I Have a parallelogram with the vertices $(1,0,1),(3,1,4),(0,2,9)$ and $(-2,1,6)$

I need to find the area of this parallelogram

My Attempt: I understand My end goal is to find the determinant of a matrix formed by these coordinates. So I first translated the whole parallelogram to the origin by subtracting each vertex by $(1,0,1).$

I am left with $(0,0,0),(2,1,3),(-1,2,8)$ and $ (-3,1,5)$ I can ignore the origin point now, but I believe I am suppose to only use 2 vertices. How do I pick these two vertices, I understand I could draw it out, but is there an easier way? Also can this question be done without translating the vertices?

  • $\begingroup$ A little bit of magic: if $\mathbf{a}$ and $\mathbf{b}$ are vectors in $\mathbb R^3$, then the area of the parallelogram formed by $0,\mathbf{a},\mathbf{b},$ and $\mathbf{a+b}$ is the magnitude of the vector product, $|\mathbf{a}\times \mathbf{b}|$. $\endgroup$ – TonyK Mar 8 '17 at 19:05
  • $\begingroup$ Choose the two which add up to the other one. $\endgroup$ – Dylan Mar 8 '17 at 19:06

Ahh Thank You for the hints above. I Was able to solve it:

Vectors (0,0,0) (2,1,3) (-1,2,8) (-3,1,5)

(-3,1,5)-(2,1,3) = (-1,2,8) So these are are two vertices needed to find the determinant.

The determinant is -2i +19j -5k. Take the magnitude of this, which is sqrt(390)

Thanks for the help!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.