If the position vectors of A, B, C, D are hat | vedclass

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

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rhombus

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(C) Let the position vectors be $\vec{a} = \hat{i}+2\hat{j}+2\hat{k}$,$\vec{b} = 2\hat{i}-\hat{j}$,$\vec{c} = \hat{i}+\hat{j}+3\hat{k}$,and $\vec{d} = 4\hat{j}+5\hat{k}$.First,find the vectors representing the sides:$\vec{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (-1-2)\hat{j} + (0-2)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k}$.$\vec{BC} = \vec{c} - \vec{b} = (1-2)\hat{i} + (1-(-1))\hat{j} + (3-0)\hat{k} = -\hat{i} + 2\hat{j} + 3\hat{k}$.$\vec{CD} = \vec{d} - \vec{c} = (0-1)\hat{i} + (4-1)\hat{j} + (5-3)\hat{k} = -\hat{i} + 3\hat{j} + 2\hat{k}$.$\vec{DA} = \vec{a} - \vec{d} = (1-0)\hat{i} + (2-4)\hat{j} + (2-5)\hat{k} = \hat{i} - 2\hat{j} - 3\hat{k}$.Note that $\vec{AB} = -\vec{CD}$ and $\vec{BC} = -\vec{DA}$,which confirms $ABCD$ is a parallelogram.Now check the lengths of the sides:$|\vec{AB}| = \sqrt{1^2 + (-3)^2 + (-2)^2} = \sqrt{1+9+4} = \sqrt{14}$.$|\vec{BC}| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1+4+9} = \sqrt{14}$.Since adjacent sides are equal $(|\vec{AB}| = |\vec{BC}| = \sqrt{14})$,the parallelogram is a rhombus.

If the position vectors of $A, B, C, D$ are $\hat{i}+2\hat{j}+2\hat{k}, 2\hat{i}-\hat{j}, \hat{i}+\hat{j}+3\hat{k}$ and $4\hat{j}+5\hat{k}$ respectively,then the quadrilateral $ABCD$ is a

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