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 parallelogram $A B C D$ but it is not a rectangle.

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Solution

Direction ratios $= A B (- 2, - 4, - 4) = (1, 2, 2);$
$B C = (3, 5, 3) C D = (2, 4, 4) = (1, 2, 2);$
$A D = (3, 5, 3)$ . We can see $A B ∥ C D$ and $B C ∥ A D$
But $- - \to A B \cdot - - \to B C = 3 + 10 + 6 = 19 \neq 0, \Rightarrow A B$ is a
not perpendicular to $B C, \Rightarrow A B C D$ is a
parallelogram but not rectangle.

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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 parallelogram ABCD but it is not a rectangle.

Direction ratios =AB(−2,−4,−4)=(1,2,2);BC=(3,5,3)CD=(2,4,4)=(1,2,2);AD=(3,5,3). We can see AB∥CD and BC∥ADBut −−→AB⋅−−→BC=3+10+6=19≠0,⇒AB is anot perpendicular to BC,⇒ABCD is a parallelogram but not rectangle.

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 parallelogram $A B C D$ but it is not a rectangle.