Build right triangle with two points and a line

[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)##

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?

 

[Jonathan Scott]

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]

 

[Kernul]

Do you remember how to tell whether two vectors are perpendicular?

Oh yes, the scalar product must be ##0##.

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.

 

[Jonathan Scott]

##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]

 

[Kernul]

##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.

 

[Jonathan Scott]

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.

 

[Jonathan Scott]

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)##