parallelogram had vertices A(3,-1,0) B(6,2,3) C(4,6,8). Find the coordinates of vertex D.
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:
$D = B + \vec{BD} = \begin{pmatrix} 6 \\ 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 1 \\ 7 \\ 8 \end{pmatrix} = \boxed{\begin{pmatrix} 7 \\ 9 \\ 11 \end{pmatrix}}$
To find the coordinates of vertex D, we know that a parallelogram has opposite sides that are equal in length and parallel to each other.
Let's denote vector AC as vector u and vector AB as vector v.
u = C - A = (4, 6, 8) - (3, -1, 0) = (1, 7, 8)
v = B - A = (6, 2, 3) - (3, -1, 0) = (3, 3, 3)
Since opposite sides of a parallelogram are parallel, we can add these vectors to any of the given vertices to find vertex D.
D = A + u + v
D = (3, -1, 0) + (1, 7, 8) + (3, 3, 3)
= (3 + 1 + 3, -1 + 7 + 3, 0 + 8 + 3)
= (7, 9, 11)
Therefore, the coordinates of vertex D are (7, 9, 11).