# Analytic geometry simple question

need help in this.

In a right triangle in the three-deminsion plane $ABC$, A=$(2,-3,4)$, B=$(1,-1,5)$. Find $C$ if its known $C$ is on the line $L: (1,5,-2)+t(3,0,-2)$.

What I did was finding the direction vector $\vec {AB}$ but saying that $\vec AB \cdot (a,b,c)=0$, as $(a,b,c)$ represents the direction vector $\vec {BC}$ (because its a right-triangle).

So I got $a=2b+c$. Why can't I choose $b$ and $c$ to have value as I like? My idea is to find the line $BC$ and then find the intersection point with $L: (1,5,-2)+t(3,0,-2)$. But I'm having trouble with finding BC's direction vector. What am I doing wrong?

Edit: An idea: Having a general point of the line L, and then finding the vector from that general point to B, and then find its dot product with $\vec AB$ (which equals 0).

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BA=i-2j-k Let C be (a,b,c). Since c is on line L a=1+3t b=5 c=-2-2t

Eliminating t, we get

2a+3c=-4 b=5

Now BA.BC must be 0 since the 2 vectors are perpendicular

BC = (a-1)i+(b+1)j+(c-5)k

doing the dot product, we get the 3rd equation a-2b-c = -2

Solving for a,b,c we get from the 3 equations

a=4, b=5, c=-4 so C is (4,5,-4)

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