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 $P (\frac{109}{9}, -\frac{19}{3}, 3)$