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Math Help - Position Vector of Reflection of a Point

  1. #1
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    Question Position Vector of Reflection of a Point

    Hello, I am currently having trouble understanding how to find the reflection point of a point with the coordiantes of that point and the vector equation of the line. I have tried finding the maginitude of the foot of the prependicular, but I still could not get the answer. I hope my doubts can be cleared.

    This is the question :
    Find the position vector of the image C' of the point C(5,2,-1) under a reflection in the line l joining the points A(3,1,3) and B(-1,-1,-3).
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  2. #2
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    Vectors

    Hello MathJunction

    Welcome to Math Help Forum!
    Quote Originally Posted by MathJunction View Post
    Hello, I am currently having trouble understanding how to find the reflection point of a point with the coordiantes of that point and the vector equation of the line. I have tried finding the maginitude of the foot of the prependicular, but I still could not get the answer. I hope my doubts can be cleared.

    This is the question :
    Find the position vector of the image C' of the point C(5,2,-1) under a reflection in the line l joining the points A(3,1,3) and B(-1,-1,-3).
    Using vectors, suppose that the position vector of A is \vec{a} = 3\vec{i} + \vec{j} + 3\vec{k}. Similarly \vec{b} = -\vec{i} -\vec{j} -3\vec{k},\, \vec{c} = 5\vec{i} + 2\vec{j} -\vec{k}.

    Then if M is the mid-point of AB, its position vector is given by \vec{m} = \tfrac12(\vec{a}+\vec{b}) = -\vec{i}.

    \Rightarrow \vec{CM} = \vec{m} - \vec{c} = -6\vec{i} -2\vec{j} +\vec{k}

    Now if C' is the reflection of C in AB, \vec{MC'}=\vec{CM}

    \Rightarrow \vec{c'} = \vec{m} + \vec{MC'} =\vec{m} + \vec{CM}, where \vec{c'} is the position vector of C'.

    Can you complete it now?

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello MathJunction

    Welcome to Math Help Forum!Using vectors, suppose that the position vector of A is \vec{a} = 3\vec{i} + \vec{j} + 3\vec{k}. Similarly \vec{b} = -\vec{i} -\vec{j} -3\vec{k},\, \vec{c} = 5\vec{i} + 2\vec{j} -\vec{k}.

    Then if M is the mid-point of AB, its position vector is given by \vec{m} = \tfrac12(\vec{a}+\vec{b}) = -\vec{i}.

    \Rightarrow \vec{CM} = \vec{m} - \vec{c} = -6\vec{i} -2\vec{j} +\vec{k}

    Now if C' is the reflection of C in AB, \vec{MC'}=\vec{CM}

    \Rightarrow \vec{c'} = \vec{m} + \vec{MC'} =\vec{m} + \vec{CM}, where \vec{c'} is the position vector of C'.

    Can you complete it now?

    Grandad
    Hello Grandad, thank you for taking some time off in helping me. I regret to tell you I am still blurred. Shouldn't vector m be 1/2(a+b) = (1,0,0) since vector AB is ( -4,-2,-6) ? Thus AM is half of AB which is (-2,-1,-3) and then OM would be vector AM + vector OA?

    Sorry to trouble you again.
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  4. #4
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    Vectors

    Hello MathJunction
    Quote Originally Posted by MathJunction View Post
    Hello Grandad, thank you for taking some time off in helping me. I regret to tell you I am still blurred. Shouldn't vector m be 1/2(a+b) = (1,0,0) since vector AB is ( -4,-2,-6) ? Thus AM is half of AB which is (-2,-1,-3) and then OM would be vector AM + vector OA?

    Sorry to trouble you again.
    You're quite right - I got a sign wrong. Sorry!

    \vec{m} = \tfrac12(\vec{a}+\vec{b})= \vec{i}

    So \vec{CM}=\vec{m}-\vec{c}=-4\vec{i} -2\vec{j} +\vec{k}

    Then continue as I said before.

    Grandad
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