11 Week 11

11.1 Vectors I

A scalar quantity has only size or magnitude (i.e. all the information is just a single number). For example: mass, time, speed \(\dots\)

A vector quantity has magnitude and direction. For example: force, velocity, acceleration, moment, \(\dots\)

A vector is represented graphically by a straight line with an arrowhead showing its direction. Often the line is drawn to scale so that the magnitude of the vector may be determined by measuring its length. To refer to a vector we give it a name. There are several different conventions used to indicate that we are referring to a vector quantity. The vector above will often be referred to in books as \(\underline{OP}\), or \(\overrightarrow{OP}\), or \(\underline p\), or \(\vec{p}\). The point at which a vector starts is called the initial point. The point at which it ends is called the final point.

The magnitude of a vector, also sometimes called the length of a vector, is denoted by enclosing the name (any name) of a vector by vertical lines (“modulus lines”; it is the same notation as used for the absolute value of a number): for example \(|\underline{OP}|\) or \(|\underline{p}|\). It is a non-negative number.

We can multiply a vector by a non-negative scalar (i.e. a non-negative number). This has the effect of leaving the direction the same, but changing the magnitude. For example, below we have the vectors \(\underline a\) and \(3\underline a\). Note that \(3\underline a\) is parallel to \(\underline a\) (has the same direction) but is \(3\) times as long. (At this point it may be helpful to also imagine that they have the same initial point, but for the sake of the picture they are drawn separately.) Multiplying a vector (say \(\underline{p}\)) by a negative scalar (say \(a<0\)) changes the direction of the vector to the opposite (imagine that it stays on the same line, the same initial point, but opposite direction). The magnitude of \(a\cdot\underline{p}\) is \(|a|\cdot |\underline{p}|\).

A position vector indicates the distance and direction that we have to travel from the origin to get to a point. We use the term displacement vector to refer to the same concept, but relative to any other point than origin.

In the diagram below, if \(O\) is the origin then \(\underline{OP}\) is the position vector of the point called \(P\). \(\underline{PQ}\) is a displacement vector from \(P\) to \(Q\). It shows the distance and direction that we have to travel from \(P\) to get to \(Q\). \(\underline{OQ}\) is the position vector of the point \(Q\).

11.1.1 Vector Algebra

We have already seen that a vector may be multiplied by a scalar. If the scalar is positive this changes the magnitude but not the direction of the vector. If the scalar is negative the vector is turned to the opposite direction, and the magnitude is changed.

We’ve seen that a vector may be multiplied by a scalar. If the scalar is positive this changes only the magnitude. If it is negative, the direction is changed to the opposite, and the magnitude is changed.

We declare two vectors \(\underline{a}\) and \(\underline{b}\) to be equal, if they have the same magnitude and the same direction. They do not actually have to start and finish at the same point. In the above diagram, \(\underline a= \underline b = \underline c\).

If a vector \(\underline b\) has the same magnitude as vector \(\underline a\), but points in the opposite direction then \(\underline b = -\underline a\). Again it doesn’t matter whether they start and finish at the same point or not.

The sum (or resultant) of two vectors is obtained by placing the initial point of the second vector in the sum at the final point of the first vector in the sum. The sum is then the vector that completes the triangle (i.e. runs from the initial point of the first to the final point of the second). This is called the triangle law of vector addition.

The difference of two vectors is given by \(\underline a + (-\underline b)\), where \(-\underline b\) points in the opposite direction to \(\underline b\). The sum of \(\underline a + (-\underline b)\) is then found from the triangle law above. Equivalently, we define \(\underline a-\underline b = \underline a + (-\underline b)\).

The sum of a set of vectors that form a closed loop is always zero. \(\underline a + \underline b + \underline c + \underline d + \underline e = \underline 0\).

Example 11.1 For the diagram below, state the magnitude of the sum of the vectors \(\underline{AB}\), \(\underline{BC}\), \(\underline{CD}\) and \(\underline{DE}\).

Solution. The sum of all the vectors starts at the initial point of the first vector and ends at the final point of the last vector. This is the vector \(\underline{AE}\). Its magnitude may be written \(|\underline{AE}|\).

If a vector joins two points \(A\) and \(B\), then it is equal to the sum of any system of vectors that form a path from \(A\) to \(B\). For example the vector \(\underline{AB}\) shown as a solid line below, is equivalent to the sum of the vectors shown dotted. This fact may be used to solve geometric problems using vectors.

Example 11.2 For the triangle \(ABC\), if \(D\) is the mid-point of \(AB\) show that \[\begin{equation} 2\underline{AB}+3\underline{BC}+\underline{CA} = 2\underline{DC}. \tag{11.1} \end{equation}\]

Solution. From the triangle \[\begin{align} \underline{AB}&=2\underline{AD} \tag{11.2}\\ \underline{BC}&=\underline{BD}+\underline{DC} \tag{11.3}\\ \underline{CA}&=\underline{CD}+\underline{DA} \tag{11.4}\\ \end{align}\] Substituting the equations (11.2), (11.3), and (11.4) for the left-hand side of (11.1) gives \[\begin{equation} 2\underline{AB}+3\underline{BC}+\underline{CA}=4\underline{AD}+3\underline{BD}+3\underline{DC}+\underline{CD}+\underline{DA}. \tag{11.5} \end{equation}\] Now \[\begin{align*} \underline{BD}&=-\underline{AD}\\ \underline{CD}&=-\underline{DC}\\ \underline{DA}&=-\underline{AD}\\ \end{align*}\] Substituting these into the right-hand side of equation (11.5) gives \[ 2\underline{AB}+3\underline{BC}+\underline{CA}=4\underline{AD}-3\underline{AD}+3\underline{DC}-\underline{DC}-\underline{AD}. \] So \[ 2\underline{AB}+3\underline{BC}+\underline{CA} = 2\underline{DC}. \]

11.2 Vectors II

11.2.1 Unit Vectors

A unit vector is a vector with magnitude of 1.

If \(\underline a\) is a vector with magnitude \(|\underline a|\), then a unit vector with the same direction is written \(\hat{\underline a}\). So \(|\hat{\underline a}| = 1\) and \[ \hat{\underline a} = \frac1{|\underline a|}{\underline a}. \]

For (3-dimensional) vector systems, there are three special unit vectors \(\underline i\), \(\underline j\) and \(\underline k\). These vectors for a set of axes for a vector system and are perpendicular to each other (and usually correspond to \(x\), \(y\) and \(z\) in Cartesian notation). Strictly speaking, they should be written \(\hat{\underline i}, \hat{\underline j}\) and \(\hat{\underline k}\) to indicate that they are unit vectors, but they are used so often that the circumflex \(\hat{}\) tends to get left off.

The axis vectors are arranged as a righthand system. This means that a screw turned from \(\underline i\) to \(\underline j\) should travel up the \(\underline k\) axis.

Any vector \(\underline{r}\) may be ‘resolved’ in terms of its components in the direction of the unit axis vectors, i.e. written as a sum of some multiples of the axis vectors.

We shall start by considering a two dimensional vector. For this we need only the \(\underline i\) and \(\underline j\) axes.

The vector \(\underline r\) makes an angle \(\theta\) with the \(\underline i\) axis and has magnitude \(|\underline r| = r\). We may say that to get to the point \(R\), we have to start at the \(\underline i\) axis, turn through an angle \(\theta\) and then travel a distance \(r\) from the origin. Alternatively, we may get to \(R\), by travelling a distance along the \(\underline i\) axis and then a distance up in the direction of the \(\underline j\) axis. The distance to be travelled along the \(\underline i\) axis is \(r\cos(\theta)\) and the distance to be travelled up in the direction of the \(\underline j\) axis is \(r\sin(\theta)\). The vector \(\underline r\) may thus be written as \(\underline r = r\cos(\theta) \underline i + r\sin(\theta)\underline j\).

When written in this form \(\underline r\) is said to be resolved into its component vectors.

A 2D vector may be given in its component form as, for example \(\underline r = 6\underline i + 5\underline j\).

If we have two 2D vectors lying in the same plane, then they are said to be co-planar vectors.

The vectors \(\underline a\) and \(\underline b\) are co-planar.

We may also write 3D vectors in terms of their unit vectors, but this time we need all three axes.

The vector \(\underline r\) projects a length \(a\) onto the \(\underline i\) axis, a length \(b\) onto the \(\underline j\) axis and a length \(c\) onto the \(\underline k\) axis, so \(\underline r\) may be written \(\underline r = a\underline i + b\underline j + c\underline k\).

11.2.2 Vector algebra with component vectors

When a vector is written in terms of its components, then the sum of two vectors is the sum of the components for each direction.

Example 11.3 If \(\underline a = 6\underline i + 2\underline j\) and \(\underline b = 5\underline i - \underline j\), calculate \(\underline{a}+\underline{b}\).

Solution. \[ \underline a + \underline b = (6 + 5)\underline i + (2 + (-1))\underline j = 11\underline i + \underline j. \]

Note that \(1\underline i\), \(1\underline j\), \(1\underline k\) are usually written as just \(\underline i\), \(\underline j\), \(\underline k\).

When a vector is written in terms of components, the difference of the two vectors is the difference of the components for each direction.

Example 11.4 If \(\underline a = 5\underline i + 4\underline j - \underline k\) and \(\underline b = 3\underline i - 7\underline j + 8\underline k\), calculate \(\underline{a}+\underline{b}\) and \(\underline{a}-\underline{b}\).

Solution. \[\begin{align*} \underline a + \underline b &= (5+3)\underline i + (4 - 7)\underline j + (-1 + 8)\underline k = 8\underline i -3\underline j + 7\underline k,\\ \underline a - \underline b &= (5 - 3)\underline i + (4 - (-7))\underline j + ((-1) - 8)\underline k = 2\underline i + 11\underline j -9\underline k. \end{align*}\]

When a vector written in terms of components is multiplied by a scalar, each component is multiplied by that scalar.

Example 11.5 If \(\underline a = 5\underline i + 4\underline j - \underline k\), calculate \(3\underline{a}\).

Solution. \[ 3\underline a = (3\times 5)\underline{i} + (3\times 4)\underline{j} + (3\times(-1))\underline{k} = 15\underline i + 12\underline j - 3\underline k. \]

11.2.3 The magnitude of a vector

Suppose \(\underline r = a\underline i + b\underline j\). From Pythagoras’ theorem, the magnitude of vector \(|\underline r| = \sqrt{a^2+b^2}\).

In three dimensions, if \(\underline r = a\underline i + b\underline j + c\underline k\), then \(|\underline r| = \sqrt{a^2+b^2+c^2}\).

11.2.4 Finding a unit vector

If \(\underline r = a\underline i + b\underline j + c\underline k\), then its magnitude (= length) is \(|r| = \sqrt{a^2+b^2+c^2}\). If we want to find a vector that has the same direction as \(\underline r\), but has length of 1, i.e. the unit vector \(\underline{\hat r}\), then we must divide \(\underline r\) by its own magnitude. So \[ \underline{\hat r} = \frac{1}{\sqrt{a^2+b^2+c^2}}(a\underline i+b\underline j+c\underline k) \] which can be also written as \[ \underline{\hat r} = \frac{a}{\sqrt{a^2+b^2+c^2}}\underline i+\frac{b}{\sqrt{a^2+b^2+c^2}}\underline j+\frac{c}{\sqrt{a^2+b^2+c^2}}\underline k \] or \[ \underline{\hat r} = \frac{a}{|\underline r|}\underline i+\frac{b}{|\underline r|}\underline j+\frac{c}{|\underline r|}\underline k. \]

Example 11.6 If \(\underline r = 5\underline i+2\underline j-\underline k\), find a unit vector that is

1. parallel to \(\underline r\),
2. has the opposite direction to \(\underline r\).

Solution. \(|\underline r| = \sqrt{5^2+2^2+1^2} = \sqrt{25+4+1} = \sqrt{30}\).

1. The unit vector in the direction of \(\underline r\) is \[ \underline{\hat r} = \frac5{\sqrt{30}}\underline i+\frac2{\sqrt{30}}\underline j-\frac1{\sqrt{30}}\underline k. \]
2. The unit vector in the opposite direction is \[ -\underline{\hat r} = -\left(\frac{5}{\sqrt{30}}\underline i+\frac{2}{\sqrt{30}}\underline j-\frac{1}{\sqrt{30}}\underline k\right) = -\frac{5}{\sqrt{30}}\underline i-\frac{2}{\sqrt{30}}\underline j+\frac{1}{\sqrt{30}}\underline k \]

11.2.5 Direction Cosines

The direction cosines of a vector are the cosines of the angles that that vector makes with each of the three axes. The angles are marked \(\alpha, \beta\) and \(\gamma\) in the figure below The direction cosines are \[ l = \cos(\alpha) = \frac{a}{|\underline r|},\;\;m = \cos(\beta) = \frac{b}{|\underline r|},\;\;n = \cos(\gamma) = \frac{c}{|\underline r|}. \]

Example 11.7 Find the direction cosines of the vector \(\underline r = 3\underline i + 4\underline j - 5\underline k\).

Solution. The magnitude of \(\underline r\) is \(|\underline r| = \sqrt{3^2+4^2+5^2} = \sqrt{9+16+25} = \sqrt{50}\).

The direction cosines of \(\underline r\) are then \[ l = \frac{3}{\sqrt{50}},\;\;m = \frac{4}{\sqrt{50}},\;\;n = \frac{-5}{\sqrt{50}} \]

11.2.6 The angle between two vectors

If we have two vectors \(\underline a\) and \(\underline b\) where \(\underline a\) has direction cosines \(l_a, m_a, n_a\) and \(\underline b\) has direction cosines \(l_b, m_b, n_b\), then the cosine of the angle \(\theta\) between the vectors is given by: \[ \cos(\theta) = l_al_b + m_am_b + n_an_b. \]

Example 11.8 Determine the angle between the two vectors and \(\underline a = 2\underline i + 3\underline j + 4\underline k\) and \(\underline b = 4\underline i - 3\underline j + 2\underline k\).

Solution. \[ |\underline a| = \sqrt{2^2+3^2+4^2} = \sqrt{29},\;\;|\underline b| = \sqrt{4^2+3^2+2^2} = \sqrt{29}. \] So \(\underline a\) has direction cosines \[ l_a = \frac{2}{\sqrt{29}},\;\;m_a = \frac{3}{\sqrt{29}},\;\;n_a = \frac{4}{\sqrt{29}} \] and \(\underline b\) has direction cosines \[ l_b = \frac{4}{\sqrt{29}},\;\;m_b = \frac{-3}{\sqrt{29}},\;\;n_b = \frac{2}{\sqrt{29}} \] The cosine of the angle \(\theta\) between the two vectors is given by \[ \cos(\theta) = \frac{2}{\sqrt{29}}\frac{4}{\sqrt{29}}+\frac{3}{\sqrt{29}}\frac{-3}{\sqrt{29}}+\frac{4}{\sqrt{29}}\frac{2}{\sqrt{29}} = \frac{8}{29}-\frac{9}{29}+\frac{8}{29} = \frac{7}{29}. \] Hence the angle is \(\theta = \cos^{-1}(7/29) = 76^\circ\).

11.3 Vectors III

11.3.1 Multiplication of vectors

In three dimensions, there are two ‘sensible’ ways to ‘multiply’ two vectors.

• “scalar” or “dot” product: The result is a number (scalar). For example, if a force (vector) is applied to an object and it moves along a displacement vector, the dot product of the two involved vectors gives the amount of work done (scalar).

• “vector” or “cross” product: The result is a vector. If a force (vector) is applied at a point whose position vector is perpendicular to it, the cross product of the two involved vectors is a moment of the force about that point (a vector).

11.3.2 The scalar or dot product

The scalar product of two vectors has a scalar quantity as its result.

If \(\underline a\) and \(\underline b\) are given in terms of component vectors as \(\underline a = a_1\underline i + a_2\underline j + a_3\underline k\) and \(\underline b = b_1\underline i + b_2\underline j + b_3\underline k\) then the dot product \(\underline{a}\cdot \underline{b}\) is defined to be \[ \underline a \cdot\underline b = a_1b_1 + a_2b_2 + a_3b_3. \]

Example 11.9 Find the scalar product of \(\underline a = 3\underline i + 2\underline j - 6\underline k\) and \(\underline b = -\underline i + 5\underline j - \underline k\).

Solution. \[ \underline a\cdot\underline b = a_1b_1 + a_2b_2 + a_3b_3 = (3)(-1)+(2)(5)+(-6)(-1) = -3+10+6=13. \] The scalar product of the vectors is 13.

If we look back at the formula for the angle between two vectors, we arrive at an alternative formula for the dot product of two vectors \(\underline a\) and \(\underline b\): \[ \underline a \cdot\underline b = |\underline a||\underline b|\cos(\theta) \] where \(\theta\) is the angle between the two vectors as shown below

Useful observations:

• If \(\underline a\) and \(\underline b\) are parallel, then \(\theta = 0^\circ\) and \(\cos(\theta) = 1\), so \(\underline a \cdot\underline b = |\underline a||\underline b|\).

• If \(\underline a\) and \(\underline b\) are perpendicular, then \(\theta = 90^\circ\) and \(\cos(\theta) = 0\) so \(\underline a \cdot\underline b = 0\).

• Having two formulas for \(\underline{a}\cdot\underline{b}\) means that we have an alternative way for finding an angle between two vectors given in terms of component vectors: \[\begin{equation} \cos(\theta) = \frac{\underline a\cdot\underline b}{|\underline a||\underline b|}. \tag{11.6} \end{equation}\]

Example 11.10 Using the vectors from the previous example, \(\underline a = 3\underline i + 2\underline j - 6\underline k\) and \(\underline b = -\underline i + 5\underline j - \underline k\), calculate the angle between them.

Solution. We already calculated that \(\underline a\cdot\underline b = 13\).

Also, \(|\underline a| = \sqrt{3^2+2^2+(-6)^2}=\sqrt{49}=7\) and \(|\underline b| = \sqrt{(-1)^2+5^2+(-1)^2} = \sqrt{27}\) so from equation (11.6) \[ \cos(\theta) = \frac{13}{7\sqrt{27}} \approx 0.3574 \text{ (4 d.p.)} \] and thus \(\theta \approx \cos^{-1}(0.3574) \approx 69^\circ\).

Example 11.11 Find the unit vector perpendicular to the vector \(4\underline i+4\underline j-7\underline k\) and to the vector \(3\underline i - 2\underline j+\underline k\).

Solution. Let the vector which is perpendicular to the two given vectors be \(p\underline i + q\underline j + r\underline k\).

If the vectors are perpendicular, then the scalar product is zero (since the angle between them is \(90^\circ\)). So \[\begin{equation} (4\underline i + 4\underline j - 7\underline k) \cdot (p\underline i + q\underline j + r\underline k) = 0 \tag{11.7} \end{equation}\] and \[\begin{equation} (3\underline i - 2\underline j + \underline k) \cdot (p\underline i + q\underline j + r\underline k) = 0. \tag{11.8} \end{equation}\] Evaluating the scalar products (11.7) and (11.8) gives \[\begin{align} 4p + 4q -7r& = 0 \tag{11.9}\\ 3p - 2q + r& = 0. \tag{11.10} \end{align}\] We now solve equations (11.9) and (11.10) simultaneously: multiplying both sides of (11.10) by \(2\) we obtain \[\begin{align*} 4p + 4q - 7r& = 0\\ 6p - 4q + 2r& = 0,\\ \end{align*}\] and then by adding the two equations, we get \(10p - 5r = 0\). Thus \[\begin{equation} r = 2p. \tag{11.11} \end{equation}\] Substituting into (11.10) gives \(3p - 2q + 2p = 0\), so \[\begin{equation} q = \frac52p. \tag{11.12} \end{equation}\] We may choose any value for \(p\) and if \(q\) and \(r\) are related by (11.11) and (11.12) with our chosen value, the resulting vector will be perpendicular to the given vectors. This is because there are infinitely many vectors that are perpendicular to any given two (non–parallel) vectors: they will all be scalar multiples of each other.

We will choose \(p = 2\). Then \(r= 4\) and \(q = 5\) so \[ p\underline i + q\underline j + r\underline k = 2\underline i + 5\underline j + 4\underline k. \] We are supposed to find a unit vector, and the one we have found so far may not be such. The magnitude of \(2\underline i + 5\underline j + 4\underline k = \sqrt{45} = 3\sqrt{5}\), so a unit vector perpendicular to the given vectors is \[ \frac{2}{3\sqrt{5}}\underline i+\frac{5}{3\sqrt{5}}\underline j+\frac{4}{3\sqrt{5}}\underline k. \] (Note that there is another solution, namely the opposite vector to the one we found, \(-\frac{2}{3\sqrt{5}}\underline i-\frac{5}{3\sqrt{5}}\underline j-\frac{4}{3\sqrt{5}}\underline k\).)

11.3.3 The vector or cross product

The result of a “vector product” of two vectors is a vector.

The vector product is defined as \(\underline a \times\underline b = |\underline a||\underline b|\sin(\theta)\underline{\hat n}\), where \(\theta\) is the angle between \(\underline{a}\) and \(\underline{b}\), and \(\underline{\hat{n}}\) is a unit vector perpendicular to both \(\underline{a}\) and \(\underline{b}\) chosen according to the right–hand rule (explained below).

(From previous slide: \(\underline a \times\underline b = |\underline a||\underline b|\sin(\theta)\underline{\hat n}\))

Unwrapping:

• The magnitude of \(\underline a\times\underline b\) is \(|\underline a||\underline b|\sin(\theta)\).
• If \(\underline{a}\) and \(\underline{b}\) are parallel, then \(\theta=0\), and so \(\underline{a}\times\underline{b} = \underline{0}\) (regardless of \(\underline{\hat{n}}\)).
• If \(\underline{a}\) and \(\underline{b}\) are not parallel, then there are two potential choices for \(\underline{\hat{n}}\). In the figure below, it is either “up” or “down”. The choice is made so that \(\underline{a}, \underline{b}, \underline{a}\times\underline{b}\) forms a right–hand system (like \(\underline{i},\underline{j},\underline{k}\)).
• It follows that \(\underline{b}\times \underline{a} = - \underline{a}\times\underline{b}\).
• If \(\underline a\) and \(\underline b\) are perpendicular (i.e. \(\theta = 90^\circ\)) then \(\sin(\theta)= 1\), and so \(\underline a \times\underline b = |\underline a||\underline b|\underline {\hat n}\).

There is also an “algebraic” formula for the cross product, when \(\underline a\) and \(\underline b\) are specified in terms of their component vectors. Supposing that \(\underline a = a_1\underline i + a_2\underline j + a_3\underline k\) and \(\underline b = b_1\underline i + b_2\underline j + b_3\underline k\), the formula can be remembered symbolically as \[ \underline a \times\underline b = \left|\begin{array}{ccc}\underline i&\underline j&\underline k\\a_1&a_2&a_3\\b_1&b_2&b_3\\\end{array}\right|. \]

Example 11.12 Find the vector product of \(\underline a=3\underline i+2\underline j-6\underline k\) and \(\underline b=-\underline i + 5\underline j-\underline k\).

Solution. \[\begin{align*} \underline a \times\underline b& = \left|\begin{array}{ccc}\underline i&\underline j&\underline k\\3&2&-6\\-1&5&-1\\\end{array}\right|\\ &= \underline i((2)(-1)-(-6)(5))-\underline j((3)(-1)-(-6)(-1))+\underline k((3)(5)-(2)(-1))\\ &=28\underline i+9\underline j+17\underline k. \end{align*}\]

Example 11.13 Find a unit vector which is perpendicular to the vector \(4\underline i+4\underline j-7\underline k\) and to the vector \(3\underline i - 2\underline j+\underline k\).

Solution. From the definition of the vector product, a vector perpendicular to both these vectors is given by \[\begin{align*} (4\underline i+4\underline j-7\underline k) \times(3\underline i - 2\underline j+\underline k)& = \left|\begin{array}{ccc}\underline i&\underline j&\underline k\\4&4&-7\\3&-2&1\\\end{array}\right|\\ &= \underline i((4)(1)-(-7)(-2))\\ &\phantom{= }-\underline j((4)(1)-(-7)(3))\\ &\phantom{= }+\underline k((4)(-2)-(4)(3))\\ &=-10\underline i-25\underline j-20\underline k. \end{align*}\] We need to find a unit vector parallel to this one; its magnitude is \(15\sqrt{5}\), and so a unit vector perpendicular to the two given ones is \[ \frac{-10}{15\sqrt{5}}\underline i-\frac{25}{15\sqrt{5}}\underline j-\frac{20}{15\sqrt{5}}\underline k = \frac{-2}{3\sqrt{5}}\underline i-\frac{5}{3\sqrt{5}}\underline j-\frac{4}{3\sqrt{5}}\underline k. \]

11.3.4 The scalar triple product

If we have three vectors \(\underline a, \underline b\) and \(\underline c\) then the scalar triple product of those vectors is defined to be \(\underline a \cdot(\underline b\times\underline c)\).

Note that the vector product must be evaluated first, then the scalar product.

The scalar triple product gives the (signed) volume of a parallelepiped with edges defined by the vectors \(\underline a, \underline b\) and \(\underline c\).

Example 11.14 Find the volume of a parallelepiped with edges \(\underline a = 2\underline i - 3\underline j + 4\underline k\), \(\underline b = \underline i + 2\underline j - \underline k\) and \(\underline c = 3\underline i - \underline j + \underline 2\).

Solution. The volume is given by \(\underline a \cdot(\underline b\times\underline c)\). First we calculate the vector product \[ \underline b \times\underline c = \left|\begin{array}{ccc}\underline i&\underline j&\underline k\\1&2&-1\\3&-1&2\\\end{array}\right| = 3\underline i-5\underline j-7\underline k. \] and now \[ \underline a \cdot(\underline b\times\underline c) = (2\underline i - 3\underline j + 4\underline k)\cdot(3\underline i-5\underline j-7\underline k) = -7. \] The volume is therefore \(7\) units (and the -ve sign tells us that the system was not right handed).

11.3.5 The vector triple product

If we have three vectors \(\underline a, \underline b\) and \(\underline c\) then the vector triple product of those vectors is defined as \[ \underline a \times (\underline b \times\underline c) \]

Example 11.15 Find the vector triple product \(\underline a \times (\underline b \times\underline c)\) where \(\underline a = 2\underline i - 3\underline j + 4\underline k\), \(\underline b = \underline i + 2\underline j - \underline k\) and \(\underline c = 3\underline i - \underline j + \underline 2\).

Solution. From the previous example, \(\underline b \times\underline c = 3\underline i-5\underline j-7\underline k\). \[ \underline a \times (\underline b \times\underline c) = \left|\begin{array}{ccc}\underline i&\underline j&\underline k\\2&-3&4\\3&-5&7\end{array}\right| = 41\underline i+26\underline j-\underline k \]