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Two points $A$ and $B$ in $\mathbb{R}^3$ with origin $O$ are given in terms of a Cartesian coordinate system by $A = (1, 2, 3)$ and $B = (4, 5, −1)$.

How do you find the point $C$, such that $OACB$ are the vertices of a parallelogram (with $A$ and $B$ diagonally opposite each other).

I have worked out $|OA| = \sqrt{14}$, $|OB| = \sqrt{42}$, $|AB| = \sqrt{34}$.

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If the points of a parallelogram have coordinate vectors $O$, $A$ and $B$ (where $O$ is the origin), then the vertex opposite $O$ has coordinate vector $A+B$. This makes sense if you think in terms of vectors: it's what you get by translating the vector $B$ along the vector $A$ so that it's based at the point with coordinate vector $A$. This is illustrated below:


[Image source: Wolfram MathWorld.]

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Put $C(x,y,z)$. We have $\overrightarrow{OA}=(1,2,3)$ and $\overrightarrow{BC}=(x - 4, y - 5,z + 1)$. OACB are the vertices of a parallelogram (with A and B diagonally opposite each other) when and only when $\overrightarrow{OA}= \overrightarrow{BC}$, and then $$x - 4 = 1, \quad y - 5 = 2, \quad z + 1 = 3.$$ Thus $C(5,7,2).$

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