1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cross Product Magnitude for Triangle Area

  1. Sep 1, 2012 #1
    1. The problem statement, all variables and given/known data

    So I have several problems on my homework which deal with the application of cross product and its magnitude to find areas. I know how to do cross product, and I know how to find the magnitude, however, no matter how many times I try to calculate it, I get the wrong answer.

    Take for example:

    Find the area of the triangle with the three vertices [0, 0, -5], [-2, 1, -4], and [-3, -1, -6].


    2. The attempt at a solution

    This is where I'm not sure if I'm doing something incorrectly. I have generally been trying to create the two vectors for the cross product by subtracting one vertex from the other two.

    So,

    [-2, 1, -4] - [0, 0, -5] = [-2, 1, 1]

    and

    [-3, -1, -6] - [0, 0, -5] = [-3, -1, -1]

    Then, computing the cross product to be [0, 5, 5]

    So the magnitude comes out to be [itex]\sqrt{50}[/itex]

    And then the area of the triangle would be [itex]\frac{1}{2}[/itex][itex]\sqrt{50}[/itex]

    Is this right? If not, where did I go wrong? Is there some proper approach to deciding which vertex to use as the starting point?

    Thanks for your help.
     
  2. jcsd
  3. Sep 1, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I don't see anything wrong with what you are doing.
     
  4. Sep 2, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    50= (2)(52) so it might be better to write it as
    [tex]\frac{\sqrt{50}}{2}= \frac{5\sqrt{2}}{2}[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Cross Product Magnitude for Triangle Area
  1. Cross Product and Area (Replies: 1)

Loading...