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Math Help - basic linear algebra questions

  1. #1
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    basic linear algebra questions

    let ABC be a triangle and let p be the midpoint of side AB
    let and

    express and in terms of and and

    hence prove that the midpoint of the hypotenuse of a right-angle triangle is equidistant from the three vertices

    ==================================================

    let p be the plane with cartesian equation 2x+y-3z=9 and let A,B be points with coordinates (1,-2,3) and (-2,1,-1)
    l is the line passing through the point A and B

    find the coordinates of the point C where the line l intersects the plane p


    i have no ideas how to find the intersection between a line and a plane

    thx
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  2. #2
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    Quote Originally Posted by Hikari Clover View Post
    let ABC be a triangle and let p be the midpoint of side AB
    let and

    express and in terms of and and

    hence prove that the midpoint of the hypotenuse of a right-angle triangle is equidistant from the three vertices
    Hi

    \overrightarrow{AP}^2 = \frac12\:\overrightarrow{AB}\cdot\frac12\:\overrig  htarrow{AB}\cdot = \frac14\:\overrightarrow{AB}\cdot\overrightarrow{A  B}

    \overrightarrow{AB} = \overrightarrow{AC} + \overrightarrow{CB} = -\vec{a} + \vec{b}

    \overrightarrow{AP}^2 = \frac14\:\left(-\vec{a} + \vec{b})^2\right) = \frac14\:\left(\vec{a}^2 + \vec{b}^2 -2 \vec{a} \cdot \vec{b}\right)

    \overrightarrow{CP} = \frac12\:\left(\overrightarrow{CA} + \overrightarrow{CB}\right) =  \frac12\:\left(\vec{a} + \vec{b}\right)

    ==================================================
    Quote Originally Posted by Hikari Clover View Post
    let p be the plane with cartesian equation 2x+y-3z=9 and let A,B be points with coordinates (1,-2,3) and (-2,1,-1)
    l is the line passing through the point A and B

    find the coordinates of the point C where the line l intersects the plane p


    i have no ideas how to find the intersection between a line and a plane

    thx
    AB coordinates are (-3,3,-4)
    Therefore one parametric equation of l is :
    x = -3t + 1
    y = 3t -2
    z = -4t +3

    C(x,y,z) is on the plane iff 2x+y-3z=9
    C is on l iff there exists t such that
    x = -3t + 1
    y = 3t -2
    z = -4t +3

    Substitute x,y,z in the Cartesian equation of the plane to get one linear equation. Solve for t and substitute in x,y,z expressions.

    Spoiler:

    2(-3t + 1)+(3t -2)-3(-4t +3)=9 gives t=2
    x = -5
    y = 4
    z = -5
    C(-5,4,-5)
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    \overrightarrow{AP}^2 = \frac12\:\overrightarrow{AB}\cdot\frac12\:\overrig  htarrow{AB}\cdot = \frac14\:\overrightarrow{AB}\cdot\overrightarrow{A  B}

    \overrightarrow{AB} = \overrightarrow{AC} + \overrightarrow{CB} = -\vec{a} + \vec{b}

    \overrightarrow{AP}^2 = \frac14\:\left(-\vec{a} + \vec{b})^2\right) = \frac14\:\left(\vec{a}^2 + \vec{b}^2 -2 \vec{a} \cdot \vec{b}\right)

    \overrightarrow{CP} = \frac12\:\left(\overrightarrow{CA} + \overrightarrow{CB}\right) = \frac12\:\left(\vec{a} + \vec{b}\right)

    ==================================================
    AB coordinates are (-3,3,-4)
    Therefore one parametric equation of l is :
    x = -3t + 1
    y = 3t -2
    z = -4t +3

    C(x,y,z) is on the plane iff 2x+y-3z=9
    C is on l iff there exists t such that
    x = -3t + 1
    y = 3t -2
    z = -4t +3

    Substitute x,y,z in the Cartesian equation of the plane to get one linear equation. Solve for t and substitute in x,y,z expressions.

    Spoiler:

    2(-3t + 1)+(3t -2)-3(-4t +3)=9 gives t=2
    x = -5
    y = 4
    z = -5
    C(-5,4,-5)



    hey , thx for ur replying

    but for the first question,did u forget to put that absolute value sign? or it doesnot matter?
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  4. #4
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    Quote Originally Posted by Hikari Clover View Post
    hey , thx for ur replying

    but for the first question,did u forget to put that absolute value sign? or it doesnot matter?
    Are you talking about the modulus ?

    ||\overrightarrow{AB}||^2 = \overrightarrow{AB}\cdot\overrightarrow{AB} = \overrightarrow{AB}^2 = AB^2
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  5. #5
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    oh i got it
    thanks so much ^_^
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