The vector product Problem set A Worked solutions

The vectors \(\mathbf{a}\) and \(\mathbf{b}\) have magnitudes 11 and 6. The angle between them is \(65^\circ\). Find \(|\mathbf{a} \times \mathbf{b}|\).

• \(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta = 11 \times 6 \times \sin 65^\circ\)
• \(= 59.8\)

The vectors \(\mathbf{a}\) and \(\mathbf{b}\) have magnitudes 5 and 12. The angle between them is \(140^\circ\). Find \(|\mathbf{a} \times \mathbf{b}|\).

• \(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta = 5 \times 12 \times \sin 140^\circ\)
• The angle is obtuse. \(\sin 140^\circ = \sin 40^\circ\), so the calculator gives the same positive value
• \(= 38.6\)

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

• \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix}(4)(2) - (-2)(-1)\\ (-2)(3) - (1)(2)\\ (1)(-1) - (4)(3)\end{pmatrix}\)
• \(= \begin{pmatrix}6\\-8\\-13\end{pmatrix}\)

Find \(\mathbf{a} \times \mathbf{b}\) when \(\mathbf{a} = \begin{pmatrix}0\\3\\-1\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}2\\-4\\5\end{pmatrix}\)

• \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix}(3)(5) - (-1)(-4)\\ (-1)(2) - (0)(5)\\ (0)(-4) - (3)(2)\end{pmatrix}\)
• \(= \begin{pmatrix}11\\-2\\-6\end{pmatrix}\)

Find \(\mathbf{a} \times \mathbf{b}\) when \(\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\) and \(\mathbf{b} = \mathbf{i} - 2\mathbf{j} + 4\mathbf{k}\)

• \(\mathbf{a} = \begin{pmatrix}2\\3\\-1\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}1\\-2\\4\end{pmatrix}\)
• \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix}(3)(4) - (-1)(-2)\\ (-1)(1) - (2)(4)\\ (2)(-2) - (3)(1)\end{pmatrix}\)
• \(= \begin{pmatrix}10\\-9\\-7\end{pmatrix}\)
• \(= 10\mathbf{i} - 9\mathbf{j} - 7\mathbf{k}\)

Find \(\mathbf{a} \times \mathbf{b}\) when \(\mathbf{a} = 5\mathbf{i} - 2\mathbf{k}\) and \(\mathbf{b} = \mathbf{i} + 3\mathbf{j}\)

• \(\mathbf{a} = \begin{pmatrix}5\\0\\-2\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}1\\3\\0\end{pmatrix}\)
• \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix}(0)(0) - (-2)(3)\\ (-2)(1) - (5)(0)\\ (5)(3) - (0)(1)\end{pmatrix}\)
• \(= \begin{pmatrix}6\\-2\\15\end{pmatrix}\)
• \(= 6\mathbf{i} - 2\mathbf{j} + 15\mathbf{k}\)

A mechanic pulls on a spanner. The spanner is 0.35 m long and the pull is a force of 80 N at \(70^\circ\) to the spanner. The turning effect (the torque) is the magnitude of the vector product of the position vector and the force. Find the torque, in N m.

• \(|\mathbf{r} \times \mathbf{F}| = |\mathbf{r}||\mathbf{F}|\sin\theta = 0.35 \times 80 \times \sin 70^\circ\)
• \(= 26.3\)

A triangular roof panel has two edges from one corner given, in metres, by the vectors \(\mathbf{a} = \begin{pmatrix}4\\0\\1\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}0\\3\\1\end{pmatrix}\). Find a vector that is perpendicular to the panel.

• \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix}(0)(1) - (1)(3)\\ (1)(0) - (4)(1)\\ (4)(3) - (0)(0)\end{pmatrix}\)
• \(= \begin{pmatrix}-3\\-4\\12\end{pmatrix}\)

The vectors are \(\mathbf{a} = \begin{pmatrix}2\\5\\-3\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}1\\-2\\4\end{pmatrix}\).

1. (a) Find \(\mathbf{a} \times \mathbf{b}\).

   • \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix}(5)(4) - (-3)(-2)\\ (-3)(1) - (2)(4)\\ (2)(-2) - (5)(1)\end{pmatrix}\)
   • \(= \begin{pmatrix}14\\-11\\-9\end{pmatrix}\)

2. (b) Show that \(\mathbf{a} \times \mathbf{b}\) is perpendicular to \(\mathbf{a}\).

   • \(\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = (2)(14) + (5)(-11) + (-3)(-9) = 0\)
   • The scalar product is 0, so the vectors are perpendicular