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Build right triangle with two points and a line

  1. Oct 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Given the points ##A (1, -1, 0)## and ##B (4, 0, 6)##, find the point ##P## of the line ##s## so that the triangle ##ABP## is a right triangle in ##B##. Calculate the area of the triangle.
    ##s : \begin{cases}
    x = 1 + 4t \\
    y = 2 - 3t \\
    z = 3
    \end{cases}##
    ##\vec v_s = (4, -3, 0)##
    2. Relevant equations


    3. The attempt at a solution
    I already know how to calculate the area of a triangle. The problem is actually finding the point that makes the triangle. I really have no idea on how to proceed in this case. I was thinking about finding the distances of the two points from the line but I don't think that would bring me anywhere. I tried finding the directional vector of the line that passes through the two points, which is ##\vec v_{AB} = (3, 1, 6)##, and find the relation between the two lines, maybe finding the intersection point too but it's useless. I really don't know where I should start. Any hint on how I should proceed?
     
  2. jcsd
  3. Oct 12, 2016 #2

    Jonathan Scott

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    Gold Member

    Do you remember how to tell whether two vectors are perpendicular? You've found a vector from A to B. You now need to find a vector from B to P and write down the condition that it is perpendicular to the vector from A to B, which will give you some equations that you can solve.

    [Edited to correct typo]
     
  4. Oct 12, 2016 #3
    Oh yes, the scalar product must be ##0##.
    Oh! So I would have something like this:
    ##(3, 1, 6) \cdot (a - 4, b - 0, c - 6) = 0## with ##P (a, b, c)##
    Am I right? But this would give me just one equation with three unknowns.
     
  5. Oct 12, 2016 #4

    Jonathan Scott

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    Gold Member

    ##P## is known to be a point on the line ##s##, so you can use the general expression for such a point, then solve for ##t##.

    [Edited again to fix a typo]
     
  6. Oct 12, 2016 #5
    I end up with ##t = \frac{25}{11}## and substituting this in the equations of ##s## I end up with this point ##(\frac{111}{11}, -\frac{53}{11}, 3)##. Is it correct? Because if I try then with those coordinates, I don't have a scalar product equal to zero.
     
  7. Oct 12, 2016 #6

    Jonathan Scott

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    Gold Member

    That was a sensible sanity check. If your scalar product isn't zero, you've obviously got an arithmetic error, as the choice of ##t## was specifically to make it zero.
     
  8. Oct 12, 2016 #7

    Jonathan Scott

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    Gold Member

    Just to help you narrow it down, I'll say that the 11 in your value 25/11 for ##t## is incorrect.
     
  9. Oct 12, 2016 #8
    Thank you! I found the mistake. It was ##t = \frac{25}{9}##.
    So the point is ##P (\frac{109}{9}, -\frac{19}{3}, 3)##
     
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