The points having position vectors 2i + 3j + | vedclass

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(C) Let the position vectors of the points be $\overrightarrow{OA} = 2i + 3j + 4k,$ $\overrightarrow{OB} = 3i + 4j + 2k,$ and $\overrightarrow{OC} = 4

Vedclass · Accepted Answer

Equilateral triangle

Vedclass · Answer

(C) Let the position vectors of the points be $\overrightarrow{OA} = 2i + 3j + 4k,$ $\overrightarrow{OB} = 3i + 4j + 2k,$ and $\overrightarrow{OC} = 4i + 2j + 3k.$First,we find the vectors representing the sides of the triangle:$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3-2)i + (4-3)j + (2-4)k = i + j - 2k.$$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (4-3)i + (2-4)j + (3-2)k = i - 2j + k.$$\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = (2-4)i + (3-2)j + (4-3)k = -2i + j + k.$Next,we calculate the lengths of the sides:$|\overrightarrow{AB}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}.$$|\overrightarrow{BC}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}.$$|\overrightarrow{CA}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}.$Since $|\overrightarrow{AB}| = |\overrightarrow{BC}| = |\overrightarrow{CA}| = \sqrt{6},$ all three sides are equal in length.Therefore,the points form an equilateral triangle.

The points having position vectors $2i + 3j + 4k,$ $3i + 4j + 2k,$ and $4i + 2j + 3k$ are the vertices of:

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