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Math Help - vectors

  1. #1
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    vectors

    I was given: a=2i - 3j - k, b=2i - j and c = -3j + k

    (a) Find a + b, |a + b| Which I'm sure I'm correct with)
    a + b = 4i - 2j - k

    |a + b| = *33 (* representing a square root)

    (b) Find the unit vector in the direction a-c (I'm more confused about this guestion)
    Here I found -(a+c), assuming it would be interpreted in the direction c towards a ??
    That is a + c = 2i - 2k
    I found unit vector then to be: - 4/*40 (2i - 6j)

    And therefore unit vector a-c: 4/*40 (-2i + 6j)

    I'm confused, because if I did it a-c I'd get 0/*8


    thanks for any help
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  2. #2
    Junior Member ignite's Avatar
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    Re: vectors

    Quote Originally Posted by cellae View Post
    I was given: a=2i - 3j - k, b=2i - j and c = -3j + k

    (a) Find a + b, |a + b| Which I'm sure I'm correct with)
    a + b = 4i - 2j - k

    |a + b| = *33 (* representing a square root)

    (b) Find the unit vector in the direction a-c (I'm more confused about this guestion)
    Here I found -(a+c), assuming it would be interpreted in the direction c towards a ??
    That is a + c = 2i - 2k
    I found unit vector then to be: - 4/*40 (2i - 6j)

    And therefore unit vector a-c: 4/*40 (-2i + 6j)

    I'm confused, because if I did it a-c I'd get 0/*8


    thanks for any help
    Are you sure your a+b is correct? a=2i-3j-k and b=2i-j gives a+b=4i-4j-k.
    a-c=2i-3j-k-(-3j+k)=2i-2k
    Unit vector in direction of a-c : \frac{1}{\sqrt{2}}(i-k)
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  3. #3
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    Re: vectors

    Oh cool !

    I then got asked: Find a unit vector to b and to a-c:

    So I got b - b.(a-c)/|a-c|

    = b - 4/4 (2j - 2k)
    b - (2j - 2k)

    thus (2i - j) - (2j - 2k)
    = 2i - 3j - 2k

    (seems wrong to me :/)
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  4. #4
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    Re: vectors

    just figured i was wrong there.. all good
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