Verify the following: (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Let the points be: A(–1, 2, 1), B(1, –2, 5), C(4, –7, 8) & D(2, –3, 4)

ABCD can be vertices of parallelogram only if opposite sides are equal.

AB = CD and BC = AD

Firstly let us calculate the distance

Calculating AB

A ≡ (– 1, 2, 1) and B ≡ (1, – 2, 5)

By using the formula,

Distance AB = 

\( \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)

So here,

x₁ = – 1, y₁ = 2, z₁ = 1

x₂ = 1, y₂ = – 2, z₂ = 5

Distance AB = 

\( \sqrt{(1 – (-1))^2 + (-2 – 2)^2 + (5 – 1)^2}\)

= \( \sqrt{(2)^2 + (-4)^2 + (4)^2}\)

= \( \sqrt{4 + 16 + 16}\)

= \( \sqrt{36}\)

= 6

Calculating BC

B ≡ (1, – 2, 5) and C ≡ (4, – 7, 8)

By using the formula,

Distance BC = 

\( \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)

So here,

x₁ = 1, y₁ = – 2, z₁ = 5

x₂ = 4, y₂ = – 7, z₂ = 8

Distance BC = 

\( \sqrt{(4 – 1)^2 + (-7 – (-2))^2 + (8 – 5)^2}\)

= \( \sqrt{(3)^2 + (-5)^2 + (3)^2}\)

= \( \sqrt{9 + 25 + 9}\)

= \( \sqrt{43}\)

Calculating CD

C ≡ (4, – 7, 8) and D ≡ (2, – 3, 4)

By using the formula,

Distance CD = 

\( \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)

So here,

x₁ = 4, y₁ = – 7, z₁ = 8

x₂ = 2, y₂ = – 3, z₂ = 4

Distance CD = 

\( \sqrt{(2 – 4)^2 + (-3 – (-7))^2 + (4 – 8)^2}\)

= \( \sqrt{(-2)^2 + (4)^2 + (-4)^2}\)

= \( \sqrt{4 + 16 + 16}\)

= \( \sqrt{36}\)

= 6

Calculating DA

D ≡ (2, – 3, 4) and A ≡ (– 1, 2, 1)

By using the formula,

Distance DA = 

\( \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)

So here,

x₁ = 2, y₁ = – 3, z₁ = 4

x₂ = – 1, y₂ = 2, z₂ = 1

Distance DA = 

\( \sqrt{(-1 – 2)^2 + (2 – (-3))^2 + (1 – 4)^2}\)

= \( \sqrt{(-3)^2 + (5)^2 + (-3)^2}\)

= \( \sqrt{9 + 25 + 9}\)

= \( \sqrt{43}\)

Since AB = CD and BC = DA (given)

So, In ABCD both pairs of opposite sides are equal.

ABCD is a parallelogram.