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AI HL · AHL 4

Vectors

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AHL 4.17

The vector product

AHL 4.17

The vector product

Target: I can say what the vector product \(\mathbf{a} \times \mathbf{b}\) is: a vector perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\). (+4 more)

By the end of this section

Key points

What the vector product is

  • The vector product \(\mathbf{a} \times \mathbf{b}\), also called the cross product, is a vector. The scalar product (dot product) is a number.
  • It is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).
  • Its direction follows the right-hand screw rule: turn a right-handed screw from \(\mathbf{a}\) towards \(\mathbf{b}\), and it moves in the direction of \(\mathbf{a} \times \mathbf{b}\).
  • \(\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})\). Swapping the vectors reverses the direction.

The magnitude

  • \(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta\), where θ is the angle between the vectors.
  • Use sin θ here, not cos θ. For the scalar product it is cos θ.
  • The magnitude is never negative. An obtuse angle works as it is, because sin θ is positive for angles up to \(180^\circ\).
  • If \(\mathbf{a}\) and \(\mathbf{b}\) are parallel, \(\sin\theta = 0\) and \(\mathbf{a} \times \mathbf{b} = \mathbf{0}\).

The components

  • \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix}a_2b_3 - a_3b_2\\a_3b_1 - a_1b_3\\a_1b_2 - a_2b_1\end{pmatrix}\)
  • The middle component runs the opposite way round from the other two. It is the one most often wrong.
  • For vectors in i, j, k form, write them as columns first, then write the answer back in i, j, k.
  • Check your answer: its scalar product with \(\mathbf{a}\) and with \(\mathbf{b}\) must be 0.

Explore: the vector product as an area

The vectors a and b lie in the plane z = 0, the page. Drag their tips. a × b is perpendicular to both, so it points straight out of the page or into it: the orange symbol shows which, and the edge-on view below the grid shows it as an arrow whose length is the area of the parallelogram. Before you swap a and b, say what will happen to the arrow.

a × b
Parallelogram (both shaded)
Triangle (the gold part)

⊙ means a × b points out of the page, towards you (+k). ⊗ means into the page (−k). Right-hand screw rule: turn a screw from a towards b (the orange arrow) and it moves the way a × b points. Click a tip, then use the arrow keys to move it one unit.

Drag a tip to change a vector.

  1. Swap a and b. What happens to a × b, to the arrow in the edge-on view and to the area? Why?
  2. Make b turn from anticlockwise of a to clockwise of a. What happens to the sign of the k part, and why does the area stay positive?
  3. Make a and b parallel. What happens to the parallelogram, to a × b and to sin θ?
  4. Make a and b perpendicular. What shape is the parallelogram, and what is |a × b| in terms of |a| and |b|?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Magnitude of a vector product

The vectors \(\mathbf{a}\) and \(\mathbf{b}\) have magnitudes 8 and 3, and the angle between them is \(35^\circ\). Find \(|\mathbf{a} \times \mathbf{b}|\).

|a| = 8|b| = 335°
Working
  1. The formula

    \(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta\)

    θ is the angle between the two vectors. The magnitude of a vector product is never negative.

  2. Substitute

    \(8 \times 3 \times \sin 35^\circ\)

    Check the calculator is in degree mode.

  3. Calculate

    \(= 13.8\)

    3 significant figures.

Video walkthrough coming
An obtuse angle

The vectors \(\mathbf{a}\) and \(\mathbf{b}\) have magnitudes 9 and 4, and the angle between them is \(115^\circ\). Find \(|\mathbf{a} \times \mathbf{b}|\).

|a| = 9|b| = 4115°
Working
  1. The formula

    \(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta\)

    θ is the angle between the two vectors. The magnitude of a vector product is never negative.

  2. Substitute

    \(9 \times 4 \times \sin 115^\circ\)

    Check the calculator is in degree mode. The angle is obtuse, and sin of an obtuse angle is still positive.

  3. Calculate

    \(= 32.6\)

    3 significant figures.

Video walkthrough coming
Two column vectors

Given \(\mathbf{a} = \begin{pmatrix}2\\-1\\3\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}4\\2\\-1\end{pmatrix}\), find \(\mathbf{a} \times \mathbf{b}\).

Working
  1. The formula

    \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix}a_2b_3 - a_3b_2\\a_3b_1 - a_1b_3\\a_1b_2 - a_2b_1\end{pmatrix}\)

    Each component uses the other two numbers of each vector. The middle one runs the opposite way round.

  2. Substitute

    \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix}(-1)(-1) - (3)(2)\\ (3)(4) - (2)(-1)\\ (2)(2) - (-1)(4)\end{pmatrix}\)

    Put brackets round negative numbers.

  3. Simplify

    \(= \begin{pmatrix}-5\\14\\8\end{pmatrix}\)

    Check: the result must be perpendicular to both a and b.

Video walkthrough coming
Vectors in i, j, k form

Given \(\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\) and \(\mathbf{b} = -\mathbf{i} + 4\mathbf{k}\), find \(\mathbf{a} \times \mathbf{b}\) and write the answer in the same form.

Working
  1. Column form

    \(\mathbf{a} = \begin{pmatrix}2\\-3\\1\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}-1\\0\\4\end{pmatrix}\)

    Write each vector as a column. A missing part is 0.

  2. Substitute

    \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix}(-3)(4) - (1)(0)\\ (1)(-1) - (2)(4)\\ (2)(0) - (-3)(-1)\end{pmatrix}\)

    Use the same formula as for columns.

  3. Simplify

    \(= \begin{pmatrix}-12\\-9\\-3\end{pmatrix}\)

  4. Back to i, j, k

    \(\mathbf{a} \times \mathbf{b} = -12\mathbf{i} - 9\mathbf{j} - 3\mathbf{k}\)

    The top, middle and bottom numbers go with i, j and k.

Video walkthrough coming
Changing the order

Given \(\mathbf{a} = \begin{pmatrix}3\\0\\-2\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}1\\4\\1\end{pmatrix}\), find \(\mathbf{b} \times \mathbf{a}\). Give the third component.

Working
  1. Put b first

    \(\mathbf{b} \times \mathbf{a} = \begin{pmatrix}(4)(-2) - (1)(0)\\ (1)(3) - (1)(-2)\\ (1)(0) - (4)(3)\end{pmatrix}\)

    The formula is the same. b now plays the part of a.

  2. Simplify

    \(= \begin{pmatrix}-8\\5\\-12\end{pmatrix}\)

  3. Compare

    \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix}8\\-5\\12\end{pmatrix} = -(\mathbf{b} \times \mathbf{a})\)

    Same length, opposite direction: \(\mathbf{b} \times \mathbf{a} = -\mathbf{a} \times \mathbf{b}\).

Video walkthrough coming