Linear Algebra Help, Area of a Parallelogram using vectors

  • #1
Axoren
17
0

Homework Statement


Find the area of the parallelogram defined by the vectors

v = {1 1 3 1}
w = {-2 -1 2 2}

Homework Equations


Area = v dot w * sin(theta)
theta = cos^-1(v dot w / |v|*|w|)

The Attempt at a Solution


Solved
General Solution:

Area of a parallelogram for non-R^3 vectors = v dot w * sin(cos^-1(v dot w / (|v|*|w|)))
 
Last edited:
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  • #2
If v and w are two vectors representing two adjacent sides of the parallelogram, then the area is the magnitude of the cross product of those two vectors.
 
  • #3
jegues said:
If v and w are two vectors representing two adjacent sides of the parallelogram, then the area is the magnitude of the cross product of those two vectors.

You can't perform cross product on vectors outside of R^3
 
  • #4
Axoren said:
You can't perform cross product on vectors outside of R^3

Whoops! Sorry, I didn't notice they were 4 dimensional vectors.
 
  • #5
I got it, some how. Updated original post.
 
  • #6
Axoren said:
I got it, some how. Updated original post.

What do you mean you got it some how... What did you do? Alegbra mistake?
 
  • #7
jegues said:
What do you mean you got it some how... What did you do? Alegbra mistake?

I didn't depend on the cross product and started using other equations around the internet.

I provided them in the original post.
 
  • #8
I don't think that's right. You can let

[tex] A=\begin{bmatrix} 1 & -2 \\ 1 & -1 \\ 3 & 2 \\ 1 & 2 \end{bmatrix} [/tex]

and then the volume is:

[tex] V = \sqrt{det \left(A^TA \right)} [/tex]

which isn't 5.

EDIT: Oh, I see you got it with some other formula. Was the answer [itex] \sqrt{131} [/itex] ?
 
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