How to find the area of a parallelogram using determinants?

Question

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?

Answer 1

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!

Answer 2

You have translated your coordinates and have a parallelogram in these translated coordinates...

$(0,0,0),(2,1,3),(−1,2,8),(−3,1,5)$

Note that $(2,1,3) + (−3,1,5) = (−1,2,8)$

Meaning that $(−1,2,8)$ is the diagonal of your parallelogram and $(2,1,3), (−3,1,5)$ are your edeges.

We can use the cross product $\|(2,1,3)\times (−3,1,5)\|$ to get the area, I think that this is a little bit easier than using determinants.

$\sqrt {(1\cdot 5 - 1\cdot 3)^2 + ((-3)\cdot 3 - 2\cdot 5)^2 + (2\cdot 1 - (-3)\cdot 1)^2}$

Answer 3

EDIT1:

If we do not know the 4 points are in a plane and belong to vertices of a parallelogram, then the following verification:

$$ A(1,0,1),B(3,1,4),C(0,2,9),D(-2,1,6) \tag 1 $$

A necessary (but not sufficient, to rule out a skewed parallelogram ) condition is that there should be a pair of opposite sides of same length. To check this, we have all 6 squared lengths of all sides/diagonals computed:

$$(BD^2,AD^2,AB^2,BC^2,CA^2,CD^2)=(29,35,14,35,64,69,14)$$

Two sides of same length are detected here. $|AB|=|DC|=\sqrt {14}$ and $|AD|= |BC|=\sqrt {35}$ arranged as follows:

However this is not yet sufficient, coplanarity should be also checked out. In order to confirm coplanarity, from the given 3d points (1) find the two center points of the two diagonals $(AC,BD) $ as :

$ (0.5,1,5)\text{ point which coincides with point } (0.5,1,5).$

So the diagonals are not skew, *the parallelogram lies in a plane.

It is direct to compute $\vec {AB},\vec {AD}$ and the cross product $|\vec {AB}\times \vec {AD}|$ for parallelogram area as suggested by user317176.

However the OP wants to use determinants for this area calculation.

EDIT2:

Since spatial orientation of the parallelogram can be calculated, we can rotate the rigid parallelogram normal parallel to $ OXY $ plane using Euler rotation angles $(\theta,\phi,\psi)$ and then use 2d determinant conforming to OP's requirement. However imho that would be a lack luster move wanting in elegance.