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Math Help - Intersection line of planes, Vector geometry

  1. #1
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    Intersection line of planes, Vector geometry

    Suppose planes P1: 4x+3y+2z+1=0 and P2: x+2y+3z+4=0 and P3: x+by+cz+7=0 possess a common line of intersection. Find b and c.
    My approach was first to perform the cross product of the normal vectors of p1 and p2. The result is the direction vector of the line of intersection of all three planes. The direction vector is (5;-10;5). Thus the cross product of P1 or P2 and P3, must have as result t(5;-10;5), too. Therefore: (1;b;c)x(1;2;3)= t(5;-10;5). Unfortunately, all the three variables (a,b,t) cancel each other out.
    Where is my mistake? How to solve this problem?
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  2. #2
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    Hi

    The three variables do not exactly cancel out
    (1;b;c)x(1;2;3)= t(5;-10;5)
    gives
    5t = 3b-2c
    -10t = c-3
    5t = 2-b
    from which you can find one relation between b and c which is c = 2b-1

    You could also find this relation by the dot product (1;b;c).(5;-10;5)=0

    Actually you have found one direction vector of the common line but this is not sufficient since all parallel lines have the same direction vector

    You should find one point of the common line (using the equation of P1 and P2)

    Then using the fact that this point is also on P3 will give you a second relation between b and c

    Solve the linear system for b and c (I can find b=3 and c=5)
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  3. #3
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    I see. but what confuses me, is that generally you have three variables and three equations, so why can you not find a value for each and every variable?
    Farther: I see, that the common direction vector does not suffice for the intersection line, but this line is not sought. As b,c do only influence the normal vector of the plane3, the only precondition for b,c would be, that the crossproduct of the normal vector of P3 and P1/P2 equal to t(5,-10,5), where exactly the line starts should be arbitrary. Why is it important to specify one point lying on the intersection line, as this line is not sought at all...

    best regards
    Last edited by Schdero; March 31st 2010 at 01:40 PM.
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  4. #4
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    Hello Schdero
    Quote Originally Posted by Schdero View Post
    Suppose planes P1: 4x+3y+2z+1=0 and P2: x+2y+3z+4=0 and P3: x+by+cz+7=0 possess a common line of intersection. Find b and c.
    My approach was first to perform the cross product of the normal vectors of p1 and p2. The result is the direction vector of the line of intersection of all three planes. The direction vector is (5;-10;5). Thus the cross product of P1 or P2 and P3, must have as result t(5;-10;5), too. Therefore: (1;b;c)x(1;2;3)= t(5;-10;5). Unfortunately, all the three variables (a,b,t) cancel each other out.
    Where is my mistake? How to solve this problem?
    I think there's a much simpler method here.

    If f(x,y,z) = 0 and g(x,y,z) = 0 are the equations of two planes, then consider the equation
    \lambda f + \mu g =0, for any values of \lambda, \mu.
    First, it is linear in x, y, z and so it represents a plane.

    Secondly, any point on the line of intersection of the two planes f=0 and g=0 lies in both planes, and so satisfies f = 0 and g = 0, and therefore satisfies \lambda f + \mu g =0.


    Therefore \lambda f + \mu g =0 represents a family of planes containing the line of intersection of f and g.


    So here we can simply say:


    x + by+cz+7= 0 has a common line of intersection with P1 and P2 if and only if:
    \lambda(4x+3y+2z+1) + \mu(x + 2y+3z+4) \equiv x + by+cz+7
    Comparing coefficients and solving for \lambda, \mu, b, c gives:
    \lambda = -\tfrac15, \mu = \tfrac95, b = 3, c = 5
    Grandad
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  5. #5
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    This seems to be a logical solution, although a little bit above my level of mathematics, as I do not see how to solve one equation with 4 unknowns...
    But what ails me more than the actual way of solving this problem, is the following:
    Be the normal vector of P1-> (nP1),
    the n. vector of P2 ->(nP2)
    the n. vector of p3 -> (nP3)
    and Direction vector of the line of intersection ->(d)


    (nP1)x(nP2) = (d)
    Thus: (nP1)/(nP2)x(nP3) = t(d)
    Even though the exact position of the line is not specified, there is only ONE direction vector of the line. So why does (nP2)x(nP3)=t(d) not lead to the correct solution but only to a correlation between b and c?
    Last edited by Schdero; April 1st 2010 at 01:44 AM.
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  6. #6
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    Hello Schdero
    Quote Originally Posted by Schdero View Post
    This seems to be a logical solution, although a little bit above my level of mathematics, as I do not see how to solve one equation with 4 unknowns...
    It's very straightforward. If you've reached vector products, you must have seen something like this before.

    The expressions on either side of the \equiv sign are identical, so we can compare coefficients of x, y, z and constant, to get:
    4\lambda+\mu=1 ...(1)

    3\lambda + 2\mu = b ...(2)

    2\lambda+3\mu = c ...(3)

    \lambda + 4\mu = 7 ...(4)
    Multiply (1) by 4, and subtract (4):
    15\lambda = -3

     \Rightarrow \lambda = -\tfrac15
    ... etc.
    But what ails me more than the actual way of solving this problem, is the following:
    Be the normal vector of P1-> (nP1),
    the n. vector of P2 ->(nP2)
    the n. vector of p3 -> (nP3)
    and Direction vector of the line of intersection ->(d)


    (nP1)x(nP2) = (d)
    Thus: (nP1)/(nP2)x(nP3) = t(d)
    Even though the exact position of the line is not specified, there is only ONE direction vector of the line. So why does (nP2)x(nP3)=t(d) not lead to the correct solution but only to a correlation between b and c?
    As running-gag explained, you have found the direction vector of the line of intersection to be
    \begin{pmatrix}5\\-10\\5\end{pmatrix}
    and the relation between b and c to be
    c = 2b-1
    What you have found so far, then, is a family of planes of the form
    x + by + (2b-1)z + 7 = 0
    whose property is that their normal is perpendicular to the vector
    \begin{pmatrix}5\\-10\\5\end{pmatrix}
    which can be demonstrated by forming the scalar (dot) product of the normal with this vector:
    \begin{pmatrix}1\\b\\2b-1\end{pmatrix}\cdot\begin{pmatrix}5\\-10\\5\end{pmatrix}=(1)(5)+b(-10)+(2b-1)(5)
    =0
    For any value of b, then, the plane satisfies all the conditions you have written down. You need to find an additional condition that specifies which particular plane contains the actual line of intersection of P1 and P2.

    I hope that may make it clearer.

    Grandad
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  7. #7
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    Quote Originally Posted by Schdero View Post
    Why is it important to specify one point lying on the intersection line, as this line is not sought at all...
    Let us take an example
    Take b=2 and c=3
    Then P1: 4x+3y+2z+1=0 and P2: x+2y+3z+4=0 and P3: x+2y+3z+7=0

    You can see that P2 and P3 are parallel
    P1 and P2 have a common line
    P1 and P3 have also a common line
    These 2 lines are parallel, they have the same direction vector, but they are not equal
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  8. #8
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    Hi both

    Thank you very, very much for your explenations. I'm preparing my final exams in mathematics (in Swiss highschool) and your tips helped me a lot! Both of you are very good at explaining!
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