If the position vectors of the vertices of a triangle are $i + 2 j + 3 k$ , $2 i + 3 j + k$ , and $3 i + j + 2 k$ , show that the triangle is an equilateral traingle.

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Solution

$A = i + 2 j + 3 k$ , $B = 2 i + 3 j + k$ , and $C = 3 i + j + 2 k$ are the position vectors of the sides of the triangle.
$A B = | ˆ i + ˆ j + - 2 ˆ k | = \sqrt{1^{2} + 1^{2} + (- 2)^{2}} = \sqrt{6}$
$B C = | ˆ i + - 2 ˆ j + ˆ k | = \sqrt{1^{2} + (- 2)^{2} + 1^{2}} = \sqrt{6}$
$C A = | - 2 ˆ i + ˆ j + ˆ k | = \sqrt{(- 2)^{2} + 1^{2} + 1^{2}} = \sqrt{6}$
Therefore, triangle is an equilateral triangle.

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