 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/11/a/solution

Δ = u \cdot (v \times w) \neq = 0 .

It is the determinant of the matrix with these columns, so this condition is equivalent to linear independence and spanning the whole space.

Take the dot product of the expansion of

y

with

v \times w

. The terms containing

v

and

w

vanish, leaving

y \cdot (v \times w) = α Δ

. Cycling the same argument yields

α = \frac{y \cdot ( v \times w )}{Δ}, β = \frac{y \cdot ( w \times u )}{Δ}, γ = \frac{y \cdot ( u \times v )}{Δ} .

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