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Math Help - Solve my vector Problems

  1. #1
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    Solve my vector Problems

    a) Show that the points (1,3,1) ,(1,1,-1),(-1,1,1) (2,2,-1) are lying on the same plane

    b) For any vector r prove that r = (r.i)i+(r.j)j+(r.k)k


    c) If a*(b*c)=(a*b)*c then prove that (c*a)*b=0
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  2. #2
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    Quote Originally Posted by Salman91 View Post
    a) Show that the points (1,3,1) ,(1,1,-1),(-1,1,1) (2,2,-1) are lying on the same plane
    Let P = (1,3,1), Q = (1,1,-1), R = (-1,1,1) and S = (2,2,-1)

    Show that
    \|\vec {PQ} \cdot (\vec{PR} \times \vec{PS}) \| = 0

    b) For any vector r prove that r = (r.i)i+(r.j)j+(r.k)k
    \vec r = \left< r_i, r_j, r_k \right> = \left< r_i, 0, 0 \right> + \left< 0, r_j, 0 \right> + \left< 0, 0, r_k \right> = \cdots

    think you can finish up?

    c) If a*(b*c)=(a*b)*c then prove that (c*a)*b=0
    what are you using * to mean?
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  3. #3
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    * = multiply ( x )
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Salman91 View Post
    * = multiply ( x )
    i assume that a, b, c are vectors here. "multiply" makes no sense. do you mean "cross product"? and they are all cross products? you have to distinguish between the dot and cross product!
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  5. #5
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    yes sir it is a cross-product

    and can you please show me the answer of part b (step by step)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Salman91 View Post
    yes sir it is a cross-product
    i will just solve this one...

    recall these properties from your text: a \times ( b \times c) = (a \cdot c)b - (a \cdot b)c, a \times b = -b \times a and a \cdot b = b \cdot a for any vectors a,b,c (of course, for cross-products we need 3-d vectors)


    Note that a \times (b \times c) = (a \cdot c)b - (a \cdot b)c

    (a \times b) \times c = -c \times (a \times b) = (-c \cdot b)a - (-c \cdot a)b and

    (c \times a) \times b = -b \times (c \times a) = (-b \cdot a)c - (-b \cdot c)a



    So, assume a \times (b \times c) = (a \times b) \times c, then

    (a \cdot c)b - (a \cdot b)c = (-c \cdot b)a - (-c \cdot a)b

    \Rightarrow (a \cdot c)b - (a \cdot b)c - (-c \cdot b)a + (-c \cdot a)b = 0

    \Rightarrow (a \cdot c)b + (-b \cdot a)c - (-b \cdot c)a - (a \cdot c)b = 0

    \Rightarrow (-b \cdot a)c - (-b \cdot c)a = 0

    \Rightarrow (c \times a) \times b = 0
    and can you please show me the answer of part b (step by step)
    there is pretty much one step left, you can't finish it? try.
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