First slide

Introduction to 3-D Geometry

Question

Three vertices of a parallelogram ABCD are A(1,2,3), B(- 1, - 2, - 1) and C(2,3,2). Find the fourth vertex D.

Easy

A

(- 4, - 7, - 6)
B

(4, 7, 6)
C

(4, 7, - 6)
D

None of these

Solution

Let the fourth vertex be (x, y, z).
We know that diagonals of a parallelogram are bisecting to each other. i.e. mid-point of a diagonals are coinciding.
Mid-point of diagonal AC = Mid-point of diagonal BD
$\begin{array}{l} ∴ (\frac{1 + 2}{2}, \frac{2 + 3}{2}, \frac{3 + 2}{2}) = (\frac{- 1 + x}{2}, \frac{- 2 + y}{2}, \frac{- 1 + z}{2}) \\ \Rightarrow (\frac{3}{2}, \frac{5}{2}, \frac{5}{2}) = (\frac{- 1 + x}{2}, \frac{- 2 + y}{2}, \frac{- 1 + z}{2}) \\ \Rightarrow \frac{3}{2} = \frac{- 1 + x}{2}, \frac{5}{2} = \frac{- 2 + y}{2}, \frac{5}{2} = \frac{- 1 + z}{2} \\ \Rightarrow x = 4, y = 7, z = 6 \end{array}$
Hence, required point is (4, 7, 6)

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