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I am given four vectors in three-dimensional space and I want to check if they form a parallelogramm. The vectors are:

$$\vec{p}=\begin{pmatrix}2 \\ 0\\3 \end {pmatrix},\vec{q}=\begin{pmatrix}8 \\ 1\\1 \end {pmatrix},\vec{r}=\begin{pmatrix}6 \\ -2\\-1 \end {pmatrix},\vec{s}=\begin{pmatrix}12 \\ -1\\-3 \end {pmatrix}$$

I read this answer where Jack D'Aurizio states:

The convex quadrilateral $ABCD$ is a parallelogram if and only if the midpoint of $AC$ is the midpoint of $BD$, too. Hence you just need to compute $\binom{4}{2}=6$ midpoints and check that two of them are the same point.

I tried to use this and computed all the line segment joining the vertices of the parallelogramm by subtracting the vectors.

$$\vec{p}-\vec{q}=(-6,-1,2)^T \\ \vec{p}-\vec{r}=(-4,2,4)^T \\ \vec{p}-\vec{s}=(-10,1,6)^T \\ \vec{q}-\vec{r}=(2,3,2)^T \\ \vec{q}-\vec{s}=(-4,2,4)^T \\ \vec{r}-\vec{s}=(-6,-1,2)^T$$

I am not sure how to continue now. Also, I wasn't sure if the order of vector subtraction matters here. Any ideas?

Nullspace
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Notice that $$\frac{\vec{p}+\vec{s}}2 = \frac{\vec{q}+\vec{r}}2$$

Hence $PQSR$ is a parallelogram.

That is we have $\vec{p}-\vec{q} = \vec{r}-\vec{s}$ and $\vec{p}-\vec{r}=\vec{q}-\vec{s}$.

Siong Thye Goh
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  • Thank you for your answer. This might be a stupid question but how do I know that:

    $$\text{PQRS is a parallelogramm} \iff \frac{\vec{p}+\vec{s}}2 = \frac{\vec{q}+\vec{r}}2$$

     – Nullspace Jan 06 '19 at 11:45
  • I will check that out and come back if I have any more questions. Thanks again. – Nullspace Jan 06 '19 at 11:50