Chapter 6.4, Problem 91E

To determine

To Proof: that the diagonals of a rhombus are perpendicular.

Explanation of Solution

Concept Involved:

Two vectors are perpendicular to each other if the dot product of their corresponding vectors is equal to 0.

  $a\cdot b=0$

Consider $ABCD$ be a rhombus. Sides of rhombus are equal to each other.

  

Here the diagonals of the rhombus are the vectors $\left(a+b\right)$ and $\left(a-b\right)$ .

So,

  $\begin{array}{c}\left(a+b\right)\cdot \left(a-b\right)=a\cdot a-b\cdot b\\ ={\Vert a\Vert }^2-{\Vert b\Vert }^2\end{array}$

Since, the sides of rhombus are equal. That is

  $a=b$

Therefore,

  $\left(a+b\right)\cdot \left(a-b\right)=0$

Conclusion:

The diagonals of a rhombus are perpendicular.

  $\left(a+b\right)\cdot \left(a-b\right)=0$