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Math Help - vectors

  1. #1
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    vectors

    how do i do this please xx

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  2. #2
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    These are very basic practice problems. You will have to show some familiarity with them before you can move on.

    Hints:

    ai) Dot Product
    aii) Cross Product

    b)
    If Z = a + bi
    Z^2 = (a^2 - b^2) + 2abi
    Z-bar = a - bi (The conjugate)

    c) Factor and use the quadratic formula.
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  3. #3
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    Hello, Amanda!

    Here's the first one . . .


    a) Condider the following position vectors: . \begin{array}{ccc}\vec{v} & = & i-j\\ \vec{w} & = & j - 2k\end{array}

    (1) Find the angle between the two vectors.
    (2) Find a vector perpendicular to both vectors.
    We have: . \begin{array}{ccc}\vec{v} & = & \langle1,\,\text{-}1,\,0\rangle \\ \vec{w} & = & \langle 0,\,1,\,\text{-}2\rangle\end{array}


    (1) We're expected to know this formula: . \cos\theta \;=\;\frac{|\vec{v}\bullet\vec{w}|}{|\vec{v}||\vec  {w}|}

    So we have: . \cos\theta \;=\;\frac{|(1)(0) + (\text{-}1)(1) + (0)(\text{-}2)|}{\sqrt{1^2+(\text{-}1)^2+0^2}\,\sqrt{0^2+1^2+(\text{-}2)^2}} \;=\;\frac{|0 -1 + 0|}{\sqrt{2}\,\sqrt{5}}\;=\;\frac{1}{\sqrt{10}}

    . . Therefore: . \theta \;=\;\cos^{-1}\!\left(\frac{1}{\sqrt{10}}\right) \;\approx\;71.6^o



    (2) We're expect to know that a vector perpendicular to two vectors
    . . .is the cross product of the two vectors.

    We have: . \vec{n} \;=\;\begin{vmatrix}i & j & k \\ 1 & \text{-}1  & 0 \\ 0 & 1 & \text{-}2\end{vmatrix} \;=\;i(2-0) - j(\text{-}2 -0) + k(1-0) \;=\;2i + 2j + k

    . . Therefore: . \vec{n} \;=\;\langle 2,\,2,\,1\rangle

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  4. #4
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    thank you both for your help, though I am finding part B rather tricky
    xx
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  5. #5
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    Let z = 2 - 3j and s = 1 + 2j

    We want:
    \frac{2 + z}{1 + s}

    = \frac{2 + (2 - 3j)}{1 + (1 + 2j)}

    = \frac{2 + 2 - 3j}{1 + 1 + 2j}

    = \frac{4 - 3j}{2 + 2j}

    Now simplify:
    = \frac{4 - 3j}{2 + 2j} \cdot \frac{2 - 2j}{2 - 2j}

    = \frac{(4 - 3j)(2 - 2j)}{(2 + 2j)(2 - 2j)}

    = \frac{8 - 14j + 6j^2}{4 - 4j^2}

    = \frac{8 - 14j - 6}{4 + 4}

    = \frac{2 - 14j}{8}

    = \frac{1 - 7j}{4}

    -Dan
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  6. #6
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    blimy didnt think it was as easy as that LOL! on b where it is Z^2 not sure on that#xx
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  7. #7
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    Quote Originally Posted by Amanda-UK View Post
    blimy didnt think it was as easy as that LOL! on b where it is Z^2 not sure on that#xx
    Z = a + bi
    Z^{2} = (a+bi)^{2} = (a+bi)(a+bi) = (a^{2} - b^{2}) + 2abi I'm confident you can do it. Show us what you get.
    Last edited by TKHunny; August 22nd 2007 at 10:22 AM. Reason: Add definition
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