Question

By vector method, prove that if the diagonals of a parallelogram intersect perpendicularly, then the parallelogram is a rhombus.

Accepted Answer

In the above diagram, a parallelogram $O P Q R$ is drawn.
Suppose $\vec{O P} = \vec{A}$ and $\vec{O R} = \vec{B}$ .

The diagonals of this parallelogram is given by,

$\vec{O Q} = \vec{A} + \vec{B}$ and $\vec{R P} = \vec{A} - \vec{B}$ , respectively.

Since, $\vec{O Q} ⊥ \vec{R P}, \vec{O Q} \cdot \vec{R P} = 0$ , substituting $\vec{O Q} = \vec{A} + \vec{B}$ and $\vec{R P} = \vec{A} - \vec{B}$ , we have, $(\vec{A} + \vec{B}) \cdot (\vec{A} - \vec{B}) = 0$ .

This gives, $\vec{A} \cdot \vec{A} - \vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{A} - \vec{B} \cdot \vec{B} = 0$ .

Since, $\vec{A} \cdot \vec{A} = | \vec{A} |^{2}, \vec{B} \cdot \vec{B} = | \vec{B} |^{2}$ and $\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$ ,

we have, $| \vec{A} |^{2} - | \vec{B} |^{2} = 0$ . Hence, $| \vec{A} | = | \vec{B} |$ . It means, $O P = O R$ .

This tells us that the given parallelogram is a rhombus.

B M Sharma Solutions for Chapter: Vectors, Exercise 41: ILLUSTRATION 3.38

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Vectors, Exercise 41: ILLUSTRATION 3.38

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