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Given three radius-vectors: $OA(5; 1; 4), OB(6;2;3), OC(4;2;4)$, find the missing vertex $D$ and calculate the area of obtained parallelogram.

My attempt: Firstly, we are to find the vectors which form the parallelogram. $$AB = OA - OB = (-1; -1; 1)$$ $$AC=OA-OC=(1; -1; 0)$$ Since the parallelogram is spanned by $AB, AC$, cross product and the norm of resulting vector should be then calculated. \begin{align*} &AB \times AC= {\begin{vmatrix} \vec{i} & \vec{j} & k \\ -1 & -1 & 1 \\ 1 & -1 & 0 \end{vmatrix}} = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6} \end{align*} $D$ is equidistant from $E$ and $B$, where $E$ is the point of diagonal's intersection. $$x_E = \frac{x_A+x_C}{2} = 9/2$$ $$y_E = \frac{y_A+y_C}{2} = 3/2$$ $$z_E = \frac{z_A+z_C}{2} = 4$$

\begin{cases} x_B - x_E = x_E - x_D\\ y_B - y_E = y_E - y_D \\ z_B - z_E = z_E - z_D \end{cases} After plugging and solving, the final answer would be $\sqrt{6}$ and $D(3;1;5).$

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As other answers point out, you’ve made some tacit assumptions in your solution. For the problem as you’ve stated it, there are three possible paralellograms, each one having a different side of $\triangle{ABC}$ as one of its diagonals. However, since each of these possibilities consists of two copies of the triangle glued together along one of its sides, all of their areas are equal, and you can compute it via the norm of a cross product as you’ve done.

To compute the position of the missing vertex $D$, a simpler way than your approach is to take advantage of the fact that opposite sides of a paralellogram are both parallel and have the same length. So, if $D$ is opposite $A$, then $\overrightarrow{AC}=\overrightarrow{BD}$ and $\overrightarrow{CD}=\overrightarrow{AB}$, and so $D$ must be at $B+\overrightarrow{AC} = C+\overrightarrow{AB}$ (and similarly for the other two possible locations of $D$). The one of the three that you’ve computed puts $D$ opposite $B$.

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A simpler way to obtain $D$ is $D = C+AB$.

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This may not have been spelled out explicitely in the problem statement, but your are missing the information in what order the vertices $A,B,C,D$ form the parallelogram. Basically the question is which are the 2 sets of opposing vertices: $\{\{A,B\}, \{C,D\}\}, \{\{A,C\}, \{B,D\}\}$ or $\{\{A,D\}, \{B,C\}\}$?

Probably owing to this, you assumed two different things for the 2 parts of the problem, for the area calculation you consider $AB$ and $AC$ neighboring sides of the parallelogram, meaning that $B$ and $C$ are opposites, as well as $A$ and $D$, respectively.

However for finding $D$, you assumed that $D$ would be opposite of $B$!

So you got two answers, each of which is correct for one choice of "layout" for the parallelogram. If you want to stick to your first layout (for the area calculation), then you can do the same as above, just consdering that the diagonal intersection point is the midpoint of both $BC$ (which you know) and of $AD$ (which you can use just as you did above to calculate the coordinates of $D$).

If you want to stick to the second layout (for the $D$ calculation, which uses the 'usual' layout of ordering $A$ to $D$ around the edges of the polygon), you'd have to calculate $\overrightarrow{AD}$ to use in the cross product with $\overrightarrow{AB}$.

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Despite this, the ideas and execution are OK, just one remark:

It doesn't matter in this case but may be vital in others: If $AB$ is supposed to be a vector (as in this case), you get it's value by subtracting the radius vector of the starting point from the ending point:

$$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (1;1;-1)$$

You got the inverse vector in your calculation. Similiarly, $\overrightarrow{AC}=(-1;1;0)$

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  • $\begingroup$ There are in fact three possible paralellograms here, but all of them have an area equal to twice the area of $\triangle{ABC}$. The “layout” doesn’t matter for that calculation. $\endgroup$ – amd Jan 21 at 21:47
  • $\begingroup$ Right, good observation! $\endgroup$ – Ingix Jan 21 at 22:29

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