Let A, B, C, D be the points in the plane wit | vedclass

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(A) The vectors representing the sides are:$\vec{AB} = \vec{b} - \vec{a} = (4 - (-2))\hat{i} + (0 - (-1))\hat{j} = 6\hat{i} + \hat{j}$$\vec{BC} = \vec

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a parallelogram which is neither a rhombus nor a rectangle

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(A) The vectors representing the sides are:$\vec{AB} = \vec{b} - \vec{a} = (4 - (-2))\hat{i} + (0 - (-1))\hat{j} = 6\hat{i} + \hat{j}$$\vec{BC} = \vec{c} - \vec{b} = (3 - 4)\hat{i} + (3 - 0)\hat{j} = -\hat{i} + 3\hat{j}$$\vec{CD} = \vec{d} - \vec{c} = (-3 - 3)\hat{i} + (2 - 3)\hat{j} = -6\hat{i} - \hat{j}$$\vec{DA} = \vec{a} - \vec{d} = (-2 - (-3))\hat{i} + (-1 - 2)\hat{j} = \hat{i} - 3\hat{j}$Since $\vec{AB} = -\vec{CD}$ and $\vec{BC} = -\vec{DA}$, $ABCD$ is a parallelogram.Calculate dot product of adjacent sides: $\vec{AB} \cdot \vec{BC} = (6)(-1) + (1)(3) = -6 + 3 = -3 
eq 0$. Thus, it is not a rectangle.Calculate magnitudes: $|\vec{AB}| = \sqrt{6^2 + 1^2} = \sqrt{37}$ and $|\vec{BC}| = \sqrt{(-1)^2 + 3^2} = \sqrt{10}$.Since $|\vec{AB}| 
eq |\vec{BC}|$, it is not a rhombus.Therefore, $ABCD$ is a parallelogram which is neither a rhombus nor a rectangle.

Let $A, B, C, D$ be the points in the plane with position vectors $\vec{a} = -2\hat{i} - \hat{j}$, $\vec{b} = 4\hat{i}$, $\vec{c} = 3\hat{i} + 3\hat{j}$, and $\vec{d} = -3\hat{i} + 2\hat{j}$ respectively. Then $ABCD$ is:

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