Homework Help: Math - Calculus

Posted by Anonymous on Monday, March 3, 2008 at 3:17pm.

The identity below is significant because it relates 3 different kinds of products: a cross product and a dot product of 2 vectors on the left side, and the product of 2 real numbers on the right side. Prove the identity below.

| a × b |² + (a • b)² = |a|²|b|²

My work, LSH:

= | a × b |² + (a • b)²

= (|a||b|sinθ)(|a||b|sinθ) + (|a||b|cosθ)(|a||b|cosθ)

= (|a|²)(|a||b|)(|a|sinθ)(|a||b|)(|b|²)(|b|sinθ)(|a| sinθ)(|b|sinθ)(sin²θ) + (|a|²)(|a||b|)(|a|cosθ)(|a||b|)(|b|²)(|b|cosθ)(|a| cosθ)(|b|cosθ)(cos²θ)

= (|a|²)(|a||b|)²(|a|sinθ)²(|b|²)(|b|sinθ)²(sin²θ) + (|a|²)(|a||b|)²(|a|cosθ)²(|b|²)(|b|cosθ)²(cos²θ)

= (|a|²|b|²(|a||b|)²) [(|a|sinθ)²(|b|sinθ)²(sin²θ) + (|a|cosθ)²(|b|²)(|b|cosθ)²(cos²θ)]

= (|a|²|b|²(|a||b|)²) [(|a|²)(sin²θ)(|b|²)(sin²θ)(sin²θ) + (|a|²)(cos²θ)(|b|²)(|b|²)(cos²θ)(cos²θ)]

= (|a|²|b|²(|a||b|)²) [(sin²θ)(sin²θ)(sin²θ) + (cos²θ)(cos²θ)(cos²θ)]

And now I don't know what else to do! Please help. Did I mess up somewhere in my steps? Or is it possible to common factor still?

Math - Calculus - Count Iblis, Monday, March 3, 2008 at 5:07pm
My work, LSH:

= | a × b |² + (a • b)²

= (|a||b|sinθ)(|a||b|sinθ) + (|a||b|cosθ)(|a||b|cosθ)

You can write this as:

|a|^2|b|^2[sin^2(theta) + cos^2(theta)]

The result follows from the fact that

sin^2(theta) + cos^2(theta) = 1

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