Let vector a, vector b, vector c, vector d be the position vectors of the distinct points A, B, C, D

Question

Let \(\vec a,\vec b,\vec c, \vec d\) be the position vectors of the distinct points A, B, C, D If \(\vec b-\vec a=\vec c-\vec d\), then show that ABCD is a parallelogram.

Answer 1

Given the position vectors of points A, B, C and D are \(\vec a,\vec b,\vec c\) and \(\vec d\) respectively.

Recall the vector \(\vec{AB}\) is given by

\(\vec{AB}\) = position vector of B - position vector of A

⇒ \(\vec{AB}\) = \(\vec b-\vec a\)

Similarly, the vector \(\vec {DC}\) is given by

\(\vec {DC}\) = position vector of C - position vector of D

⇒ \(\vec {DC}\) = \(\vec c-\vec d\)

But, it is given that \(\vec b-\vec a=\vec c-\vec d\)

⇒ \(\vec {AB}=\vec{DC}\)

Two vectors are equal only when both their magnitudes and directions are equal.

This means that the opposite sides in quadrilateral ABCD are parallel and equal.

Thus, ABCD is a parallelogram.