# calculate the determinant

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hello

in math, all we know that if we want to proof that the two vectors AB & AC are parallele between them, we can use the determinant of them

if det(AB, AC) = 0 then (AB) // (AC), my question how to do it for vectors of 3D type, i know how to do it for 2D vectors

if AB(x, y) & AC(x', y') then det(AB, AC) = x * y' - y * x'

my email = abdellahgogop@hotmail.com

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Absolute value of dot product is 1 edit: if normalised, product of lengths otherwise. Cross product will have length zero. Up to accuracy limits of course.

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what is the determinant of : AB(x,y,z) & AC(x',y',z') , what is the math operation?

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You can only take the determinant of square matrices.

thank's

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And a determinant of 0 means the rows and columns are linearly dependent. Which for a 2x2 matrix means they are colinear.

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The condition

|x1 y1| = 0
|x2 y2|

means that the two vectors (x1,y1) and (x2,y2) are contained in the same line (containing also the origin).

The generalization to 3D is that the condition

|x1 y1 z1|
|x2 y2 z2| = 0
|x3 y3 z3|

means the three vectors that form the rows are contained in the same plane (again, containing also the origin).

There are other related things you can do with the determinant. In 2D you can tell if three points are collinear by checking if

|x1 y1 1|
|x2 y2 1| = 0
|x3 y3 1|

Similarly, in 3D you can tell if four points lie on the same plane by checking if

|x1 y1 z1 1|
|x2 y2 z2 1| = 0
|x3 y3 z3 1|
|x4 y4 z4 1|

EDIT: Oh, to actually answer your question, two vectors (x1,y1,z1) and (x2,y2,z2) are proportional if and only if the matrix

(x1 y1 z1)
(x2 y2 z2)

has rank less than 2. In particular, that means that you can check two 2x2 minors and they both should be 0. So you can check if the vectors are proportional by checking if

|x1 y1| = 0
|x2 y2|

and

|x1 z1| = 0
|x2 z2|

FURTHER EDIT: Actually, you also have to check

|y1 z1| = 0
|y2 z2|

Otherwise you would think that (0,1,0) and (0,0,1) are proportional. Edited by Álvaro

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Just one clarfication to Alvaro's otherwise fine answer: both for 2x2 matrix and 3x3 matrix determinants equal to 0, it doesn't necessarily mean the actual lines or planes contain the origin.

However, as vectors alone don't have information about their placement, you can think of your vectors as being placed in the origin if you like, for these calculations at least.

Or maybe I misunderstood the notes between parentheses.

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Just one clarfication to Alvaro's otherwise fine answer: both for 2x2 matrix and 3x3 matrix determinants equal to 0, it doesn't necessarily mean the actual lines or planes contain the origin.

However, as vectors alone don't have information about their placement, you can think of your vectors as being placed in the origin if you like, for these calculations at least.

Or maybe I misunderstood the notes between parentheses.

Let me try to be more clear. When talking about a vector space and its subspaces, a subspace always contains the vector 0 (perhaps that's a better name than "origin"). So the 2x2 determinant being 0 means the two row vectors don't span the whole 2-dimensional space, so they are both contained in a subspace of dimension less than 2. That's why I said that they are both contained in a line that also contains the vector 0 (that was my attempt at describing vector subspaces without introducing a bunch of definitions).

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It was the origin term what mislead me. Everything is clear now. Thanks for the explanation.

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