, my teacher is very unclear about these things, he puts questions on his sheets but doesn't always give the answers during class)[/STRIKE]

    edit: cross-product = 0
     
    Last edited: Oct 2, 2011
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