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Math Help - position vectors

  1. #1
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    position vectors

    how do i tackle these :
    1A)
    Consider the following position vectors, v=i-j and w=j-2k

    find i) the angle between the above vectors
    ii) a vector which perpendicular on both vectors

    b) Let z=2-3j and s=1+2j find

    2+z
    Z^2-Z and 1+s

    c) solve the following equation:

    z^3 - z^2 +z = 0


    amanda xx
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Amanda-UK View Post
    how do i tackle these :
    1A)
    Consider the following position vectors, v=i-j and w=j-2k

    find i) the angle between the above vectors
    ii) a vector which perpendicular on both vectors
    i) v \cdot w = |v| |w| cos(\theta)

    ii) The vector  v \times w is perpendicular to both v and w.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Amanda-UK View Post
    b) Let z=2-3j and s=1+2j find

    2+z
    Z^2-Z and 1+s
    \frac{2 + z}{z^2 - z} = \frac{2 + (2 - 3j)}{(2 - 3j)^2 - (2 - 3j)}

    = \frac{4 - 3j}{4 - 12j + 9j^2 - 2 + 3j}

    = \frac{4 - 3j}{2 - 9j - 9} = \frac{4 - 3j}{-7 - 9j}

    Of course, you need to simplify this expression...

    (Do you really need someone to do 1 + s for you??)

    -Dan
    Last edited by topsquark; August 17th 2007 at 07:36 AM.
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  4. #4
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    Quote Originally Posted by topsquark View Post
    i) v \cdot w = |v| |w| cos(\theta)

    ii) The vector  v \times w is perpendicular to both v and w.

    -Dan
    how would i then go on from there to get the answer? x
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Amanda-UK View Post
    c) solve the following equation:

    z^3 - z^2 +z = 0
    Factor:
    z^3 - z^2 + z = 0

    z(z^2 - z + 1) = 0

    I leave the rest to you.

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    i) v \cdot w = |v| |w| cos(\theta)

    ii) The vector  v \times w is perpendicular to both v and w.

    -Dan
    Quote Originally Posted by Amanda-UK View Post
    how would i then go on from there to get the answer? x
    i) This is a dot product:
    v \cdot w = v_x w_x + v_y w_y + v_z w_z

    |v| = \sqrt{v_x^2 + v_y^2 + v_z^2}

    ii) v \times w = \text{det} \left [ \begin{matrix} i & j & k \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{matrix} \right ]

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    \frac{2 + z}{z^2 - z} = \frac{2 + (2 - 3j)}{(2 - 3j)^2 - (2 - 3j)}

    = \frac{4 - 3j}{4 - 6j + 9j^2 - 2 + 3j}

    = \frac{4 - 3j}{2 - 3j - 9} = \frac{4 - 3j}{-11 - 3j}

    Of course, you need to simplify this expression...

    (Do you really need someone to do 1 + s for you??)

    -Dan
    sorry i wrote in eqution wrong ( not to sure how to write maths language on here: should be this:


    Let z = 2-3j and s= 1+2j find:
    .
    Z^2 - Z (there should be a line above the last Z but not lettin me do it)

    and

    2+z
    1+s

    Let z = 2-3j and s= 1+2j find:
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  8. #8
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    Quote Originally Posted by topsquark View Post
    i) This is a dot product:
    v \cdot w = v_x w_x + v_y w_y + v_z w_z

    |v| = \sqrt{v_x^2 + v_y^2 + v_z^2}

    ii) v \times w = \text{det} \left [ \begin{matrix} i & j & k \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{matrix} \right ]

    -Dan
    would you be able to show me this method more please x
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Amanda-UK View Post
    sorry i wrote in eqution wrong ( not to sure how to write maths language on here: should be this:


    Let z = 2-3j and s= 1+2j find:
    .
    Z^2 - Z (there should be a line above the last Z but not lettin me do it)

    and

    2+z
    1+s

    Let z = 2-3j and s= 1+2j find:
    Just as well, I goofed in the denominator when I posted anyway.

    Okay, so the first one is
    z^2 - \bar{z} = (2 - 3j)^2 - (2 + 3j)

    = (4 - 12j + 9j^2) - (2 + 3j)

    = 4 - 12j - 9 - 2 - 3j

    = -7 - 15j

    The second one is now \frac{2 + z}{1 + s} ?

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

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

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

    Because this is so simple up to here I presume you need help simplifying it. j = \sqrt{-1}, so we have a radical in the denominator that we wish to get rid of. Now, recall the identity:
    (a + b)(a - b) = a^2 - b^2

    Thus if we have 2 + 2\sqrt{-1} we can multiply it by 2 - 2\sqrt{-1} to get rid of the radical. But what we do to the denominator we must do to the numerator...
    = \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}{2^2 - (2j)^2}

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

    = \frac{2 - 14j}{4 + 4}

    = \frac{2 - 14j}{8} <-- Now cancel the common factor of 2

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

    -Dan
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    i) This is a dot product:
    v \cdot w = v_x w_x + v_y w_y + v_z w_z

    |v| = \sqrt{v_x^2 + v_y^2 + v_z^2}

    ii) v \times w = \text{det} \left [ \begin{matrix} i & j & k \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{matrix} \right ]

    -Dan
    Quote Originally Posted by Amanda-UK View Post
    would you be able to show me this method more please x
    If you don't know what I'm talking about here then you must be using another method in your class. Can you tell me what you've done on this?

    -Dan
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