Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 6

(a \times b) \cdot c = det (a, b, c) .

A cyclic permutation of its three columns is an even permutation, so it leaves the determinant unchanged. Thus

det (a, b, c) = det (b, c, a) = a \cdot (b \times c)

, proving the identity.

ii

(a \times b) \times c = b (a \cdot c) - a (b \cdot c), a \times (b \times c) = b (a \cdot c) - c (a \cdot b) .

Subtracting cancels the common term, yielding

(a \times b) \times c - a \times (b \times c) = c (a \cdot b) - a (b \cdot c) .

It is generally nonzero: the cross product does not satisfy the associative property.

With

s = sin θ_{0}

and

c = cos θ_{0}

, conversion from spherical coordinates gives

a = r_{0} (s, 0, c)

b = r_{0} (- s /2, 3 s /2, c)

and

c = r_{0} (- s /2, - 3 s /2, c)

V = ∣ det (a, b, c) ∣ = \frac{3 3}{2} r_{0}^{3} sin^{2} θ_{0} ∣ cos θ_{0} ∣.

The absolute value matters when the polar angle is in the lower hemisphere. At the poles or equator the three edge vectors are linearly dependent and the volume vanishes.

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