# [SOLVED] Vector Calculus: Parallelogram Problem

• Aug 31st 2009, 03:45 PM
Alterah
[SOLVED] Vector Calculus: Parallelogram Problem
Hello, I am stuck on a problem that at the conceptual level makes perfect sense. Here's the problem:

Use vectors to show that the diagonals of a parallelogram have the same length if and only if the parallelogram is a rectangle. (Hint: let a and b be vectors along two sides of the parallelogram. Express vectors running along the diagonals in terms of a and b.)

I have attached a picture. I can let:
d1 = a + b
d2 = b - a

I know that I am trying to prove that:
$\displaystyle \|d1\| = \|d2\|$
When $\displaystyle \cos\theta = 0$ for the angle between a and b.

Most of my work so far essentially amounts to me plugging in for the Angle Between Two Vectors formula and coming out with a dotted with b = 0. This occurs if either is the 0 vector or if they are perpendicular. However, this doesn't help me with the length of d1 and d2. I am not sure how I can relate the two together. Thanks for any assistance.
• Aug 31st 2009, 04:11 PM
Plato
Suppose that $\displaystyle \overrightarrow a ~\&~\overrightarrow b$ are the two adjacent vectors determining the parallelogram.
The diagonals of the parallelogram are: $\displaystyle \left( {\overrightarrow a + \,\overrightarrow b } \right)\;\& \,\left( {\overrightarrow a - \overrightarrow b } \right)$.
Look at the following.
$\displaystyle \left( {\overrightarrow a + \,\overrightarrow b } \right)\; \cdot \,\left( {\overrightarrow a - \overrightarrow b } \right) = \overrightarrow a \overrightarrow { \cdot a} - \overrightarrow b \overrightarrow b$
• Aug 31st 2009, 04:34 PM
Alterah
$\displaystyle \left( {\overrightarrow a + \,\overrightarrow b } \right)\; \cdot \,\left( {\overrightarrow a - \overrightarrow b } \right) = \overrightarrow a \overrightarrow { \cdot a} - \overrightarrow b \overrightarrow b$

I am not sure I understand the significance in dotting the diagonals together. The expression shows that the diagonals dotted together equal the length of a squared minus the length of b squared. Thanks for the help so far, I feel like I am missing something that should be very obvious to me.
• Aug 31st 2009, 05:09 PM
luobo
$\displaystyle \|\vec{d_1}\|^2 = (\vec{a} +\vec{b}) \cdot (\vec{a} +\vec{b}) = \|\vec{a}\|^2+2 \vec{a} \cdot \vec{b} + \|\vec{b}\|^2$
$\displaystyle \|\vec{d_2}\|^2 = (\vec{a} -\vec{b}) \cdot (\vec{a} -\vec{b}) = \|\vec{a}\|^2-2 \vec{a} \cdot \vec{b} + \|\vec{b}\|^2$
$\displaystyle \|\vec{d_1}\|^2=\|\vec{d_2}\|^2\Leftrightarrow \vec{a} \cdot \vec{b}=0\Leftrightarrow \cos \theta=0$
• Aug 31st 2009, 05:13 PM
Alterah
Luobo...thank you. That is what I was missing. That makes complete sense. I suppose was too tired to think clearly. I was hoping I could get a nudge, and that's what Plato tried, but I guess the roadblock in my mind just wouldn't allow me go past it. Thanks again!