Prove that the triangle formed by joining the three points whose coord

1.	Prove that the triangle formed by joining the three points whose coordinates are (1, 2, 3), (2, 3, 1) and (3, 1, 2) is an equilateral triangle.
Answer» Given: Points are A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2) To prove: the triangle formed by given points is an equilateral triangle An equilateral triangle is a triangle whose all sides are equal So we need to prove AB = BC = AC Formula used: The distance between any two points (a, b, c) and (m, n, o) is given by, \(\sqrt{(a−m)^2+(b−n)^2+(c−0)^2}\) Therefore, The distance between A(1, 2, 3) and B(2, 3, 1) is AB = \(\sqrt{(1-2)^2+(2-3)^2+(3-1)^2} \) = \(\sqrt{(-1)^2+(−1)^2+4^2} \) = \(\sqrt{1+1+4} \) = \(\sqrt{6}\) Distance between B(2, 3, 1) and C(3, 1, 2) is BC = \(\sqrt{(1-3)^2+(2-1)^2+(3-2)^2} \) = \(\sqrt{(-2)^2+1^2+1^2} \) = \(\sqrt{1+1+4} \) = \(\sqrt{6}\) Clearly, AB = BC = AC Thus, Δ ABC is a equilateral triangle Hence Proved

Prove that the triangle formed by joining the three points whose coordinates are (1, 2, 3), (2, 3, 1) and (3, 1, 2) is an equilateral triangle.

Answer»

Given: Points are A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2)

To prove: the triangle formed by given points is an equilateral triangle

An equilateral triangle is a triangle whose all sides are equal

So we need to prove AB = BC = AC

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

\(\sqrt{(a−m)^2+(b−n)^2+(c−0)^2}\)

Therefore,

The distance between A(1, 2, 3) and B(2, 3, 1) is AB

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

= \(\sqrt{(-1)^2+(−1)^2+4^2} \)

= \(\sqrt{1+1+4} \)

= \(\sqrt{6}\)

Distance between B(2, 3, 1) and C(3, 1, 2) is BC

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

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

= \(\sqrt{1+1+4} \)

= \(\sqrt{6}\)

Clearly, AB = BC = AC

Thus, Δ ABC is a equilateral triangle

Hence Proved