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Help with a relatively simple linear algebra proof

  1. Feb 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the lines given by the equation ax + by + c = 0 and bx - ay + d = 0 (where a, b, c, d are in R) are perpendicular by finding a vector in the direction of each line and showing that these vectors are orthogonal. (Hint: Watch out for the cases in which a or b equals zero.)


    2. Relevant equations
    Vectors a and b are orthogonal if the dot product of a and b are 0.

    3. The attempt at a solution

    I have spent a long time trying to figure out this problem but I don't even know how to start. I don't know how to create a vector in the same direction as a line unless I know 2 points on the line or the slope of the line. In this case I don't know either because the equation is so general. That is, unless choosing my own values of a, b, and c is acceptable (Is it?).

    I cannot use many techniques to prove this since not much has been introduced in the course. Only basic vector properties, the dot product, and projections have been introduced so far.
     
  2. jcsd
  3. Feb 5, 2009 #2

    Dick

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    Homework Helper

    Choosing a, b and c is not acceptable. Choosing a value of x or y is. Since those equations are supposed to be true for all x and y. Pick x=0 and x=1. Or y=0 and y=1. Now you have two points.
     
  4. Feb 5, 2009 #3
    Thank you, I'm not sure why I didn't think of it like that before...
     
  5. Feb 6, 2009 #4
    Sorry, I didn't have time to actually do the problem before. Now that I'm looking at it again, I'm still confused. How do I relate the first equation to the second equation?

    For example, if I choose x=0 y=0 for the first equation, what do I do with that? I get c = 0, but that doesn't tell me anything about the second equation. That I can do x=1 y=1 and get c = -(a+b) giving me the vector [1,1] for the first equation but again I don't see what that tells me. I can't just plug in he same values of x and y for the second equation, because obviously that will just give me the same vector.

    Sorry if that doesn't make sense...I feel like I'm missing something obvious, but I don't know what it is.
     
  6. Feb 6, 2009 #5
    Start with the line ax + by + c = 0. To get a vector in the direction of this line, you need two points. I think this is where you are getting confused.

    Take the line y = 3x + 2 as an example. How do you get a point on this line? Well x is the independent variable so you can choose any number. Choose x=0 for simplicity. Then you get that y = 2. So a point on this line is (0,2)

    Going back to ax + by + c = 0, you cannot pick x=0 and y=0 at the same time. The value of y depends on what you choose for x (or vice versa). So if you choose x=0, find the y that satisfies the equation a*0 + by + c = 0.
     
  7. Feb 6, 2009 #6
    Ok...put the points for y won't be numbers, but variables.

    if x = 0, y = -(c/b) giving point (0, -c/b)
    if x = 1, y = (-c-a)/b giving point (1, -c/b - a/b)

    Now I have a vector: [1, -a/b]

    using y = 0 and y = 1 for the second equation will give me [a/b, 1]

    These are orthogonal! Thank you!
    Now all that is left is the hint part of the problem. b cannot equal 0, or it is invalid. Is there anything special I have to add to account for that?
     
  8. Feb 6, 2009 #7
    go back to the equations for the lines ax + by + c = 0 and bx - ay + d = 0. If you substitute b=0 into both equations, what are the lines that do get?
     
  9. Feb 6, 2009 #8
    I worked on it a bit more and figured everything out. Thanks for your help.
     
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