The scalar product Problem set B Worked solutions

The diagram shows two vectors \(\mathbf{a}\) and \(\mathbf{b}\) that leave the same point, with \(|\mathbf{a}|\) = 15, \(|\mathbf{b}|\) = 4 and an angle of 65° between them. Find \(\mathbf{a}\cdot\mathbf{b}\), to 3 significant figures.

• \(\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta = 15\times4\times\cos 65^\circ\)
• \(= 25.4\) (positive, because the angle is acute)

The diagram shows two vectors \(\mathbf{a}\) and \(\mathbf{b}\) that leave the same point, with \(|\mathbf{a}|\) = 5, \(|\mathbf{b}|\) = 14 and an angle of 110° between them. Find \(\mathbf{a}\cdot\mathbf{b}\), to 3 significant figures.

• \(\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta = 5\times14\times\cos 110^\circ\)
• \(= -23.9\) (negative, because the angle is obtuse)

Find the scalar product of each pair of vectors.

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

   • \(\mathbf{a}\cdot\mathbf{b} = (3)(2) + (1)(5) + (4)(-1)\)
   • \(= 6 + 5 - 4 = 7\)

2. (b) \(\mathbf{a} = \begin{pmatrix}0\\-6\\2\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}-4\\3\\7\end{pmatrix}\)

   • \(\mathbf{a}\cdot\mathbf{b} = (0)(-4) + (-6)(3) + (2)(7)\)
   • \(= 0 - 18 + 14 = -4\)

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

   • \(\mathbf{a}\cdot\mathbf{b} = (2)(5) + (-3)(3) + (1)(-1)\)
   • \(= 10 - 9 - 1 = 0\)

Find the scalar product of each pair of vectors.

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

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

2. (b) \(\mathbf{a} = -5\mathbf{i} + 2\mathbf{j} + 3\mathbf{k},\ \mathbf{b} = \mathbf{i} + 6\mathbf{j} - 2\mathbf{k}\)

   • \(\mathbf{a} = \begin{pmatrix}-5\\2\\3\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}1\\6\\-2\end{pmatrix}\)
   • \(\mathbf{a}\cdot\mathbf{b} = (-5)(1) + (2)(6) + (3)(-2) = 1\)

A tug applies a force \(\mathbf{F} = \begin{pmatrix}-8\\9\\5\end{pmatrix}\) N to a load, and the load moves through the displacement \(\mathbf{d} = \begin{pmatrix}4\\1\\6\end{pmatrix}\) m. The work done is \(\mathbf{F}\cdot\mathbf{d}\) joules. Find the work done.

• \(\mathbf{F}\cdot\mathbf{d} = (-8)(4) + (9)(1) + (5)(6)\)
• \(= 7\)
• Work done 7 J.

A handle pulls a trolley with a force of 18 N. The handle makes an angle of 52° with the floor and the trolley moves 9 m along the floor. The work done is the scalar product of the force and the displacement. Find the work done, in joules, to 3 significant figures.

• \(\text{Work} = |\mathbf{F}||\mathbf{d}|\cos\theta = 18\times9\times\cos 52^\circ\)
• \(= 99.7\) J

The points P, Q and R have coordinates P(−2, 0, 3), Q(1, 4, −1) and R(3, −3, 2). Find \(\overrightarrow{PQ}\cdot\overrightarrow{PR}\).

• \(\overrightarrow{PQ} = \begin{pmatrix}1\\4\\-1\end{pmatrix} - \begin{pmatrix}-2\\0\\3\end{pmatrix} = \begin{pmatrix}3\\4\\-4\end{pmatrix}\)
• \(\overrightarrow{PR} = \begin{pmatrix}3\\-3\\2\end{pmatrix} - \begin{pmatrix}-2\\0\\3\end{pmatrix} = \begin{pmatrix}5\\-3\\-1\end{pmatrix}\)
• \(\overrightarrow{PQ}\cdot\overrightarrow{PR} = (3)(5) + (4)(-3) + (-4)(-1) = 7\)

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

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

   • \(\mathbf{a}\cdot\mathbf{b} = (3)(-2) + (2)(4) + (-1)(1) = 1\)
   • \(\mathbf{a}\cdot\mathbf{c} = (3)(5) + (2)(-3) + (-1)(2) = 7\)

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

   • \(\mathbf{b}+\mathbf{c} = \begin{pmatrix}3\\1\\3\end{pmatrix}\)
   • \(\mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) = (3)(3) + (2)(1) + (-1)(3) = 8\)

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

   • \(\mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c} = 1 + (7) = 8\)
   • This equals 8, the value from part (b).