Triple scalar product/coplanarity of

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SUMMARY

The discussion centers on proving that if the triple scalar product (a × b) · c = 0 for nonparallel nonzero vectors a, b, and c, then vector c can be expressed as a linear combination of vectors a and b. The participants emphasize avoiding geometric interpretations and instead focus on algebraic methods using vector identities. The solution involves setting up the vectors in coordinate form and calculating the scalar product to demonstrate coplanarity algebraically.

PREREQUISITES
  • Understanding of vector algebra and operations, specifically cross products and dot products.
  • Familiarity with linear combinations of vectors.
  • Knowledge of vector identities and their applications in proofs.
  • Basic proficiency in manipulating vectors in coordinate form.
NEXT STEPS
  • Study vector identities and their applications in algebraic proofs.
  • Learn about the properties of the triple scalar product in vector calculus.
  • Explore methods for expressing vectors as linear combinations in higher dimensions.
  • Practice problems involving coplanarity and vector relationships in 3D space.
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This discussion is beneficial for students studying linear algebra, particularly those focusing on vector calculus, as well as educators seeking to reinforce algebraic proof techniques in vector analysis.

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Homework Statement


Suppose that a,b,c are nonparallel nonzero vectors, and that ( a \times b) \cdot c = 0. Show that c is expressible as a linear combination of a and b. Avoid geometric arguments (that is, try to stick to vector algebra and symbols in the proof).

Homework Equations


The Attempt at a Solution


The geometric interpretation is that because the triple scalar product of a,b,c is 0, the three vectors are coplanar. Thus, c lies on the plane P determined by a,b. In other words, c is an element of the set of vectors given by the parametrization of the plane P, namely, P = (t_{1}a + t_{2}b: t_{1},t_{2} \epsilon ℝ) And thus, c is expressible as a linear combination of a and b.

But I'm trying to prove this algebraically using vector identities, matrices, and the like. No geometry. Any ideas? I appreciate all help thanks!

BiP
 
Last edited:
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So you have to do this the hard way? Then let a= <a, b, c>, b= <p, q, r>, and c= <x, y, z>. Calculate the coordinates of (a\times b)\cdot c and set them equal to 0.
 
HallsofIvy said:
So you have to do this the hard way? Then let a= <a, b, c>, b= <p, q, r>, and c= <x, y, z>. Calculate the coordinates of (a\times b)\cdot c and set them equal to 0.

Thanks Ivy. Is there any faster way around it, that involves using vector identities and the like?

BiP
 

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