Triple scalar product/coplanarity of

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The discussion focuses 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 a and b. The geometric interpretation suggests that the vectors are coplanar, with c lying in the plane defined by a and b. The user seeks an algebraic proof using vector identities rather than geometric arguments. Suggestions include representing the vectors in coordinate form and calculating the scalar product to demonstrate the relationship. The conversation emphasizes the desire for a more efficient algebraic approach to the proof.
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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
 
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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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