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

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Parallelogram but not a square

Vedclass · Answer

(D) The position vectors are $\vec{p} = -2\hat{i} - \hat{j}$,$\vec{q} = 4\hat{i}$,$\vec{r} = 3\hat{i} + 3\hat{j}$,and $\vec{s} = -3\hat{i} + 2\hat{j}$.Calculate the vectors representing the sides:$\vec{PQ} = \vec{q} - \vec{p} = (4 - (-2))\hat{i} + (0 - (-1))\hat{j} = 6\hat{i} + \hat{j}$.$\vec{QR} = \vec{r} - \vec{q} = (3 - 4)\hat{i} + (3 - 0)\hat{j} = -\hat{i} + 3\hat{j}$.$\vec{RS} = \vec{s} - \vec{r} = (-3 - 3)\hat{i} + (2 - 3)\hat{j} = -6\hat{i} - \hat{j}$.$\vec{SP} = \vec{p} - \vec{s} = (-2 - (-3))\hat{i} + (-1 - 2)\hat{j} = \hat{i} - 3\hat{j}$.Since $\vec{PQ} = -\vec{RS}$ and $\vec{QR} = -\vec{SP}$,the opposite sides are parallel and equal in magnitude,so $PQRS$ is a parallelogram.Check for perpendicularity: $\vec{PQ} \cdot \vec{QR} = (6)(-1) + (1)(3) = -6 + 3 = -3 
eq 0$. Thus,it is not a rectangle.Check side lengths: $|\vec{PQ}| = \sqrt{6^2 + 1^2} = \sqrt{37}$ and $|\vec{QR}| = \sqrt{(-1)^2 + 3^2} = \sqrt{10}$.Since the adjacent sides are not equal and the angle is not $90^{\circ}$,it is a parallelogram but not a rhombus or square.

Let $P, Q, R,$ and $S$ be points on a 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. What type of quadrilateral is $PQRS$?

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