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Equation of 3d vector question

  1. Dec 27, 2006 #1
    1. The problem statement, all variables and given/known data

    Find an equation of the line through the point A (4,5,5) which meets the line (x-11)/3 = (y+8)/-1 = z-4 at right angles.

    2. Relevant equations

    3. The attempt at a solution

    well i know that the direction of the given line is d(3,-1,1) and the original line crosses through the point (11,-8,4)

    The direction of the new line must be able to dot with the old line to get 0 (thus right angles).

    I dont know how to find a direction that i know will pass through A.

    Help or hint please
  2. jcsd
  3. Dec 28, 2006 #2


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    You can write any point P on the given line in parametric form, e.g. (x = t, y = a function of t, z = another function of t). Then you can write the direction of AP in terms of t.
  4. Dec 28, 2006 #3
    So, you have point A and point P (P on the line). what can you do with the orthogonal projection of AP onto the given line? from there, how can you find a vector that points from the line to A and that is orthogonal to <3,-1,1> ?
  5. Dec 28, 2006 #4


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    There are two common ways to write a vector equation of a line:

    i. Given two points, say P and Q, an equation of the line through P and Q (in that order as t ranges over [0,1]) is given by

    [tex]\vec{r}(t)=\vec{Q}t+\vec{P}(1-t)[/tex] or [tex]\vec{r}(t)=\vec{P}+\left(\vec{Q}-\vec{P}\right) t [/tex]​

    notice that [tex]\vec{Q}-\vec{P}[/tex] is a vector in the direction of the line, hence

    ii. Given a point and a vector in the direction of a line, say point P and vector [tex]\vec{v}[/tex], we have an equation of the line through P and parallel to [tex]\vec{v}[/tex] is given by

    [tex]\vec{r}(t)=\vec{P}+\vec{v} t [/tex]​
  6. Dec 28, 2006 #5


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    Any vector, <a,b,c>, perpendicular to <3, -1, 1> must satisfy 3a-b+ c= 0 or c= b- 3a. Such a line passing through (4, 5, 5) must be of the form (x-4)/a= (y- 5)/b= (z-5)/(b- 3a). Now, what are the conditions on a and b, in order to ensure that (x-4)/a= (y- 5)/b= (z-5)/(b- 3a) intersects (x-11)/3 = (y+8)/-1 = z-4 ?
  7. Jan 1, 2007 #6
    couple questions

    Any vector, <a,b,c>, perpendicular to <3, -1, 1> must satisfy 3a-b+ c= 0

    shouldnt there be a D value such that 3a-b+c+d=0

    and how do i make sure they intersect
  8. Jan 3, 2007 #7

    cant I just make up any A and B values and it will work?
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