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Equation of plane containing 2 lines.

  1. Apr 2, 2012 #1
    Find the equation of the plane containing each set of lines with symmetric equations below, if no equation exists, prove it.
    1. (x-3)/2 = (y+1)/4 = (z-2)/-1

    (x-3)/2 = (y+1)/-2 = (z-3)/2

    2. (x-1) = (y+3)/2 = (z+3)/3

    (x-3)/3 = (y-1)/6 = (z-3)/9

    3. (x-2)/2 = y/5 = (z+3)/6

    (x-4)/4 = (y+1)/10 = (z+1)/12


    I put the equations in parametric form
    1. x = 2t + 3 y = 4t - 1 z = -t +2
    x = 5s +3 y = -2s - 1 z = 2s + 3

    2. x = t + 1 y = 2t - 3 z = 3t -3
    x = 3s + 3 y = 6s +1 z = 9s + 3

    3.x = 2t + 2 y = 5t z = 6t - 3
    x = 4s + 4 y = 10s - 1 z = 12s - 1



    Number 1

    using both equations for y and z i solved for s and t. s = -2/3 and t = 1/3. But when s and t are subsituted in the x parmetric equations the values of x are not equal. x = 11/3 for one and 1/3 for another. My question is would this mean the lines are skew and if the lines are skew does that mean there is no equation of a plane that contains both lines.

    Number 2

    When I equate the eqauation for y to eachother and the eqautions for z to eachother and then solve for one of the variables s or t I run into a problem. The equated values to solve for one variable are equal. For example when solving for s i end up with 9s+6 = 9s+6 which is equivalent to 1 = 1. Unsure what that meant i used the dot product of the v vector for each line and found that the angle between the lines is 90 degrees making them perpendicular. My question is there an equation for the plane containing both of these lines. I dont think there is because taking the cross product of the v vectors to find the normal results in all zeros [0,0,0]. That means for any r0 the plane is 0 = 0.

    Number 3

    I equate the equations of y's together and the equations of z together to find values for t and s. When I would solve for a variable, the variable would cancel out of the equation. For example when solving for s i end up with 12s - 6/5 = 12s +4. This cancels out the s and leaves me with -6/5 = 4. I am not sure how to interpret what that means. So i used the dot product to find the angle between the lines and found that it was 0 degrees. I am pretty sure the lines would intersect at every point then. I can easily find the normal by taking the cross product of the v vectors but what i need to also find for the equation of a plane is a point on that plane. So my question is how would I find a point that lies on the plane that contains both of these lines.

    Any help is greatly appreciated because i am stumped on this problem. Thank you in advance.
     
  2. jcsd
  3. Apr 2, 2012 #2
    1. Yes, the lines do not intersect given your parametric equations, because the intersection in the y-z project does not correspond to the same x value. Because the lines intersect in the y-z projection, they are not parallel, therefore they are skew. There is probably a typo in your second set of original equalities for this problem, should "(x-3)/2" be "(x-3)/5"?

    2. In that situation, the lines are coincident as far as the y-z projection is concerned. The inferred relationship between t and s can be assessed in the remaining coordinate. I don't find them to be perpendicular.

    3. The absence of a solution in that case means that the lines do not meet in the y-z projection (and therefore the 3-space lines also do not meet). They are not necessarily parallel in 3-space though - the x ordinate can be checked against y or z to establish that.
     
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