Let P, Q, R and S be the points on the plane | vedclass

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(A) The position vectors of the vertices are $\vec{p} = -2\hat{i} - \hat{j}$,$\vec{q} = 4\hat{i}$,$\vec{r} = 3\hat{i} + 3\hat{j}$,and $\vec{s} = -3\ha

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parallelogram,which is neither a rhombus nor a rectangle

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(A) The position vectors of the vertices are $\vec{p} = -2\hat{i} - \hat{j}$,$\vec{q} = 4\hat{i}$,$\vec{r} = 3\hat{i} + 3\hat{j}$,and $\vec{s} = -3\hat{i} + 2\hat{j}$.First,we check the midpoints of the diagonals $PR$ and $QS$:Midpoint of $PR = \frac{\vec{p} + \vec{r}}{2} = \frac{(-2\hat{i} - \hat{j}) + (3\hat{i} + 3\hat{j})}{2} = \frac{\hat{i} + 2\hat{j}}{2} = \frac{1}{2}\hat{i} + \hat{j}$.Midpoint of $QS = \frac{\vec{q} + \vec{s}}{2} = \frac{(4\hat{i}) + (-3\hat{i} + 2\hat{j})}{2} = \frac{\hat{i} + 2\hat{j}}{2} = \frac{1}{2}\hat{i} + \hat{j}$.Since the midpoints of the diagonals coincide,$PQRS$ is a parallelogram.Next,we calculate the side vectors:$\vec{PQ} = \vec{q} - \vec{p} = 4\hat{i} - (-2\hat{i} - \hat{j}) = 6\hat{i} + \hat{j}$.$\vec{PS} = \vec{s} - \vec{p} = (-3\hat{i} + 2\hat{j}) - (-2\hat{i} - \hat{j}) = -\hat{i} + 3\hat{j}$.Check for rectangle (dot product of adjacent sides):$\vec{PQ} \cdot \vec{PS} = (6\hat{i} + \hat{j}) \cdot (-\hat{i} + 3\hat{j}) = (6)(-1) + (1)(3) = -6 + 3 = -3 
eq 0$.Since the dot product is not zero,the sides are not perpendicular,so it is not a rectangle.Check for rhombus (lengths of adjacent sides):$|\vec{PQ}| = \sqrt{6^2 + 1^2} = \sqrt{36 + 1} = \sqrt{37}$.$|\vec{PS}| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$.Since $|\vec{PQ}| 
eq |\vec{PS}|$,the sides are not equal,so it is not a rhombus.Thus,$PQRS$ is a parallelogram,which is neither a rhombus nor a rectangle.

Let $P, Q, R$ and $S$ be the points on the plane with position vectors $-2 \hat{i}-\hat{j}, 4 \hat{i}, 3 \hat{i}+3 \hat{j}$ and $-3 \hat{i}+2 \hat{j}$ respectively. The quadrilateral $PQRS$ must be a

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