The points whose position vectors are 2i+3j+4 | vedclass

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(C) Let the position vectors of the vertices be $\vec{a} = 2i+3j+4k$,$\vec{b} = 3i+4j+2k$,and $\vec{c} = 4i+2j+3k$. The side vectors are: $\vec{AB} =

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an equilateral triangle

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(C) Let the position vectors of the vertices be $\vec{a} = 2i+3j+4k$,$\vec{b} = 3i+4j+2k$,and $\vec{c} = 4i+2j+3k$. The side vectors are: $\vec{AB} = \vec{b} - \vec{a} = (3-2)i + (4-3)j + (2-4)k = i + j - 2k$. $\vec{BC} = \vec{c} - \vec{b} = (4-3)i + (2-4)j + (3-2)k = i - 2j + k$. $\vec{CA} = \vec{a} - \vec{c} = (2-4)i + (3-2)j + (4-3)k = -2i + j + k$. The lengths of the sides are: $|\vec{AB}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1+1+4} = \sqrt{6}$. $|\vec{BC}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$. $|\vec{CA}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$. Since $|\vec{AB}| = |\vec{BC}| = |\vec{CA}| = \sqrt{6}$,all three sides are equal in length. Therefore,the triangle is an equilateral triangle.

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

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