Cross products

\( \textbf{(1)} \left| \left| \vec{a} \times \vec{b} \right| \right|^2 = (a_2 b_3 - a_3 b_2 )^2 + (a_3 b_1 - a_1 b_3)^2 + (a_1 b_2 - a_2 b_1)^2 \)

(2)=a22b32−2a2a3b2b3+a32b22+a32b12−2a1a3b1b3+a12b32+a12b22−2a1a2b1b2+a22b12 \textbf{(2)} =a_{2}^{2} b_{3}^{2} - 2a_2 a_3 b_2 b_3 + a_{3}^{2} b_{2}^{2} + a_{3}^{2} b_{1}^{2} - 2a_1 a_3 b_1 b_3 + a_{1}^{2} b_{3}^2 + a_{1}^{2}b_{2}^{2} - 2a_1 a_2 b_1 b_2 + a_{2}^{2} b_{1}^{2}

(3)=(a12+a22+a32)(b12+b22+b32)−(a1b1+a2b2+a3b3)2 \textbf{(3)} =(a_{1}^2 + a_{2}^2 + a_{3}^2 ) (b_{1}^2 + b_{2}^2 + b_{3}^2 ) - (a_1 b_1 + a_2 b_2 + a_3 b_3)^2

And so on.

How do we go from step (2) to step (3), algebraically?

#Algebra

Note by Hobart Pao
4 years, 4 months ago

No vote yet
1 vote

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Comments

Brahmagupta-Fibonacci identity.

I fail to see why you'd torment yourself in such a way; an easier way would be to start with the Binet-Cauchy identity and the vector triple product...

A Former Brilliant Member - 2 years, 9 months ago
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