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Math Help - Supposingly Easy Vector Question

  1. #1
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    Supposingly Easy Vector Question

    In triangle ABC, the point X divides BC internally in the ration m:n, where m+n = 1. Express (AX)^2 in terms of AB , BC,CA,m and n.

    My attempt :

    (AX)^2 = (AB)^2 - (BX)^2 ----1
    (AX)^2 = (CA)^2 - (XC)^2 ----2

    1+2 , 2(AX)^2 = (AB)^2 + (CA)^2 - (m BC)^2 - (n BC)^2
    (AX)^2 = [(AB)^2 +(CA)^2 - (m BC)^2 - (n BC)^2]/2
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by ose90 View Post
    In triangle ABC, the point X divides BC internally in the ration m:n, where m+n = 1. Express (AX)^2 in terms of AB , BC,CA,m and n.

    My attempt :

    (AX)^2 = (AB)^2 - (BX)^2 ----1
    (AX)^2 = (CA)^2 - (XC)^2 ----2

    1+2 , 2(AX)^2 = (AB)^2 + (CA)^2 - (m BC)^2 - (n BC)^2
    (AX)^2 = [(AB)^2 +(CA)^2 - (m BC)^2 - (n BC)^2]/2
    I don't quite understand your 1 & 2 equations...

    What exactly do you want ? Is it \vec{AX} ? Or the length AX^2 ?

    Be careful, \vec{AX}=\vec{AB} {\color{red}+} \vec{BX} and \vec{AX}=\vec{AC}+\vec{CX}
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    I don't quite understand your 1 & 2 equations...

    What exactly do you want ? Is it \vec{AX} ? Or the length AX^2 ?

    Be careful, \vec{AX}=\vec{AB} {\color{red}+} \vec{BX} and \vec{AX}=\vec{AC}+\vec{CX}
    Sorry for the carelessly constructed post.

    I want to find AX^2, thanks.
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