Cross Product - Vertices

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  • #1
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Homework Statement


Let A = (-5, -2, -5), B = (-7, -7, -6), C = (-3, -3, 0), and D = (-5, -8, -1). Find the area of the parallelogram determined by these four points.


Homework Equations


Area of Parallelogram = ||a x b||

The Attempt at a Solution


I drew the parallelogram and decided to use CA and CD as my two vectors for the cross product. Finding the vector for each came from the following:

CA = C - A = <-3, -3, 0> - <-5, -2, -5> = <2, -1, 5>
CD = C - D = <-3, -3, 0> - <-5, -8, -1> = <-2, -5, -1>

Then I took the cross product:

CA x CD = <26, -8, -12>

To which I tried to find the magnitude:
||CA x CD|| = √( 26^2 + (-8)^2 + (-12)^2) = √995

I submitted this to Webwork but it says this answer is wrong; can someone explain to me what went wrong?
 

Answers and Replies

  • #2
SammyS
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Homework Statement


Let A = (-5, -2, -5), B = (-7, -7, -6), C = (-3, -3, 0), and D = (-5, -8, -1). Find the area of the parallelogram determined by these four points.


Homework Equations


Area of Parallelogram = ||a x b||

The Attempt at a Solution


I drew the parallelogram and decided to use CA and CD as my two vectors for the cross product. Finding the vector for each came from the following:

CA = C - A = <-3, -3, 0> - <-5, -2, -5> = <2, -1, 5>
CD = C - D = <-3, -3, 0> - <-5, -8, -1> = <-2, -5, -1>

Then I took the cross product:

CA x CD = <26, -8, -12>

To which I tried to find the magnitude:
||CA x CD|| = √( 26^2 + (-8)^2 + (-12)^2) = √995

I submitted this to Webwork but it says this answer is wrong; can someone explain to me what went wrong?
Your cross product is correct.

Check your arithmetic on 262 + (-8)2 + (-12)2 .
 
  • #3
LCKurtz
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Everything looks good except the last ##\sqrt{995}##.

[Edit]: Good timing Sammys.
 

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