1. ## Parallelogram and Vectors

A Parallelogram with sides of equal length is called a rhombus. Show that the diagonals of a rhombus are perpendicular.

So, I start with v and u which are perpendicular vectors. v + w is a diagonal of the rhombus.

I am not sure how to get the other one, or to solve this question, really.

I know that in order for two vectors to be perpendicular then the dot product must be 0.

so I guess, I need both diagonals and to see if their dot product is zero?

2. Originally Posted by zodiacbrave
A Parallelogram with sides of equal length is called a rhombus. Show that the diagonals of a rhombus are perpendicular.
Suppose that $\displaystyle \overrightarrow a \;\& \,\overrightarrow b$ are adjacent sides of thee rhombus.
Then the diagonals are $\displaystyle \overrightarrow a + \overrightarrow b \;\& \,\overrightarrow a - \overrightarrow b$.
Now look at the dot product, $\displaystyle \left( {\overrightarrow a + \overrightarrow b } \right) \cdot \left( {\overrightarrow a - \overrightarrow b } \right) = ?$

I know (a + b) is just a regular vector but the fact that we are labeling it as the sum of two other vectors is confusing to me. What I mean is, if vector a has components (a1, a2) and vector b has components (b1, b2) then won't the dot product of (a + b) and (a - b)
equal (a1)^2 - (b1)^2 + (a2)^2 - (b2)^2

I guess, I am not understanding the actual method of proving it algebraically.

4. Originally Posted by zodiacbrave
I know (a + b) is just a regular vector but the fact that we are labeling it as the sum of two other vectors is confusing to me. What I mean is, if vector a has components (a1, a2) and vector b has components (b1, b2) then won't the dot product of (a + b) and (a - b)
equal (a1)^2 - (b1)^2 + (a2)^2 - (b2)^2
Recall that $\displaystyle \vec{p}\cdot\vec{p}=\left\| p \right\|^2$

$\displaystyle (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=\vec{a}\cdot\vec{a}-\vec{b}\cdot\vec{b}=0$