Results 1 to 4 of 4

Math Help - Equation of the Line of Intersection of two Planes Help

  1. #1
    Junior Member Morphayne's Avatar
    Joined
    May 2008
    Posts
    31

    Exclamation Equation of the Line of Intersection of two Planes Help

    Problem:

    Determine a vector equation of the line of intersection of the planes 3x-y+4z-2=0 and x+6y+10z+8=0.

    Answer(from textbook):

    Answers may vary.
    [x,y,z]=[2,0,-1]+t[34,26,-19]<br />

    Comments:

    Please help me. The questions involving planes are really frustrating me.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Taking the cross product of their normals will give you a vector parallel to the line of intersection, i.e. your direction vector.

    Find a position vector that lies on the line of intersection (i.e. that satisfies both equations of the plane). This can be pretty much anything and is why your answer will vary. What you can do is fix a value for x (or y or z) and solve for remaining simpler system of equations for the other 2 variables to get a point common to both planes. This point will form the position vector needed to define your line.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,683
    Thanks
    615
    Hello, Morphayne!

    o_O has the best method.

    I'll show you another . . .


    Determine the equations of the line of intersection of the planes
    . . 3x-y+4z-2\:=\:0 and x+6y+10z+8\:=\:0.

    Answer(from textbook):

    Answers may vary. . . . . . .
    Finally ... a book that gives fair warning!

    . . [x,y,z]\:=\:[2,0,-1]+t[34,26,-19]

    We have: . \begin{array}{cccc}3x - y + 4z &=& 2 & {\color{blue}[1]} \\ x + 6y + 10z &=& \text{-}8 & {\color{blue}[2]} \end{array}

    \begin{array}{cccc}\text{Multiply {\color{blue}[1]} by 6:} & 18x - 6y + 24z &=& 12 \\ \text{Add {\color{blue}[2]}:} & x + 6y + 10z &=& -8 \\<br />
\text{and we have:} & 19x + 34z &=& 4\end{array}

    . . Hence: . x \:=\:\frac{4}{19} - \frac{34}{19}z

    Substitute into [2]: . \left(\frac{4}{19} - \frac{34}{19}z\right) + 6y + 10z \:=\:\text{-}8 \quad\Rightarrow\quad y \:=\:\text{-}\frac{2}{19} - \frac{26}{19}z


    We have: . \begin{Bmatrix}x &=& \frac{4}{19} - \frac{34}{19}z \\ \\[-3mm] y &=& \text{-}\frac{2}{19} - \frac{26}{19}z \\ \\[-3mm] z &=& z \end{Bmatrix}

    On the right side, replace z with the parameter t.

    And we have: . \begin{Bmatrix}x &=& \frac{4}{19} - \frac{34}{19}t \\ \\[-3mm] y &=& \text{-}\frac{2}{19} - \frac{26}{19}t \\ \\[-3mm] z &=& 0 + t \end{Bmatrix}


    Therefore: . [x,y,z] \;=\;\left[\frac{4}{19},\:\text{-}\frac{2}{19},\:0\right] + t\left[\text{-}\frac{34}{19},\:\text{-}\frac{26}{19},\:1\right]

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member Morphayne's Avatar
    Joined
    May 2008
    Posts
    31

    Lightbulb

    Quote Originally Posted by Soroban View Post
    Hello, Morphayne!

    o_O has the best method.

    I'll show you another . . .



    We have: . \begin{array}{cccc}3x - y + 4z &=& 2 & {\color{blue}[1]} \\ x + 6y + 10z &=& \text{-}8 & {\color{blue}[2]} \end{array}

    \begin{array}{cccc}\text{Multiply {\color{blue}[1]} by 6:} & 18x - 6y + 24z &=& 12 \\ \text{Add {\color{blue}[2]}:} & x + 6y + 10z &=& -8 \\<br />
\text{and we have:} & 19x + 34z &=& 4\end{array}

    . . Hence: . x \:=\:\frac{4}{19} - \frac{34}{19}z

    Substitute into [2]: . \left(\frac{4}{19} - \frac{34}{19}z\right) + 6y + 10z \:=\:\text{-}8 \quad\Rightarrow\quad y \:=\:\text{-}\frac{2}{19} - \frac{26}{19}z


    We have: . \begin{Bmatrix}x &=& \frac{4}{19} - \frac{34}{19}z \\ \\[-3mm] y &=& \text{-}\frac{2}{19} - \frac{26}{19}z \\ \\[-3mm] z &=& z \end{Bmatrix}

    On the right side, replace z with the parameter t.

    And we have: . \begin{Bmatrix}x &=& \frac{4}{19} - \frac{34}{19}t \\ \\[-3mm] y &=& \text{-}\frac{2}{19} - \frac{26}{19}t \\ \\[-3mm] z &=& 0 + t \end{Bmatrix}


    Therefore: . [x,y,z] \;=\;\left[\frac{4}{19},\:\text{-}\frac{2}{19},\:0\right] + t\left[\text{-}\frac{34}{19},\:\text{-}\frac{26}{19},\:1\right]

    That is the answer I got! I know it says "Answers may vary", but I thought I had to find and equation without the fractions in the direction vector. Thanks anyway though.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Angle between planes and line of intersection of planes.
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 6th 2011, 12:08 PM
  2. Replies: 3
    Last Post: September 22nd 2010, 02:46 AM
  3. Line of intersection of planes
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: July 5th 2010, 03:48 PM
  4. Replies: 2
    Last Post: May 10th 2010, 02:32 AM
  5. Replies: 7
    Last Post: November 1st 2008, 05:57 PM

Search Tags


/mathhelpforum @mathhelpforum