Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 14 / a

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- Solution a

Solution

If either vector is zero, both sides vanish. Otherwise let

θ

be the angle between them. The magnitudes of the cross product and dot product are described by

∣ a \times b ∣ = ∣ a ∣ ∣ b ∣ sin θ, a \cdot b = ∣ a ∣ ∣ b ∣ cos θ .

Squaring and using

sin^{2} θ = 1 - cos^{2} θ

proves Lagrange's identity:

∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2} .

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