Build right triangle with two points and a line

In summary: My apologies for the long summary, but the conversation involves solving a problem involving finding the point of a line to create a right triangle and calculating its area. The conversation also discusses using the scalar product and finding the intersection point of two lines. The final answer is found to be ##P (\frac{109}{9}, -\frac{19}{3}, 3)##.
  • #1
Kernul
211
7

Homework Statement


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)##

Homework Equations

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?
 
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  • #2
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]
 
  • #3
Jonathan Scott said:
Do you remember how to tell whether two vectors are perpendicular?
Oh yes, the scalar product must be ##0##.
Jonathan Scott said:
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]
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.
 
  • #4
##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]
 
  • #5
Jonathan Scott said:
##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]
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.
 
  • #6
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.
 
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  • #7
Just to help you narrow it down, I'll say that the 11 in your value 25/11 for ##t## is incorrect.
 
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  • #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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Related to Build right triangle with two points and a line

1. How do I find the third point to complete the triangle?

The third point can be found by drawing a perpendicular line from one of the given points to the given line. The intersection of this perpendicular line and the given line will be the third point.

2. Can I use any two points and any line to create a right triangle?

Yes, as long as the given line is not parallel to the line connecting the two points, a right triangle can be constructed by drawing a perpendicular line from one of the given points to the given line.

3. How do I know if the triangle I've constructed is a right triangle?

A right triangle has one angle that measures 90 degrees. You can use a protractor to measure the angles of the triangle and determine if one of them is 90 degrees.

4. What if the two given points are on the given line?

In this case, it is not possible to construct a right triangle with those two points and the given line. The third point would have to be located on the given line, which would result in a degenerate triangle.

5. Is there a formula for finding the length of the third side of the right triangle?

Yes, the Pythagorean theorem can be used to find the length of the third side of a right triangle. The formula is c = √(a^2 + b^2), where c is the length of the hypotenuse and a and b are the lengths of the other two sides.

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