 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2014-resit/11/i/solution

Put

μ = ∣ b ∣^{2} > 0

and

λ = a \cdot b

. Expanding the nonnegative squared Euclidean norm of

c = μ a - λb

gives

0 \leq ∣ c ∣^{2} = μ^{2} ∣ a ∣^{2} - 2 μ λ^{2} + λ^{2} ∣ b ∣^{2} = μ (∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}) .

Dividing by

μ

∣ a \cdot b ∣ \leq ∣ a ∣ ∣ b ∣.

Equality holds exactly when

c = 0

, equivalently when the nonzero vectors

a, b

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