Show that the points A(1, 2, 3), B(-1, -2, -1), C(2, 3, 2) and D (4, 7, 6) are the vertices of a parallelogram ABCD, but not a rectangle.

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Solution

AB = $\sqrt{(- 1 - 1)^{2} + (- 2 - 2)^{2} + (- 1 - 3)}$

= $\sqrt{4 + 16 + 16}$

=6 units

BC = $\sqrt{(2 + 1)^{2} + (3 + 2)^{2} + (2 + 1)^{2}}$

$= \sqrt{9 + 25 + 9}$

$= \sqrt{43}$ units

CD = $\sqrt{(4 - 2)^{2} + (7 - 3)^{2} + (6 - 2)^{2}}$

$\sqrt{4 + 16 + 16}$

6 units

DA = $\sqrt{(1 - 4)^{2} + (2 - 7)^{2} + (3 - 6)^{2}}$

$\sqrt{9 + 25 + 9}$

$\sqrt{43}$ units

AC = $\sqrt{(2 - 1)^{2} + (3 - 2)^{2} + (2 - 3)^{2}}$

$\sqrt{1 + 1 + 1}$

$\sqrt{3}$ units

BD = $\sqrt{(4 + 1)^{2} + (7 + 2)^{2} + (6 + 1)^{2}}$

$\sqrt{25 + 81 + 49}$

$\sqrt{155}$ units

Since,

AB = DC and BC = DA

So,

ABCD is a parallelogram

$A B^{2} + B C^{2} = 36 + 43 = 97$

$A C = \sqrt{3}$

$∴ A B^{2} + B C^{2} \neq A C^{2}$ i.e. $∠ B$ is not a right

angle. ABCD is not a rectangle.

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