The points whose position vectors are 2hat{i} | vedclass

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

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an equilateral triangle

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(C) Let the position vectors of the vertices be $\vec{a} = 2\hat{i}+3\hat{j}+4\hat{k}$, $\vec{b} = 3\hat{i}+4\hat{j}+2\hat{k}$, and $\vec{c} = 4\hat{i}+2\hat{j}+3\hat{k}$.The side vectors are:$\vec{AB} = \vec{b} - \vec{a} = (3-2)\hat{i} + (4-3)\hat{j} + (2-4)\hat{k} = \hat{i} + \hat{j} - 2\hat{k}$.$\vec{BC} = \vec{c} - \vec{b} = (4-3)\hat{i} + (2-4)\hat{j} + (3-2)\hat{k} = \hat{i} - 2\hat{j} + \hat{k}$.$\vec{CA} = \vec{a} - \vec{c} = (2-4)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k} = -2\hat{i} + \hat{j} + \hat{k}$.Calculating the lengths of the sides:$|\vec{AB}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1+1+4} = \sqrt{6}$.$|\vec{BC}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.$|\vec{CA}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$.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 $2\hat{i}+3\hat{j}+4\hat{k}$, $3\hat{i}+4\hat{j}+2\hat{k}$, and $4\hat{i}+2\hat{j}+3\hat{k}$ are the vertices of

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