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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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2 Answers 2

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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I agree with Babji's result, and we can demonstrate this this is a correct position for point $ \ C \ $ , first because we find that

$$ \vec{BC} \ \centerdot \ \vec{BA} \ = \ \langle 4 - 1 \ , \ 5 - (-1) \ , \ (-4) - 5 \rangle \ \centerdot \ \langle 1 \ , \ -2 \ , \ -1 \rangle $$

$$ = \ \langle 3 \ , \ 6 \ , \ -9 \rangle \ \centerdot \ \langle 1 \ , \ -2 \ , \ -1 \rangle \ = \ 3 - 12 + 9 \ = \ 0 \ \ , $$

and that we can also confirm that the three points form a right triangle, since

$$ \parallel \vec{BA} \parallel^2 + \ \parallel \vec{BC} \parallel^2 = \ [ \ 1^2 + (-2)^2 + (-1)^2 \ ] \ + \ [ \ 3^2 + 6^2 + (-9)^2 \ ] \ = \ 6 \ + \ 126 \ = \ 132 $$

$$ = \ (4 - 2)^2 \ + \ (5 - [-3])^2 \ + \ ([-4] - 4)^2 \ = \ 2^2 \ + \ 8^2 \ + \ (-8)^2 \ = \ \parallel \vec{AC} \parallel^2 \ \ . $$

Having said this, it should also be noted that the statement of the problem doesn't indicate whether the right angle is located at vertex $ \ B \ $ , as is assumed in the existing answer, or at vertex $ \ A \ . $ So a second position for the third vertex, which we'll designate as $ \ C' \ $ , is also possible. The vector from $ \ A \ $ to the unknown point on line $ \ L \ $ is then

$$ \vec{AC'} \ = \ \langle \ (1 + 3t) - 2 \ , \ 5 - (-3) \ , \ (-2 - 2t) - 4 \ \rangle \ = \ \langle -1 + 3t \ , \ 8 \ , \ -6 - 2t \ \rangle \ . $$

We want to arrange that $ \ \vec{AC'} \ \perp \ \vec{AB} \ , $ so we require that

$$ \vec{AC'} \ \centerdot \vec{AB} \ = \ \langle -1 + 3t \ , \ 8 \ , \ -6 - 2t \ \rangle \ \centerdot \ \langle -1 \ , \ 2 \ , \ 1 \rangle $$

$$ = \ (1 - 3t) \ + \ 16 \ + \ (-6 - 2t) \ = \ 11 \ - \ 5t \ = \ 0 \ \ \Rightarrow \ \ t \ = \ \frac{11}{5} \ \ . $$

So $ \ C' \ $ is located at $ \ \left( \ 1 + 3 \cdot \frac{11}{5} \ , \ 5 \ , \ -2 - 2 \cdot \frac{11}{5} \ \right) \ = \ \left( \frac{38}{5} \ , \ 5 \ , \ - \frac{32}{5} \right) \ , $ producing the vector $ \ \vec{AC'} \ = \ \langle \ \frac{38}{5} - 2 \ , \ 5 - (-3) \ , \ - \frac{32}{5} - 4 \ \rangle \ = \ \langle \ \frac{28}{5} \ , \ 8 \ , \ - \frac{52}{5} \ \rangle \ . $

We can check this result, observing that

$$ \vec{AC'} \ \centerdot \vec{AB} \ = \ \langle \ \frac{28}{5} \ , \ 8 \ , \ - \frac{52}{5} \ \rangle \ \centerdot \ \langle \ -1 \ , \ 2 \ , \ 1 \ \rangle \ = \ -\frac{28}{5} + \ 16 \ - \frac{52}{5} $$

$$ = \ \frac{-28+80-52}{5} \ = \ 0 $$

and

$$ \parallel \vec{AB} \parallel^2 + \ \parallel \vec{AC'} \parallel^2 = \ 6 \ + \ \left[ \ \left( \frac{28}{5} \right)^2 + 8^2 + \left( -\frac{52}{5} \right)^2 \ \right] \ = \ 6 \ + \ 203.52 \ = \ 209.52 $$

$$ = \ \left( \frac{38}{5} - 1 \right)^2 \ + \ (5 - [-1])^2 \ + \ ( \ [-\frac{32}{5}] - 5 \ )^2 $$

$$ = \ \left( \frac{33}{5} \right)^2 \ + \ 6^2 \ + \ \left( -\frac{57}{5} \right)^2 \ = \ \parallel \vec{BC'} \parallel^2 \ \ . $$

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