# Simple Linear Algebra Problem

I am working on a question from the first chapter of a Linear Algebra textbook I'm reading.

Let $A=(1,1,-1)$, $B=(-3,2,-2)$, and $C=(2,2,-4)$. Prove that $\Delta ABC$ is a right-angled triangle.

I know that the angle between $\overrightarrow{AB}$ and $\overrightarrow{AC}$ must be $90°$. In other words, $\overrightarrow{AB} \cdot \overrightarrow{AC}=0$. My trouble is I cant translate these given points into vectors in order to show the necessary calculations. I'm having the same trouble with other questions. Any help would be appreciated.

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Shift things so that the right angle is at the origin before taking dot products. – J. M. Jul 15 '12 at 18:00

$$\vec{AB} = (-3,2,-2) - (1,1,-1) = (-4,1,-1)$$ $$\vec{BC} = (2,2,-4) - (-3,2,-2) = (5,0,-2)$$ $$\vec{CA} = (1,1,-1) - (2,2,-4) = (-1,-1,3)$$ Now look at $\vec{AB} \cdot \vec{CA}$ to conclude that the right angle is at $A$.
If $U$ and $V$ are two different vectors, then the vector from $U$ to $V$ is given by $V-U$.