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

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

The scalar triple product is a determinant:

(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

Solution

Use the vector triple product identity twice:

(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.

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