# Build right triangle with two points and a line

• Kernul
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)##.
Kernul

## 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)##

## 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?

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]

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.

##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]

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.

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.

Kernul
Just to help you narrow it down, I'll say that the 11 in your value 25/11 for ##t## is incorrect.

Kernul
Thank you! I found the mistake. It was ##t = \frac{25}{9}##.
So the point is ##P (\frac{109}{9}, -\frac{19}{3}, 3)##

Jonathan Scott

## 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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