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Intersection of 2 lines

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the vector equation of the line through 4,5,5 that meets (x-11)/3=7+8=z+4 at right angles

    2. Relevant equations

    3. The attempt at a solution
    i have no idea where to start.
  2. jcsd
  3. Jan 21, 2009 #2


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    Come on. You must have SOME idea. For example, what does the equation represent? A line, a plane, or what? Try and figure that out first. I'm not totally sure you posted it correctly.
  4. Jan 21, 2009 #3
    it was a question from a test that i got 0 on, the exact wording is as follows:

    find a vector equ ation of the line through the point (4,5,5) that meets the line (x-11)/3=(y+8)/1=(z-4)/1 at right angles

    i understand that the dot product of the vector i have to find and of the line given will be equal to zero. other than that, im completely lost.
  5. Jan 21, 2009 #4
    i dont know if this makes sense, but i tried to first came up with an expression for the vector i am trying to find

    4-(11+3t), 5-(-8+t), 5-(4+t) (starting point of the vector subtract a point on the line)

    then using dot product equation, i isolated t to -7/11

    is there anything flawed in this logic?
  6. Jan 21, 2009 #5


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    Yeah, I don't think it's quite right. A general point on the line is, fixing z=t. y=t-4 and x=3t+23. i.e. (3t+23,t-4,t). Or did I do that wrong? I would then take the difference between that and (4,5,5) and minimize the distance wrt to t. That would give you two points on the line, yes?
  7. Jan 22, 2009 #6
    Well, from the given question it is evident that the equation which to be found is of a line. Now, the equation,
    [tex]\frac{x-11}{3}=\frac{y+8}{1}=\frac{z-4}{1} [/tex]
    directly gives you the direction ratios of the required line, isnt it? What would they be?
  8. Jan 22, 2009 #7
    it does not give me the vectors to the line...
    there are an infinite number of possible normals to the original line
  9. Jan 22, 2009 #8
    Well, (3,1,1) are the direction ratios! And if this line passes through (4,5,5), substitue it in the general equation,
    \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} [/tex]

    where, (a,b,c) are the direction ratios and [tex](x_1,y_1,z_1)[/tex] is the point through which the line passes!
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