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?
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
It seems like you made a mistake along the way. Let's start fresh and go step by step to prove the identity.
Given the identity: |a × b|² + (a • b)² = |a|²|b|²
First, we can expand the cross product: |a × b|² = (|a||b|sinθ)², where θ is the angle between vectors a and b.
Next, let's expand the dot product: (a • b)² = (|a||b|cosθ)²
Now, let's substitute these expansions back into the original identity:
|a × b|² + (a • b)² = (|a||b|sinθ)² + (|a||b|cosθ)²
We can notice that both terms have a factor of (|a||b|)², so let's factor that out:
= (|a||b|)²[(sinθ)² + (cosθ)²]
Now, recall the trigonometric identity: (sinθ)² + (cosθ)² = 1
= (|a||b|)²(1)
= |a|²|b|²
Therefore, we have successfully proven the identity: |a × b|² + (a • b)² = |a|²|b|².