Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV / 1 / a

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- Solution a

Solution

a, b, c \in R^{3}

, the scalar product, cross product and scalar triple product are respectively

a \cdot b a \land b (a \land b) \cdot c = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}, = (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1}), = det a_{1} b_{1} c_{1} a_{2} b_{2} c_{2} a_{3} b_{3} c_{3} .

Expanding the dot product gives

(a \land b) \cdot a = a_{1} a_{2} b_{3} - a_{1} a_{3} b_{2} + a_{2} a_{3} b_{1} - a_{1} a_{2} b_{3} + a_{1} a_{3} b_{2} - a_{2} a_{3} b_{1} = 0.

Interchanging

a

and

b

changes the sign of their cross product, so its dot product with

b

is also zero. Thus the cross product is orthogonal to both vectors, including when it is the zero vector.

The displayed scalar triple product is a determinant. A square matrix has nonzero determinant exactly when its rows are linearly independent. Hence

(a \land b) \cdot c \neq = 0 ⟺ a, b, c are linearly independent .

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