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Math Help - vector2

  1. #1
    Faz
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    vector2

    Let
    -> -> -> ->
    u = 3 i -2j + k

    and
    -> -> -> ->
    v = 2i + j - k
    -> ->
    a)calculate the dot product of u and v

    -> ->
    a)calculate the cross product of u and v
    -> ->
    c)calculate 5 u - v
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  2. #2
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    Quote Originally Posted by Faz View Post
    Let
    -> -> -> ->
    u = 3 i -2j + k

    and
    -> -> -> ->
    v = 2i + j - k
    -> ->
    a)calculate the dot product of u and v

    -> ->
    a)calculate the cross product of u and v
    -> ->
    c)calculate 5 u - v

    hey mate,

    In relation to (a), (b), for any two 3 Dimensional vectors of the form
    v1 = a1*i + b1*j + c1*k, v2 = a2*i + b2*j + c2*k

    Their Euclidean Inner or 'Dot' Product is calculated using either
    (i) v1 . v2 = a1*a2 + b1*b2 + c1*c2
    (ii) v1 . v2 = |v1|*|v2|*cos(y) where y is the angle between v1 and v2
    Since the angle is unknown I would recommend using (i)

    Their Cross Product is defined as,

    v1 X v2 = det(A), where A is a 3x3 matrix where the first row of A = (i,j,k), the second row is v1 and the third is v2

    For (c)
    As before for a given three dimension vector of the form v1 (as defined previously)
    Scalar Multiplication -
    m*v1 = m(a1*i + b1*j + c1*k) = m*a1*i + m*b1*j + m*c*k
    Addition
    v1 + v2 = (a1 + a2)*i + (b1+b2)*j + (c1 + c2)*k


    Hope this helps,

    Let me know if you require any further assistance,

    Regards,

    David
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  3. #3
    Faz
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    vector2

    hi

    i got these answer plse check for me.

    a)6i-2j-k
    b)

    i j k 1st row
    3 -2 1 second row
    2 1 -1 3rd row in matrix form

    c)17i-9j-4k


    thks.
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  4. #4
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    Hello, Faz!

    These are straight forward . . . exactly where is your difficulty?


    Given: . \begin{array}{ccccc}<br />
\vec u &=& 3\vec i - 2\vec j + \vec k &=& \langle 3,\text{-}2,1\rangle \\<br />
\vec v &=& 2\vec i + \vec j - \vec k &=& \langle 2,1,\text{-}1\rangle \end{array}

    a) Find: . \vec u \cdot \vec v

    \vec u\cdot \vec v \;=\;\langle3,\text{-}2,1\rangle\cdot\langle 2,1,\text{-}1\rangle \;=\;(3)(2) + (\text{-}2)(1) + (1)(\text{-}1) \;=\;6-2-1 \;=\;\boxed{3}




    b) Find: .  \vec u \times \vec v

    \vec u \times \vec v \;=\;\begin{vmatrix}\vec i & \vec j & \vec k \\ 3 & \text{-}2 & 1 \\ 2 & 1 & \text{-}1 \end{vmatrix} \;=\;\vec i(2-1) - \vec j(\text{-}3-2) + \vec k(3+4)\;=\;\boxed{\vec i + 5\vec j + 7\vec k}




    c) Find: . 5\vec u - \vec v

    5\vec u - \vec v \;=\;5(3\vec i - 2\vec j + \vec k) - (2\vec i + \vec j - \vec k)

    . . . . . =\;15\vec i - 10\vec j + 5\vec k - 2\vec i - \vec j + \vec k

    . . . . . = \;\boxed{13\vec i - 11\vec j + 6\vec k}

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  5. #5
    Faz
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    vectors

    Thank you guys...u are the best....i understand it very clearly.
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