Three consecutive vertices of parallelogram ABCD are given…

Three consecutive vertices of parallelogram $ABCD$ are given: $A(1,-2,0)$, $B(2,1,3)$, $C(-2,0,5)$. Find the fourth vertex $D$ and the length of diagonal $BD$.

Step 1: Property used

In a parallelogram $\overrightarrow{AD} = \overrightarrow{BC}$ (equivalently $\overrightarrow{AB} = \overrightarrow{DC}$), and the diagonals bisect each other.

Step 2: Method 1: $\overrightarrow{AD} = \overrightarrow{BC}$

\[ \overrightarrow{BC} = (-2-2,\ 0-1,\ 5-3) = (-4,-1,2) \]

\[ D = A + \overrightarrow{BC} = (1,-2,0) + (-4,-1,2) = (-3,-3,2) \]

\[D(-3,\,-3,\,2)\]

Step 3: Method 2: common midpoint of the diagonals

\[ M = \text{mid}(AC) = \left(-\frac{1}{2},\,-1,\,\frac{5}{2}\right), \qquad D = 2M - B = (-1-2,\ -2-1,\ 5-3) = (-3,-3,2) \;\checkmark \]

Step 4: Verification

\[ \overrightarrow{AB} = (1,3,3) = \overrightarrow{DC}, \qquad \overrightarrow{AD} = (-4,-1,2) = \overrightarrow{BC} \;\checkmark \]

Step 5: Diagonal $BD$

\[ \overrightarrow{BD} = (-3-2,\ -3-1,\ 2-3) = -5\vec{i}-4\vec{j}-\vec{k} \]

\[|\overrightarrow{BD}| = \sqrt{25+16+1} = \sqrt{42} \approx 6.481\]

For comparison, $\overrightarrow{AC} = -3\vec{i}+2\vec{j}+5\vec{k}$ and $|AC| = \sqrt{38}$; the diagonals are unequal, so $ABCD$ is not a rectangle. Sides: $|AB| = \sqrt{19}$, $|AD| = \sqrt{21}$.