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anonymous

  • 5 years ago

this question is crushing my brain space!!! . Consider the vectors p = xi + 5j + yk and q = 2i 4j + 3k. a) For what values of x and y are the vectors p and q parallel? b) If x = 7 for what value of y are the vectors p and q perpendicular? Given that x and y takes these values nd a third vector that is perpendicular to both pand q.

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  1. anonymous
    • 5 years ago
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    Which parts are confusing to you?

  2. anonymous
    • 5 years ago
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    Did you attempt anything so far?

  3. anonymous
    • 5 years ago
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    i did, and got something with a) if parallel then they are scalar multiples of each other... so ap=q then axi = 2j a5=-4 ay= 3 q=(2 -4 3) p= (x 5 y) a=-4/5 x = -0.4 y=-4/15

  4. anonymous
    • 5 years ago
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    does that mean i have done a) and then im stuck on b?

  5. anonymous
    • 5 years ago
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    a is not quite right.

  6. anonymous
    • 5 years ago
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    wait, is the q vector 2i -4j + 3k ?

  7. anonymous
    • 5 years ago
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    yes

  8. anonymous
    • 5 years ago
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    Oh, then yes that's right

  9. anonymous
    • 5 years ago
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    So for b) you know that the dot product of two vectors is 0 if and only if they are perpendicular. So if they are perpendicular you know that the dot product of the two must be 0.

  10. anonymous
    • 5 years ago
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    That should give you another nice system you can solve.

  11. anonymous
    • 5 years ago
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    haha ok, i might be back to confirm in a few minutes (maybe 10) :)

  12. anonymous
    • 5 years ago
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    does that mean i should be able to find j and k directions' magnitude when x=-7 by comparing the other values i got previously?

  13. anonymous
    • 5 years ago
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    No, b has nothing to do with a.

  14. anonymous
    • 5 years ago
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    just plug in the given value for x and solve \[p \cdot q = 0\]

  15. anonymous
    • 5 years ago
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    oh no, so i'm meant to do this?: -7i+5j +yk . q... cos 90 = 0

  16. anonymous
    • 5 years ago
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    no.

  17. anonymous
    • 5 years ago
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    The dot product of two vectors: \[<a,b,c> \cdot <e,f,g>\ = ae + bf + cg\]

  18. anonymous
    • 5 years ago
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    In your case you have x=7 so: p = <7,5,y> q = <2,-4,3> \(\implies p \cdot q = 14 - 20 + 3y = 0\)

  19. anonymous
    • 5 years ago
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    oh, turns out i didn't know how to use dot product nor do i understand it.

  20. anonymous
    • 5 years ago
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    What class is this for?

  21. anonymous
    • 5 years ago
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    http://www.maths.uq.edu.au/courses/MATH1050/

  22. anonymous
    • 5 years ago
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    This is something you probably would have covered in a linear algebra class or some other introduction to vectors

  23. anonymous
    • 5 years ago
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    The dot product of two vectors is the sum of the product of their components.

  24. anonymous
    • 5 years ago
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    gah sorry, i have no memory of doing vectors in high school, and this is 8 years later now

  25. anonymous
    • 5 years ago
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    so y should be 11.3 recurring?

  26. anonymous
    • 5 years ago
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    np. Basically you just multiply each component pair, and add all those products. So if p = i + 2j + 3k and q = 4i + 3j - 2k Then \(p \cdot q = (1\cdot4) + (2\cdot 3) + (3 \cdot -2) = 4 + 6 -6 = 4\) The notable thing about the dot product is that if the two vectors are orthogonal (perpendicular) then the dot product will equal 0.

  27. anonymous
    • 5 years ago
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    If 14−20+3y=0 then 3y = 6 so y = 2

  28. anonymous
    • 5 years ago
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    oh right, but you made x=-7 into x=7 there :)

  29. anonymous
    • 5 years ago
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    Well your original post said x = 7.

  30. anonymous
    • 5 years ago
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    I can't help it if I solve the problem you give me ;p

  31. anonymous
    • 5 years ago
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    so it does, but thats not what the actual question says on the pdf i copied :) damn

  32. anonymous
    • 5 years ago
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    Ok well then it'd be -14 - 20 + 3y = 0 so 3y = 34 so y = 34/3

  33. anonymous
    • 5 years ago
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    so to finish the question i have ot take the cross product of q and latest version of p?

  34. anonymous
    • 5 years ago
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    Yes. That will give you a vector that is orthogonal to each. You recall how to take the cross product?

  35. anonymous
    • 5 years ago
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    add the sum of the products of the components of the two vectors multiplied by the sine of the angle between them ... but that doesnt seem like it'll work cause sine 0 is 0 so i will end up with 0 which doesn't help

  36. anonymous
    • 5 years ago
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    the angle between them is 90 actually

  37. anonymous
    • 5 years ago
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    damn

  38. anonymous
    • 5 years ago
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    0 would mean they were parallel.

  39. anonymous
    • 5 years ago
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    thats cool then, now ill only have 4 more questions like that to get through today

  40. anonymous
    • 5 years ago
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    =)

  41. anonymous
    • 5 years ago
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    so just to make sure im not making more silly mistakes, is the new vector for b) -14i - 20j + 34k ?

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