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Ncert Solutions Maths class 11th Maths Class 11th Introduction to Three Dimensional Geometry

15. Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex.

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Payal Gupta | Contributor-Level 10

4 months ago

15. Let D(x, y, z) be the fourth vertex of the parallelogram ABCD.

In a parallelogram, the diagonal AC and BD bisects each other at point say O.

=> $(\frac{3 - 1}{2}, \frac{- 1 + 1}{2}, \frac{2 + 2}{2}) = (\frac{1 + x}{2}, \frac{2 + y}{2}, \frac{- 4 + z}{2})$

=> (1, 0, 2) = $(\frac{1 + x}{2}, \frac{2 + y}{2}, \frac{- 4 + z}{2})$

Equating the coordinates we get,

$\frac{1 + x}{2}$ = 1

=> $1 + x = 2$

=> $x = 1$

And

$\frac{2 + y}{2}$ = 0

=> $y = - 2$

And

$\frac{- 4 + z}{2}$ = 2

=> $- 4 + z = 4$

=> $z = 4 + 4$

=> $z = 8$

So, coordinates of fourth vertex is (1, –2, 8)

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15. Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex.

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