Form a Parallelogram by 4 Points

Question

This is a question from my school. The following is the whole question.

The vertices of a triangle $A$, $B$ and $C$ are given by the points $(-1, 0, 2)$, $(0, 1, 0)$, $(1, -1, 0)$ respectively. Find point $D$ so that the figure $ABCD$ forms a plane parallelogram.

I have no idea with points in 3D.

Would you mind helping me solve the question and explain to me in detailed.

Thank you for your attention.

Answer 1

There are three possible responses, depending on the choice of the first point in the next procedure:

Let's say we choose point $A$ to start:

choose one of the other points, $B$ or $C$. It doesn't matter as the result will be the same whatever is the choice. Let's say you choose point $B$. Then calculate:

$V = B - A$. (With this $V$ must be parallel and same length as the "hidden" side of the parallelogram).

Then add this vector $V$ to the other point ($C$ in this case). This will be the point $D$ you are looking for, so: $$ D = C + V = C + B - A\tag{1} $$ If you choose $C$ instead of $B$ you would have: \begin{alignat}{2} V &= C - A\\ D &= B + V &&{}= B + C - A \end{alignat} as you can see same result as in (1).

You can repeat it starting with $B$ and $C$ points.

So, you have three possible solutions: \begin{align} D_1 &= B + C - A\\ D_2 &= A + C - B\\ D_3 &= A + B - C \end{align}

Answer 2

I assume that you never saw a definition of sum beetween vectors through the parallelogram rule. That, together with this operation's properties, would give you an immediate answer.

But also if you don't know nothing about it, you can answer this question by giving some conditions to your fourth point $D$. First you should require $ABCD$ to be on the same plane; the cartesian equation of a generic plane is given by: $$ax+by+cz=d.$$ You know that $A,B,C$ belong to this plane and this is enough to find $a,b,c,d$. The second condition is that segment $AB=CD$ and segment $AC=BD$, that is$$(x-x_C)^2+(y-y_C)^2+(z-z_C)^2=(x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2,$$ $$(x-x_B)^2+(y-y_B)^2+(z-z_B)^2=(x_C-x_A)^2+(y_C-y_A)^2+(z_C-z_A)^2.$$ Thinking about the problem geometrically, you should see that there's only a point $D:(x,y,z)$ that can satisfy the above three equations, so you can simply check that the point: $$D:(x_C+x_B-x_A,y_C+y_B-y_A,z_C+z_B-z_A),$$ does (first you have to find the parameters $a,b,c$ of the plane equation).

A look at the parallelogram rule will show you why this works.