Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 15 / b

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- Solution b

Solution

Expand the same double cross product in two ways. First expanding with

c \times d

as the inner cross product gives

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

Alternatively, use

(a \times b) \times w = (a \cdot w) b - (b \cdot w) a

with

w = c \times d

to get

(a \times b) \times (c \times d) = [a, c, d] b - [b, c, d] a .

Equating these expressions and moving every term to one side proves

[b, c, d] a - [a, c, d] b + [a, b, d] c - [a, b, c] d = 0.

This four-vector scalar triple product identity is valid without assuming that any three of the four vectors are independent.

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