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Vector Algebra

Section: Mathematics

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Question : 32 of 38

Marks: +1, -0

Let P, Q, R and S be the points on the plane with position vectors

-\hat{2i}-\hat{j}

\hat{4i},\hat{3i}+\hat{3j}

-\hat{3i}+\hat{2j}

, respectively. The quadrilateral PQRS must be a

[JEE Adv 2010 P1]

parallelogram, which is neither a rhombus nor a rectangle.
square.
rectangle, but not a square.
rhombus, but not a square.

Solution:

Evaluating midpoint of PR and QS which gives M =

\left[\frac{\hat{i}}{2}+\hat{j}\right]

same for both.

\vec{PQ}

=

\vec{SR}

=

\hat{6i}+\hat{j}

\vec{PS}

=

\vec{QR}

=

-\hat{i}+\hat{3j}

⇒

\vec{PQ}\cdot\vec{PS}

≠ 0

\vec{PQ}

||

\vec{SR},\vec{PS}

||

\vec{QR}

and

\left|\vec{PQ}\right|

=

\left|\vec{SR}\right|,\left|\vec{PS}\right|

=

\left|\vec{QR}\right|

Hence, PQRS is a parallelogram but not rhombus or rectangle.

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