You are here:

Advanced Math/Parallela pipe

Advertisement


Question
My name is Andy and I have a Parallela pipe.
A (3,1,0)
B (5,1,0)
C (4,5,0)
D (3,1,4)

I need to find the area of it. How should I start it off?

Answer
Hi, Charlie,

One of the major problems I face is that everyone assumes I am familiar with the specialized vocabulary used by their own teachers.  I am not familiar with the term Parallela Pipe.  (Actually I thought that was the name of an actress from the 40's and 50's who specialized in movies of ill repute.)

So please define all your terms exactly, assuming I know nothing of your teacher's or textbook's vocabulary.  Copy out the definitions and the problems exactly, word for word.

However, I can make this suggestion, since you seem to be doing something in 3 dimensions, suggesting vector algebra.

There is an operation called the Vector Product (a.k.a. the 'cross product' of two vectors.  It has the property that given two vectors  <AB> and <AC>, if you find the cross product and take its absolute value, you get the area of the parallelogram spanned by the two vectors, like this:

    C ------------->
   /              /
  /              /
 /              /
/              /
/------------->/
A             B

This concept might help you in finding areas.  

Advanced Math

All Answers


Answers by Expert:


Ask Experts

Volunteer


Paul Klarreich

Expertise

I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.

Experience

I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.

Education/Credentials
-----------

©2012 About.com, a part of The New York Times Company. All rights reserved.