The scalar product Problem set A One question at a time

The diagram shows two vectors \(\mathbf{a}\) and \(\mathbf{b}\) that leave the same point, with \(|\mathbf{a}|\) = 7, \(|\mathbf{b}|\) = 12 and an angle of 40° between them. Find \(\mathbf{a}\cdot\mathbf{b}\), to 3 significant figures.

• \(\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta = 7\times12\times\cos 40^\circ\)
• \(= 64.3\) (positive, because the angle is acute)

The diagram shows two vectors \(\mathbf{a}\) and \(\mathbf{b}\) that leave the same point, with \(|\mathbf{a}|\) = 10, \(|\mathbf{b}|\) = 6 and an angle of 150° between them. Find \(\mathbf{a}\cdot\mathbf{b}\), to 3 significant figures.

• \(\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta = 10\times6\times\cos 150^\circ\)
• \(= -52.0\) (negative, because the angle is obtuse)

Find the scalar product of each pair of vectors.

1. (a) \(\mathbf{a} = \begin{pmatrix}2\\3\\1\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}4\\-1\\5\end{pmatrix}\)

   • \(\mathbf{a}\cdot\mathbf{b} = (2)(4) + (3)(-1) + (1)(5)\)
   • \(= 8 - 3 + 5 = 10\)

2. (b) \(\mathbf{a} = \begin{pmatrix}-3\\0\\6\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}5\\7\\-2\end{pmatrix}\)

   • \(\mathbf{a}\cdot\mathbf{b} = (-3)(5) + (0)(7) + (6)(-2)\)
   • \(= -15 + 0 - 12 = -27\)

3. (c) \(\mathbf{a} = \begin{pmatrix}1\\-4\\2\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}8\\1\\-2\end{pmatrix}\)

   • \(\mathbf{a}\cdot\mathbf{b} = (1)(8) + (-4)(1) + (2)(-2)\)
   • \(= 8 - 4 - 4 = 0\)

Find the scalar product of each pair of vectors.

1. (a) \(\mathbf{a} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k},\ \mathbf{b} = \mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\)

   • \(\mathbf{a} = \begin{pmatrix}3\\1\\-2\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}1\\4\\5\end{pmatrix}\)
   • \(\mathbf{a}\cdot\mathbf{b} = (3)(1) + (1)(4) + (-2)(5) = -3\)

2. (b) \(\mathbf{a} = 6\mathbf{i} - \mathbf{j} + 2\mathbf{k},\ \mathbf{b} = -\mathbf{i} + 7\mathbf{j} + 3\mathbf{k}\)

   • \(\mathbf{a} = \begin{pmatrix}6\\-1\\2\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}-1\\7\\3\end{pmatrix}\)
   • \(\mathbf{a}\cdot\mathbf{b} = (6)(-1) + (-1)(7) + (2)(3) = -7\)

A crane applies a force \(\mathbf{F} = \begin{pmatrix}15\\-6\\4\end{pmatrix}\) N to a load, and the load moves through the displacement \(\mathbf{d} = \begin{pmatrix}2\\5\\3\end{pmatrix}\) m. The work done is \(\mathbf{F}\cdot\mathbf{d}\) joules. Find the work done.

• \(\mathbf{F}\cdot\mathbf{d} = (15)(2) + (-6)(5) + (4)(3)\)
• \(= 12\)
• Work done 12 J.

A rope pulls a sled with a force of 85 N. The rope makes an angle of 30° with the ground and the sled moves 40 m along the ground. The work done is the scalar product of the force and the displacement. Find the work done, in joules, to 3 significant figures.

• \(\text{Work} = |\mathbf{F}||\mathbf{d}|\cos\theta = 85\times40\times\cos 30^\circ\)
• \(= 2940\) J

The points A, B and C have coordinates A(1, 2, −1), B(4, −2, 3) and C(2, 5, 0). Find \(\overrightarrow{AB}\cdot\overrightarrow{AC}\).

• \(\overrightarrow{AB} = \begin{pmatrix}4\\-2\\3\end{pmatrix} - \begin{pmatrix}1\\2\\-1\end{pmatrix} = \begin{pmatrix}3\\-4\\4\end{pmatrix}\)
• \(\overrightarrow{AC} = \begin{pmatrix}2\\5\\0\end{pmatrix} - \begin{pmatrix}1\\2\\-1\end{pmatrix} = \begin{pmatrix}1\\3\\1\end{pmatrix}\)
• \(\overrightarrow{AB}\cdot\overrightarrow{AC} = (3)(1) + (-4)(3) + (4)(1) = -5\)

Let \(\mathbf{a} = \begin{pmatrix}2\\-1\\3\end{pmatrix}\), \(\mathbf{b} = \begin{pmatrix}4\\0\\-2\end{pmatrix}\) and \(\mathbf{c} = \begin{pmatrix}-1\\5\\1\end{pmatrix}\).

1. (a) Find \(\mathbf{a}\cdot\mathbf{b}\) and \(\mathbf{a}\cdot\mathbf{c}\).

   • \(\mathbf{a}\cdot\mathbf{b} = (2)(4) + (-1)(0) + (3)(-2) = 2\)
   • \(\mathbf{a}\cdot\mathbf{c} = (2)(-1) + (-1)(5) + (3)(1) = -4\)

2. (b) Find \(\mathbf{b}+\mathbf{c}\), and then \(\mathbf{a}\cdot(\mathbf{b}+\mathbf{c})\).

   • \(\mathbf{b}+\mathbf{c} = \begin{pmatrix}3\\5\\-1\end{pmatrix}\)
   • \(\mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) = (2)(3) + (-1)(5) + (3)(-1) = -2\)

3. (c) Show that \(\mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) = \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c}\).

   • \(\mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c} = 2 + (-4) = -2\)
   • This equals −2, the value from part (b).