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.
A. parallelogram, which is neither a rhombus not a rectangle
B. Square
C. Rectangle, but not a square
D. Rhombus, but not a square
VerifiedHint: To solve this question, we need to know the basic theory related to the vectors. As we know that Vector describes the movement of an object from one point to another. Here we have a total of four position vectors $-2 \hat{i}-\hat{j}, 4 \hat{i}, 3 \hat{i}+3 \hat{j}$ and $-3 \hat{i}+2 \hat{j}$ on the plane. So, we will calculate length by using distance formula and also check angle so that we must tell about quadrilateral PQRS.
Complete step-by-step answer:
As we are given $(-2 \hat{i}-\hat{j}),(4 \hat{i}),(3 \hat{i}+3 \hat{j}),(-3 \hat{i}+2 \hat{j})$
On converting above vectors into Cartesian coordinates we get,
$P(-2,-1), Q(4,0), R(3,3), S(-3,2)$
By distance formula, $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
$(\mathrm{PQ})_{\mathrm{d}}=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
$=\sqrt{(4+2)^{2}+(0+1)^{2}}$
$=\sqrt{(6)^{2}+(1)^{2}}$
$=\sqrt{36+1}$
$=\sqrt{37}$
Similarly, we will calculate-
$(\mathrm{QR})$
$=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
$=\sqrt{(4-3)^{2}+(0-3)^{2}}$
$=\sqrt{10}$
$(\mathrm{RS})_{\mathrm{d}}$
$=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
$=\sqrt{(3+3)^{2}+(3-2)^{2}}$
$=\sqrt{37}$
$(\mathrm{SP})_{\mathrm{d}}$
$=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
$=\sqrt{(-2+3)^{2}+(-1-2)^{2}}$
$=\sqrt{10}$
Therefore, $(\mathrm{PQ})_{\mathrm{d}}=(\mathrm{RS})_{\mathrm{d}}$ and $(\mathrm{QR})_{\mathrm{d}}=(\mathrm{SP})_{\mathrm{d}}$
Hence,PQRS is parallelogram
Now,
Slope $=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$(\mathrm{PQ})_{\text {slope }}=\dfrac{0-(-1)}{4-(-2)}=\dfrac{1}{6}$
Similarly,
$(\mathrm{QR})_{\text {slope }}$
$=\dfrac{0-(3)}{4-(3)}$
$=-3$
$(\mathrm{RS})_{\text {slope }}$
$=\dfrac{3-(2)}{3-(-3)}$
$=\dfrac{1}{6}$
$(\mathrm{SP})_{\text {slope }}$
$=\dfrac{2-(-1)}{-3-(-2)}$
$=-3$
Therefore,
$(\mathrm{PQ})_{\text {slope }}=(\mathrm{RS})_{\text {slope }}$ this implies $\mathrm{PQ} \| \mathrm{RS}$ and $(\mathrm{QR})_{\text {slope }}=(\mathrm{SP})_{\text {slope }}$ this implies $\mathrm{QR} \| \mathrm{SP}$ $(\mathrm{PQ})_{\text {slope }} \neq(\mathrm{QR})_{\text {slope }}$ i.e. they aren't parallel $(\mathrm{PQ})_{\text {slope }} \times(\mathrm{QR})_{\text {slope }} \neq-1$ i.e. they aren't perpendicular.
Therefore PQRS is a parallelogram but it is neither a rhombus nor a rectangle.
Hence option $A$ is the correct answer.
Note: A position vector is a vector whose initial point is fixed at the origin so that each point corresponds to $\mathrm{P}=$ Since a position vector cannot be translated, it is technically not a vector, so it instead should be considered as a means of using arithmetic of vectors with points in the plane. In this case, any two given points on a line will have its own position vector.
