 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013/3/solution

Write all vectors in Cartesian components. The first component of the nested cross product 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) .

The missing

a_{x}

terms cancel as

- b_{x} a_{x} c_{x} + c_{x} a_{x} b_{x} = 0

. Computing the

y

and

z

components by cyclic permutation gives the same form, proving the vector triple product identity

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

Apply this expression to the three cyclic terms. Their sum is

(a \cdot c) b - (a \cdot b) c + (b \cdot a) c - (b \cdot c) a + (c \cdot b) a - (c \cdot a) b = 0.

Each coefficient cancels because the dot product is symmetric. This is the Jacobi identity for the cross product.

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