Why is this theorem on coplanar vectors true (LINEAR ALGEBRA)?

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SUMMARY

The theorem on coplanar vectors states that three vectors A, B, and C are coplanar if and only if the determinant of the matrix formed by these vectors equals zero. This is mathematically expressed as A · (B × C) = 0, indicating that vector A can be represented as a linear combination of B and C. The volume of the parallelepiped defined by these vectors is zero when they are coplanar, confirming that at least one vector is linearly dependent. The determinant's value serves as a definitive test for linear independence among the vectors.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically vector operations.
  • Familiarity with determinants and their properties in matrix theory.
  • Knowledge of the cross product and dot product of vectors.
  • Basic comprehension of linear combinations and vector spaces.
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  • Study the properties of determinants in matrix algebra.
  • Learn about the geometric interpretation of the cross product in three-dimensional space.
  • Explore the concept of linear independence and its implications in vector spaces.
  • Investigate the relationship between coplanarity and the volume of geometric shapes formed by vectors.
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This discussion is beneficial for students and professionals in mathematics, particularly those studying linear algebra, as well as educators teaching vector theory and its applications in geometry and physics.

BlueRope
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A necessary and sufficient condition for three vectors to be coplanar is the equality is that the determinant of the matrix equals zero.
 
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Basically that just says that A\cdot (B\times C)=0 (or any reordering thereof). BXC is a vector that is perpendicular to B and C. If A is coplanar with B and C, then it can be expressed as a linear combination of the two, i.e. A=bB+cC where b and c are real numbers. In that case, then it's obvious that A dotted into this vector which is perpendicular to both B and C would be 0.

Another way to think about it is to note that the above triple product has a value which is the volumn of the parallelepiped defined by A, B and C. If A, B and C are coplanar, then the parallelepiped has 0 volume.
 
What Matterwave said is correct it follows from a theorem that says the determinant of a matrix is non zero if and only if the vectors which make it up are all linearly independent. So if the vectors are coplanar obviously one the vectors is linearly dependent so the determinant of the matrix they form must be zero.
 
do this, find this determinant:

\begin{vmatrix}x_1&y_1&ax_1+by_1\\x_2&y_2&ax_2+by_2\\x_3&y_3&ax_3+by_3 \end{vmatrix}

the results should be enlightening.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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