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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.

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प्रश्न

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.

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उत्तर

Let A (1, 2, 3) , B (2, 3, 1) and C (3, 1, 2) are the coordinates of the triangle \[\bigtriangleup ABC\]

AB =\[\sqrt{\left( 2 - 1 \right)^2 + \left( 3 - 2 \right)^2 + \left( 1 - 3 \right)^2}\]

\[= \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( - 2 \right)^2}\]
\[ = \sqrt{1 + 1 + 4}\]
\[ = \sqrt{6}\] 

BC =\[\sqrt{\left( 3 - 2 \right)^2 + \left( 1 - 3 \right)^2 + \left( 2 - 1 \right)^2}\] 

\[= \sqrt{\left( 1 \right)^2 + \left( - 2 \right)^2 + \left( 1 \right)^2}\]
\[ = \sqrt{1 + 4 + 1}\]
\[ = \sqrt{6}\] 
 AC =\[\sqrt{\left( 3 - 1 \right)^2 + \left( 1 - 2 \right)^2 + \left( 2 - 3 \right)^2}\]
\[= \sqrt{\left( 2 \right)^2 + \left( - 1 \right)^2 + \left( - 1 \right)^2}\]
\[ = \sqrt{4 + 1 + 1}\]
\[ = \sqrt{6}\]
Now, AB = BC = AC

Therefore, it is an equilateral triangle.
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अध्याय 28: Introduction to three dimensional coordinate geometry - Exercise 28.2 [पृष्ठ ९]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 28 Introduction to three dimensional coordinate geometry
Exercise 28.2 | Q 8 | पृष्ठ ९

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