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Find plane, given a line & perpendicular plane

  1. Nov 8, 2006 #1
    Please help me! I have been sitting with this problem for god knows how long, and I just can't figure it out. I've tried re-reading the problem text, re-reading the chapter, reading alternative explanations on the web, drawing the problem on paper -- heck, I've even tried shouting at it -- but no luck. Can someone please give me some pointers?

    The problem?
    Find the equation of the plane that satisfies the given conditions:
    Passing through the line x+y=2, y-z=3, and perpendicular to the plane 2x+3y+4z=5.

    Firstly, I'm not even sure if I'm reading it correctly. Does passing through mean that the plane contains the line, or that it passes through a point in the line? Is it one line (x+y=2, y-z=3) or two lines (x+y=2 and y-z=3)?I haven't seen it in that form before. Is it the wanted plane or the line that is perpendicular to the given plane? Secondly, how do I solve it?

    I have a feeling that this should be easy, so this is really bad for my self-confidence. :-(
    Last edited: Nov 8, 2006
  2. jcsd
  3. Nov 8, 2006 #2


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    Since just being told a line that passes through the plane and that it is perpendicular to another plane would not be enough to identify the plane, this must mean that the line lies in the plane- every point of the line is in the plane. And "x+ y= 2, y- z= 3" gives a single line, not 2. In 3 dimensions, a single equation is not sufficient to define a line. In effect, a single equation defines a plane. Saying "x+ y= 2, y- z= 3" means that the line is the intersection of those planes.
    To find a plane, it is sufficient to determine a single point in the plane, (a, b, c), and a vector, Ai+ Bj+ Ck, perpendicular to the plane. Then the equation of the plane is A(x-a)+ B(y-b)+ C(z-c)= 0. In particular, the plane 2x+3y+4z=5 is perpendicular to the vector 2i+ 3y+ 4k. Since the plane you seek is perpendicular to the given plane, that perpendicular, 2i+ 3y+ 4k, must be in the plane. Knowing that the line x+ y= 2, y- z= 3 is in the plane tells us that a vector in its direction must also be in the plane. We can write x= 2- y, z= y- 3 so we can take y itself as parameter: parametric equations for this line are x= 2- t, y= t, z= t- 3. The coefficients for x, y, z, -1, 1, 1, respectively, tell us that the vector -i+ j+ k is in the direction of the line and so also in the plane. You now know two vectors in the plane and you should know that their cross product is perpendicular to the plane itself. Now you know a vector perpendicular to the plane. All you need is a single point in the plane which you can get by plugging any value of t into the parametric equations for the given line.
  4. Nov 8, 2006 #3
    Thank you so much for that great explanation, HallsofIvy! It made it clear for me. I can finally continue. :-)
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