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Intersection points of two lines in two-space

  • Thread starter adrimare
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Homework Statement



Show that the following pairs of lines intersect. Determine the coordinates of the point of intersection.

L1: r= (-3,-1) + t(3,4)

L2: r= (6,2) + s(3,-2)

Homework Equations



?

The Attempt at a Solution



I know that eventually the two lines will reach the same point at which they will intersect, but I'm not really sure how to get to that point. The x and y of L1 has to equal the x and y of L2, right? Do I have to use the direction vectors somehow or guess and check?
 

Answers and Replies

  • #2
gabbagabbahey
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. The x and y of L1 has to equal the x and y of L2, right?
Right.:approve:

You can combine those into a single equation very easily. You are given an expression for the position vector of any point an both lines. At the point point of intersection, those position vectors must be the same (otherwise they represent different points). That means, that for some specific values of [itex]s[/itex] and [itex]t[/itex], you must have

[tex](-3,-1)+t(3,4)=(6,2)+s(3,-2)[/itex]

Solve that equation for [itex]s[/itex] and [itex]t[/itex].
 
  • #3
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How would you solve for that, though? I have made them into parametric equations and tried to solve so that both x's and y's have the same s and t values. But there are two variables. Without guess and check, how can I solve for these values? The answer is supposed to be (1.5,5) according to the answer section of the book.
 
  • #4
gabbagabbahey
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Each component gives you an equation, so you will have two equations and two unknowns. Solving that system of equations is something you should have learned how to do in High School algebra.
 
  • #5
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Am I wrong in using parametric equations? So far, I have tried the substitution method because I can't think of another way to do it. For the x-value of the first line,
I got -3 + 3t = 6 + 3s.
I simplified to get t = (9 + 3s)/3, which turns into t = 3 + s. I went to the y-value of the first line. This has to have the same t-value as the x-value, right? I got -1 + 4t = 2- 2s, which turns into 4t = 3 - 2s, which becomes t = (3-2s)/4. The t-values are not the same. Did I do something wrong or am I completely off track here?
 
  • #6
gabbagabbahey
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The t-values are not the same.
They will be the same for a specific value of [itex]s[/itex]...set them equal to each other and solve for that value.
 

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