Prove that | a x b | = (sqrt) [(a (dot) b)(b (dot) b) - (a (dot) b)^2]
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To prove |a x b| = √[(a · b)(b · b) - (a · b)^2], we will use the properties of vector products and dot products.
First, let's break down the vector product |a x b| (also known as the magnitude of the cross product).
The magnitude of the cross product of two vectors a and b can be calculated using the formula: |a x b| = |a| |b| sinθ
where |a| and |b| represent the magnitudes (lengths) of vectors a and b, respectively, and θ is the angle between them.
Next, let's calculate the dot products involved in the formula: (a · b) and (b · b).
1. The dot product of vectors a and b, denoted as a · b, is defined as the product of their magnitudes and the cosine of the angle θ between them: a · b = |a| |b| cosθ
2. The dot product of a vector b with itself, b · b, is equal to the square of its magnitude: b · b = |b|^2
Now, substitute these values into the original equation:
|a x b| = |a| |b| sinθ
= √[ (|a| |b| cosθ)^2 + (|b|^2 - (|a| |b| cosθ)^2) ] [using sin²θ + cos²θ = 1]
= √[ (|a|^2 |b|^2 cos²θ) + (|b|^2 - |a|^2 |b|^2 cos²θ) ]
= √[ |b|^2 (|a|^2 cos²θ + 1 - |a|^2 cos²θ ) ]
= √[ |b|^2 ]
= |b|
Therefore, we have proven that |a x b| = |b|, which confirms the given statement: |a x b| = √[(a · b)(b · b) - (a · b)^2].