Finding the Fourth Vertex of a Parallelogram in 3D

The problem asks us to find the coordinates of the fourth vertex, \(D\), of a parallelogram \(ABCD\), given the coordinates of three consecutive vertices: \(A(1, -1, 0)\), \(B(-2, 1, 8)\), and \(C(-1, 2, 7)\).

Understanding Parallelogram Properties

A key property of a parallelogram is that its diagonals bisect each other. This means the midpoint of diagonal \(AC\) is the same as the midpoint of diagonal \(BD\). We can use this property, or vector properties, to find the coordinates of vertex \(D\).

Method 1: Using the Midpoint Formula

Let the coordinates of the fourth vertex \(D\) be \((x, y, z)\).

Step 1: Find the midpoint of diagonal AC.

The midpoint formula for two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)\).

Midpoint of \(AC = \left(\frac{1 + (-1)}{2}, \frac{-1 + 2}{2}, \frac{0 + 7}{2}\right)\)

Midpoint of \(AC = \left(\frac{0}{2}, \frac{1}{2}, \frac{7}{2}\right) = \left(0, \frac{1}{2}, \frac{7}{2}\right)\)

Step 2: Find the midpoint of diagonal BD.

Midpoint of \(BD = \left(\frac{-2 + x}{2}, \frac{1 + y}{2}, \frac{8 + z}{2}\right)\)

Step 3: Equate the midpoints to find D.

Since the midpoints are the same:

Equating the x-coordinates: \(\frac{-2 + x}{2} = 0 \implies -2 + x = 0 \implies x = 2\)

Equating the y-coordinates: \(\frac{1 + y}{2} = \frac{1}{2} \implies 1 + y = 1 \implies y = 0\)

Equating the z-coordinates: \(\frac{8 + z}{2} = \frac{7}{2} \implies 8 + z = 7 \implies z = -1\)

Therefore, the coordinates of vertex \(D\) are \((2, 0, -1)\).

Method 2: Using Vector Equality

In a parallelogram \(ABCD\), the vector from \(A\) to \(B\) is equal to the vector from \(D\) to \(C\). That is, \(\vec{AB} = \vec{DC}\).

Step 1: Calculate the vector \(\vec{AB}\).

\(\vec{AB} = B - A = (-2 - 1, 1 - (-1), 8 - 0) = (-3, 2, 8)\)

Step 2: Express the vector \(\vec{DC}\).

Let \(D = (x, y, z)\).

\(\vec{DC} = C - D = (-1 - x, 2 - y, 7 - z)\)

Step 3: Equate the vectors \(\vec{AB}\) and \(\vec{DC}\).

Equating the components:

x-component: \(-3 = -1 - x \implies x = -1 + 3 \implies x = 2\)

y-component: \(2 = 2 - y \implies y = 2 - 2 \implies y = 0\)

z-component: \(8 = 7 - z \implies z = 7 - 8 \implies z = -1\)

Again, we find that the coordinates of vertex \(D\) are \((2, 0, -1)\).

Alternatively, we can use the property \(\vec{AD} = \vec{BC}\).

Step 1: Calculate the vector \(\vec{BC}\).

\(\vec{BC} = C - B = (-1 - (-2), 2 - 1, 7 - 8) = (1, 1, -1)\)

Step 2: Express the vector \(\vec{AD}\).

Let \(D = (x, y, z)\).

\(\vec{AD} = D - A = (x - 1, y - (-1), z - 0) = (x - 1, y + 1, z)\)

Step 3: Equate the vectors \(\vec{AD}\) and \(\vec{BC}\).

Equating the components:

x-component: \(1 = x - 1 \implies x = 1 + 1 \implies x = 2\)

y-component: \(1 = y + 1 \implies y = 1 - 1 \implies y = 0\)

z-component: \(-1 = z \implies z = -1\)

This method also yields \(D = (2, 0, -1)\).

Conclusion

Both methods confirm that the fourth vertex \(D\) of the parallelogram \(ABCD\) is located at the coordinates \((2, 0, -1)\).