\( \DeclareMathOperator{cosec}{cosec} \)

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International Baccalaureate Mathematics

Geometry and Trigonometry

Syllabus Content

The definition of the vector product of two vectors.
Properties of the vector product.
Geometric interpretation of |v×w|

Here is an exam-style questions on this statement:

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Furthermore

Official Guidance, clarification and syllabus links:

The vector product is also known as the “cross product”.

\( v×w=|v||w| \sin \theta \; n\), where \( \theta \) is the angle between \(v\) and \(w\), and \(n\) is the unit normal vector whose direction is given by the right-hand screw rule.

\(v×w=-w×v\);

\(u×(v+w)=u×v+u×w\);

\((kv)×w=k(v×w)\);

\(v×v=0\).

For non-zero vectors \(v×w=0\) is equivalent to the vectors being parallel.

Use of \( |v×w| \) to find the area of a parallelogram (and hence a triangle).


Formula Booklet AHL 3.16:

Vector product

\( \mathbf{v \times w} = \begin{pmatrix} v_2w_3-v_3w_2 \\ v_3w_1-v_1w_3 \\ v_1w_2-v_2w_1 \end{pmatrix}. \quad \text{where} \quad \mathbf{v} =\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}, \quad \mathbf{w} = \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix} \)

\( \mathbf{|v \times w|} = \mathbf{|v||w|} \sin{ \theta} \), where \( \theta \) is the angle between \( \mathbf{v} \) and \( \mathbf{w} \)

Area of a parallelogram

\(A=|\mathbf{v \times w|}\) where \( \mathbf{v} \) and \( \mathbf{w} \) form two adjacent sides of a parallelogram


The vector product of two vectors is a binary operation that produces a vector perpendicular to both of the input vectors. It is also known as the cross product.

The formula for the vector product of two vectors \( \vec{a} \) and \(\vec{b} \) is given by:

$$\vec{a} \times \vec{b} = \begin{pmatrix} a_{1} \\ a_{2} \\ a_{3} \end{pmatrix} \times \begin{pmatrix} b_{1} \\ b_{2} \\ b_{3} \end{pmatrix} = \begin{pmatrix} a_{2}b_{3} - a_{3}b_{2} \\ a_{3}b_{1} - a_{1}b_{3} \\ a_{1}b_{2} - a_{2}b_{1} \end{pmatrix}$$

For example, let \( \vec{a} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} \) and \( \vec{b} = \begin{pmatrix} 4 \\ -1 \\ 5 \end{pmatrix} \). Then:

$$\vec{a} \times \vec{b} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} \times \begin{pmatrix} 4 \\ -1 \\ 5 \end{pmatrix} = \begin{pmatrix} (3)(5) - (1)(-1) \\ (1)(4) - (2)(5) \\ (2)(-1) - (3)(4) \end{pmatrix} = \begin{pmatrix} 16 \\ -6 \\ -14 \end{pmatrix}$$

Therefore, \( \vec{a} \times \vec{b} = \begin{pmatrix} 16 \\ -6 \\ -14 \end{pmatrix} \).


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