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.

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)

Solution:

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