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Homework Help: Orthogonal equation plane help

  1. Sep 10, 2005 #1
    Find an equation of a plane through the point (-1, -2, -3) which is orthogonal to the line x=5+2t,y=-3-5t,z=2-2t
    in which the coefficient of x is 2.

    ______________________________=0

    i dont get this problem at all, but here's what i came up with after sitting here at the computer for a long time attempting to do this problem.

    okay i know that by theorem, The vector a X b is orthogonal to both a and b.

    a = <-1,-2,-3>
    b= <2,-5,2>

    that's all i came up, i attempted many different ways, but it doesnt make sense at all, can someone lend me a hand?
     
  2. jcsd
  3. Sep 10, 2005 #2

    quasar987

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    Ok, here's a plan: Find out the general form of a vector parallel to the line. (it will be a function of two parameters; say you labeled them t_1 and t_2.)Write out the form of a vector that lies in the plane (it will be a function of x,y & z). Then take the dot product of those two vectors and set it equal to 0. This is the equation of your plane. Now chose a particular vector parrarel to the line such that the coeff. of x is 2... that is to say, set t_1 and t_2 such that the coeff. of x is 2.

    I understand that this might sound very confusing but try your best to progress. If you get stuck, or if after thinking very hard about it, you still can't find a way, I'll help you some more. Good luck!!
     
  4. Sep 10, 2005 #3
    N*r = n*r_0 is the equation im looking for right?

    i am totally lost here, book doesnt help much either, can you help me start it.
     
  5. Sep 10, 2005 #4

    quasar987

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    What are N,r,n and r_0 ?!

    Have you done more simple similar problems before? If you're new to this kind of problem then this one is a rought introduction IMO!

    Let's start with the first step of my plan. Let's find the general form of a vector parallel to the line. The equation of the line is written in parametric form. It means that for any value of t, the point (x,y,z)=(5+2t,-3-5t,-2-2t) is on the line. Ok, so let's find two points on the line by setting t = t_1 and t = t_2. Our two points are [itex]\vec{P}_1 = (5+2t_1,-3-5t_1,-2-2t_1)[/itex] and [itex]\vec{P}_2 = (5+2t_2,-3-5t_2,-2-2t_2)[/itex]. By means of a simple drawing, you can convince yourself that [itex]\vec{P}_1 - \vec{P}_2 = (2(t_1-t_2), -5(t_1-t_2), -2(t_1-t_2))[/itex] is a vector parallel to the line (it is in fact IN/MERGED WITH the line).

    Now, make sure you understand every step in this paragraph and do step 2 on your own. Supposing (x,y,z) is a point in the plan, use the same method as above (substraction) to find a vector that is parallel to the plan.
     
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