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The question is

Three vertices of a parallelogram ABCD are A(3,-1,2), B(1,2,-4) and C(-1,1,2). Find the coordinate of the fourth vertex.

To get the answer I tried the distance formula, equated AB=CD and AC=BD.

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  • $\begingroup$ Doesn't this have more than one solution, depending on the order of the vertices in the parallelogram or is D meant to be connected to A and C? $\endgroup$ Commented Dec 19, 2012 at 8:35
  • $\begingroup$ A upper left, B upper right, C lower right, D lower left. $\endgroup$
    – chndn
    Commented Dec 19, 2012 at 8:38
  • $\begingroup$ $AC$ and $BD$ must have the same mid point. The mid point of $AC$ is $(1,0,2)$ and hence $D$ is $(1, -2, 8)$ $\endgroup$
    – user348749
    Commented Aug 9, 2016 at 6:18

3 Answers 3

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If you have a parallelogram ABCD, then you know the vectors $\vec{AB}$ and $\vec{DC}$ need to be equal as they are parallel and have the same length. Since we know that $\vec{AB}=(-2,\,3,-6)$ you can easily calculate $D$ since you (now) know $C$ and $\vec{CD}(=-\vec{AB})$. We get for $\vec{0D}=\vec{0C}+\vec{CD}=(-1,\,1,\,2)+(\,2,-3,\,6)=(\,1,-2,\,8)$ and hence $D(\,1,-2,\,8)$.

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You don't need to consider distance. You need to compute the difference vectors between adjacent sides. For instance, compute the difference vector from $B$ to $A$ and then add that to $C$.

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use mid point formulae let point be $d(a,b,c)$ so, $$\frac{(a+1)}2 = \frac{(3-1)}2$$ so $a=1$ similarly find c ad b as well

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