 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2/11/a/solution

Expand the vector triple product in Cartesian components. Its first component is

[a \times (b \times c)]_{x} = a_{y} (b_{x} c_{y} - b_{y} c_{x}) - a_{z} (b_{z} c_{x} - b_{x} c_{z}) = b_{x} (a_{y} c_{y} + a_{z} c_{z}) - c_{x} (a_{y} b_{y} + a_{z} b_{z}) = b_{x} (a \cdot c) - c_{x} (a \cdot b),

where adding

a_{x} b_{x} c_{x}

to both products introduces cancelling terms. Likewise the other two components are

[a \times (b \times c)]_{y} [a \times (b \times c)]_{z} = a_{z} (b_{y} c_{z} - b_{z} c_{y}) - a_{x} (b_{x} c_{y} - b_{y} c_{x}) = b_{y} (a \cdot c) - c_{y} (a \cdot b), = a_{x} (b_{z} c_{x} - b_{x} c_{z}) - a_{y} (b_{y} c_{z} - b_{z} c_{y}) = b_{z} (a \cdot c) - c_{z} (a \cdot b) .

Together these prove

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

Now use the cyclic property of the scalar triple product and the just-proved identity:

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

This proves Lagrange's identity in the requested form,

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

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