parallelogram had vertices A(3,-1,0) B(6,2,3) C(4,6,8). Find the coordinates of vertex D.

Answer 1

We can use the fact that opposite sides of a parallelogram are parallel and have the same length.

First, we can find the vectors representing two of the sides:

$\vec{AB} = \begin{pmatrix} 6-3 \\ 2-(-1) \\ 3-0 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix}$

$\vec{AC} = \begin{pmatrix} 4-3 \\ 6-(-1) \\ 8-0 \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 8 \end{pmatrix}$

To find $\vec{AD}$, we need to add one of these vectors to the vector representing the diagonal $\vec{BD}$. Since $\vec{AB}$ and $\vec{CD}$ are parallel, we can find the vector $\vec{CD}$ by adding $\vec{AB}$ to $\vec{AC}$:

$\vec{CD} = \vec{AC} + \vec{AB} = \begin{pmatrix} 1 \\ 7 \\ 8 \end{pmatrix} + \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 10 \\ 11 \end{pmatrix}$

Now we can find $\vec{BD}$ by subtracting $\vec{AB}$ from $\vec{CD}$:

$\vec{BD} = \vec{CD} - \vec{AB} = \begin{pmatrix} 4 \\ 10 \\ 11 \end{pmatrix} - \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 8 \end{pmatrix}$

Finally, we can find the coordinates of vertex D by adding $\vec{BD}$ to the coordinates of B:

$D = B + \vec{BD} = \begin{pmatrix} 6 \\ 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 1 \\ 7 \\ 8 \end{pmatrix} = \boxed{\begin{pmatrix} 7 \\ 9 \\ 11 \end{pmatrix}}$