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Finding Equations of Vector Line Perpendicular to another Vector Line

  1. Aug 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Line 1's equation is given as r(t) = <t, t, 1-t>. Line 2 passes through the origin and is perpendicular to line 1. Find the equation of line 2.


    2. Relevant equations



    3. The attempt at a solution
    If it is a 2-D case then it would be a lot easier, since m1 x m2 = -1. However, this is a 3-D case.

    Anyway, here is my attempt.
    I know that the dot product of two vector lines equals 0 iff they are perpendicular.
    Line 1, r(t) = <t, t, 1-t> = <0, 0, 1> + t <1, 1, -1>
    Line 2, s(t) = <0, 0, 0> + u <a, b, c>.
    So I need to find the value of a, b and c.
    <1, 1, -1> dot <a, b, c> = 0
    a + b - c = 0.
    Let a = b = 1, so c = 2.
    So the equation of line 2 is s(t) = <0, 0, 0> + u <1, 1, 2>.

    The answer and method seems plausible to me.

    However, I remember seeing somewhere in the book where in order to solve this kind of problem, we need to find the equation of the plane that contains the first line. Then, find the equation of another plane and the intersection of this 2 planes is the equation of the line 2. This is done by using the cross product of 2 normal lines.

    Thanks.
     
  2. jcsd
  3. Aug 19, 2008 #2

    HallsofIvy

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    Science Advisor

    I started to say you were completely correct and then had to stop and think!
    You made a very lucky (or intelligent) choice of a and b! Yes, two vectors are perpendicular if and only if their dot product is 0 so you are correct that <1,1,2> is perpendicular to <1, 1, -1>. But in order that two lines be perpendicular they must also intersect! Do the lines given by <t, t, 1-t> and <u, u, 2u> intersect? Look at t= u, t= u, 1-t= 2u. Obviously, from the first two equations, t= u and so the third becomes 1-t= 2t so 1= 3t and t= u= 1/3. Yes, the two lines intersect at <1/3, 1/3, 2/3> so your answer is correct.

    But do you see that two lines may be "skew"? For example, the line given by <0, 0, t> (the z-axis) and the line given by <u+1, u, 0> are perpendicular- their direction vectors are <0, 0, t> and <1, 1, 0>. But they do not intersect.

    A more general method is to recognize that there is, as you say a whole plane of lines perpendicular to any given line, through a given point. From you dot product, a+ b- c= 0, you chose a= 1, b= 1 and so c= 0. In fact, c= a+b so the vector <a, b, a+b> is perpendicular to <1, 1, -1> for any a,b and the line given by <au, bu, (a+b)u> is perpendicular to <t, t, -t> for all a and b. (Note that a plane is two dimensional. We need the two parameters, a and b, to describe the plane.)

    Now, which of the lines given by <au, bu, (a+b)u> will intersect <t, t, -t>? Again, we must have au= t, bu= t, and (a+b)u= -t. From the first equation, t= au so the second equation becomes bu= au and we must have b= a for that to be true. In that case, a+b= 2a and the third equation becomes 2au= -au so, again, u= 1/3. The line given by <au, au, 2au>, for any a, will pass through (0,0,0) and be perpendicular to <t, t, -t> (and in particular, b= a= 1).
     
  4. Aug 19, 2008 #3
    Oh yeah... They must intercept... I miss out that one. Let me try again one more time.
     
  5. Aug 19, 2008 #4
    Nice explanation, HallsofIvy. Now let me try do solve it and rephrase it in my own understanding.

    Let the equation of line 2 be <a, b, c>. So, <a, b, c> dot <1, 1, -1> = 0. So we have a + b - c = 0. And then c = a + b. This gives us u<a, b, a+b>. Another vector is t<1, 1, -1>.

    Since the two lines intercept each other, we also have u<a, b, a+b> = t<1, 1, -1>.
    We have ua = t, ub = t, u(a+b) = -t
    ua = ub => u(a-b) = 0. u cannot equal to 0, so a - b = 0 => a = b
    u(a + b) = -t
    u(2a) = -ua
    2ua + ua = 0
    3ua = 0
    Oops, since that either u = 0 or a = 0.
    So, since u can't be 0, we need a = 0, which also implies b = 0.
    I am getting nothing here. Hm..
     
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