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Math Help - Vectors help!

  1. #1
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    Vectors help!

    Hello. I'm currently having some problems on this question on vectors. If you could help me, it'll be great.

    -----

    Vector AC is [2,4,-3] and it forms the hypotenuse of a right angled triangle of trangle ABC. Vector AB is parallel to vector [1,1,0]. Find AB.



    Well, I've done a little diagram to show you what it looks like. For my working out, I deduced that vector AB would be [k,k,0] for some scalar k because it's parallel to [1,1,0]. Is that the right way to go about it?

    First, I tried finding the angle between A and C. I managed to get Cosθ = 6k/squareroot(58k). From there, I deduced that AB = 6k and AC = sqaureroot(58k). However, that doesn't really seem to help me much at all.

    I also tried using vector subtraction so vector BC = vector AC - vecor AB but I'm not too sure how to go about that. Also, right-angled triangles seem to indicate Pythagoras so maybe that has something to do with it?

    Anyway, can somebody please show me how to solve this problem or perhaps provide a starting point?

    -----

    Thankyou. Any help is of course, appreciated.
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  2. #2
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    \overrightarrow {BC}  = \overrightarrow {AC}  - \overrightarrow {AB}

    \left\| {\overrightarrow {AC} } \right\|^2  = \left\| {\overrightarrow {AB} } \right\|^2  + \left\| {\overrightarrow {BC} } \right\|^2
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  3. #3
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    Alright, I managed to get BC as [2-k,4-k,-3].

    From there, I used Pythagoras like you mentioned:

    (√29) = (√2k) + (√2k - 12k + 29)
    29 = 2k + 2k - 12k + 29
    4k - 12k = 0
    k - 3k = 0

    k = 0 & 3

    So therefore, my answer is 3 meaning vector AB = [3,3,0]?
    I cannot use 0 because a zero vector won't be possible.

    -----

    It looks right but there's still something that's bothering me...

    BC = AC - AB right?
    So if I simply pop the vectors back in and replace k with 3, it'll work out:

    [2-3,4-3,-3] = [2,4,-3] - [3,3,0]
    [-1,1,-3] = [-1,1,-3]

    But however, this seems to work with ANY integer. For example, if I take 8 and replace all k's with 8:

    [2-8,4-8,-3] = [2,4,-3] - [8,8,0]
    [-6,-4,-3] = [-6,-4,-3]

    So does that mean [8,8,0] is also an answer as for [k,k,0] where k is ANY integer? Or am I just complicating things more and doing it wrong?

    I thought there's only suppose to be one answer...

    Thanks!
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  4. #4
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    Quote Originally Posted by sqleung View Post
    ...

    But however, this seems to work with ANY integer. For example, if I take 8 and replace all k's with 8:

    [2-8,4-8,-3] = [2,4,-3] - [8,8,0]
    [-6,-4,-3] = [-6,-4,-3]

    ...
    Keep in mind that \overline{AC} is the hypotenuse in a right triangle which is the longest of the sides of the triangle.

    Since |\overrightarrow{AC}| = \sqrt{29} the sum of the squares of the 2 other legs can't exceed 29.
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  5. #5
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    Quote Originally Posted by earboth View Post
    Keep in mind that \overline{AC} is the hypotenuse in a right triangle which is the longest of the sides of the triangle.

    Since |\overrightarrow{AC}| = \sqrt{29} the sum of the squares of the 2 other legs can't exceed 29.
    Ah, okay. Thankyou very much for that
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