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Miscellaneous Examples · Example 7

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

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✓ Free question

The four points form a parallelogram because the midpoints of the diagonals coincide, and it is not a rectangle because the adjacent sides are not perpendicular — their dot product is non-zero.

We need to check two things: first, that these four points are the vertices of a parallelogram (in the order A→B→C→DA \to B \to C \to D), and second, that this parallelogram is not a rectangle.


1. The key idea — what makes a parallelogram?

A quadrilateral ABCDABCD is a parallelogram if and only if its diagonals bisect each other. That is, the midpoint of diagonal ACAC must equal the midpoint of diagonal BDBD. This is often simpler than checking that opposite sides are parallel and equal — one calculation instead of four.

Tip

The midpoint condition works for any order of vertices, but here the order is given as A,B,C,DA, B, C, D. So we check diagonals ACAC and BDBD.


2. Find the midpoints

Midpoint of ACAC:

MAC=(1+22,2+32,3+22)=(32,52,52)M_{AC} = \left( \frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2} \right) = \left( \frac{3}{2}, \frac{5}{2}, \frac{5}{2} \right)

Midpoint of BDBD:

MBD=(−1+42,−2+72,−1+62)=(32,52,52)M_{BD} = \left( \frac{-1+4}{2}, \frac{-2+7}{2}, \frac{-1+6}{2} \right) = \left( \frac{3}{2}, \frac{5}{2}, \frac{5}{2} \right)

They are identical. So the diagonals bisect each other, and ABCDABCD is a parallelogram.


3. Now check if it is a rectangle

A rectangle is a parallelogram with all angles 90∘90^\circ. For that, we need adjacent sides to be perpendicular. Let’s take sides ABAB and BCBC.

Vectors:

AB⃗=B−A=(−1−1,−2−2,−1−3)=(−2,−4,−4)\vec{AB} = B - A = (-1-1, -2-2, -1-3) = (-2, -4, -4)

BC⃗=C−B=(2−(−1),3−(−2),2−(−1))=(3,5,3)\vec{BC} = C - B = (2-(-1), 3-(-2), 2-(-1)) = (3, 5, 3)

Perpendicularity check: two vectors are perpendicular if their dot product is zero.

AB⃗⋅BC⃗=(−2)(3)+(−4)(5)+(−4)(3)=−6−20−12=−38\vec{AB} \cdot \vec{BC} = (-2)(3) + (-4)(5) + (-4)(3) = -6 -20 -12 = -38

Since −38≠0-38 \neq 0, the sides are not perpendicular. So the angle at BB is not 90∘90^\circ, and the parallelogram is not a rectangle.

Watch out

A common mistake is to check only one pair of adjacent sides — that is enough. If any angle is not 90∘90^\circ, the figure cannot be a rectangle. You do not need to check all four angles.


4. Could it be a square or rhombus?

No need — the question only asks to show it is a parallelogram but not a rectangle. We have done both.


✓Final answer

The points form a parallelogram (diagonals bisect each other), but it is not a rectangle because AB⃗⋅BC⃗=−38≠0\vec{AB} \cdot \vec{BC} = -38 \neq 0.

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