Triangle Vertices on Two Line Segments

We first note that segments A and B are perpendicular, since the dot product $\langle 1,2,3 \rangle \cdot \langle -4,-1,2 \rangle = 0$ . Letting $P$ and $Q$ be the points chosen on A , and $R$ the point chosen on B , we get the diagram below:

Then the triangle's area is clearly $\dfrac{1}{2} (PQ)(OR)$ . Segments A and B have lengths $\sqrt{14}$ and $\sqrt{21}$ respectively. Since $R$ is chosen uniformly on B , the expected length of $OR$ is clearly $\dfrac{\sqrt{21}}{2}$ . We claim that the expected length of $PQ$ is one-third of the length of A , i.e. $\dfrac{\sqrt{14}}{3}$ , by the argument below.

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Suppose we uniformly choose three random points $X, Y$ and $Z$ on a line segment of length $l$ , and let the expected distance between $X$ and $Y$ be $d$ ; then the probability $Z$ lies between $X$ and $Y$ is $\dfrac{d}{l}$ . But the probability that $Z$ lies between $X$ and $Y$ must be $\dfrac{1}{3}$ , as the possibilities that $Z$ lies between $X$ and $Y$ , $X$ lies between $Z$ and $Y$ , and $Y$ lies between $X$ and $Z$ , must all be equally likely. Thus $\dfrac{d}{l} = \dfrac{1}{3}$ , i.e. $d = \dfrac{l}{3}$ .

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Therefore the expected area of triangle $PQR$ is $\dfrac{1}{2} (PQ)(OR) = \dfrac{1}{2} \left( \dfrac{\sqrt{14}}{3} \right) \left( \dfrac{\sqrt{21}}{2} \right) = \boxed{\dfrac{7 \sqrt{6}}{12}}$