Title: Default for 9.4 The Vector Product
Course: HELM Workbooks 2023-2024

The right-handed screw rule

To understand how the vector product is formed it is helpful to consider first the right-handed screw rule. Consider the two vectors \(\underline{a}\) and \(\underline{b}\) shown in the following figure.

[Image: vectors a and b with coinciding tails and the angle theta between them]

The two vectors lie in a plane; this plane is shaded in the figure above. The following figure shows the same two vectors and the plane in which they lie together with a unit vector denoted \(\hat{\underline{e}}\), which is perpendicular to this plane. Imagine turning a right-handed screw, aligned along \(\underline{\hat{e}}\), in the direction from \(\underline{a}\) towards \(\underline{b}\) as shown. A right-handed screw is one which when turned clockwise enters the material into which it is being screwed (most screws are of this kind). You will see from the figure below that the screw will advance in the direction of \(\hat{\underline{e}}\).

[Image: clockwise turning of the screw sitting along e]

On the other hand, if the right-handed screw is turned from \(\underline{b}\) towards \(\underline{a}\) the screw will retract in the direction of \(\underline{\hat{f}}\) as shown in the following figure.

[Image: anticlockwise turning of the screw sitting along e]

We are now in a position to describe the vector product.

Definition of the vector product

We define the vector product of \(\underline{a}\) and \(\underline{b}\), written \(\underline{a}\times\underline{b}\), as \[\qquad\underline{a}\times\underline{b}=|\underline{a}|\,|\underline{b}|\,\sin\theta\,\underline{\hat{e}}\] By inspection of this formula note that this is a vector of magnitude\(|\underline{a}|\,|\underline{b}|\,\sin\theta\) in the direction of the vector \(\underline{\hat{e}}\), where \(\hat{\underline{e}}\) is a unit vector perpendicular to the plane containing \(\underline{a}\) and \(\underline{b}\) in the sense defined by the right-handed screw rule. The quantity \(\underline{a}\times\underline{b}\) is read as ‘‘\(\underline{a}\) cross \(\underline{b}\)’’ and is sometimes referred to as the cross product. The angle is chosen to lie between \(0\) and \(\pi\). See the figure below.

[Image]

Formally we have

Key Point

The vector product of \({\underline a}\) and \({\underline b}\) is: \(\underline{a}\times \underline{b} = |\underline{a}|\,|\underline{b}|\,\sin \theta\,\hat{\underline{e}}\)

The modulus of the vector product is: \(|\underline{a}\times \underline{b}| = |\underline{a}|\,|\underline{b}|\,\sin \theta\)

Note that \(|\underline{a}|\,|\underline{b}|\,\sin \theta\) gives the modulus of the vector product whereas \(\hat{\underline{e}}\) gives its direction.

Now study the figure below which is used to illustrate the calculation of \(\underline{b}\times \underline{a}\). In particular note the direction of \(\underline{b}\times\underline{a}\) arising through the application of the right-handed screw rule.

We see that \(\underline{a}\times \underline{b}\) is not equal to \(\underline{b}\times \underline{a}\) because their directions are opposite. In fact \(\underline{a}\times \underline{b}=-\underline{b}\times\underline{a}\).

[Image]

Example

If \(\underline{a}\) and \(\underline{b}\) are parallel, show that \(\underline{a}\times \underline{b}=\underline{ 0}\).

Solution

If \(\underline{a}\) and \(\underline{b}\) are parallel then the angle \(\theta\) between them is zero. Consequently \(\sin\theta=0\) from which it follows that \(\underline{a}\times \underline{b}= |\underline{a}|\,|\underline{b}|\,\sin \theta \,\hat{\underline{e}}=\underline{0}\). Note that the result, \(\underline{0}\), is the zero vector.

Note in particular the important results which follow:

Key Point

\(\underline{i}\times \underline{i} = \underline{0}\qquad \underline{j}\times \underline{j}=\underline{0}\qquad\underline{k}\times \underline{k}=\underline{0}\)

Example

Show that \(\underline{i}\times \underline{j}=\underline{k}\) and find expressions for \(\underline{j}\times \underline{k}\) and \(\underline{k}\times \underline{i}\).

Solution

Note that \(\underline{i}\) and \(\underline{j}\) are perpendicular so that the angle between them is \(90^\circ\). Also, the vector \(\underline{k}\) is perpendicular to both \(\underline{i}\) and \(\underline{j}\). Using what we learned from previous keypoints, the modulus of \(\underline{i}\times \underline{j}\) is \((1)(1)\sin90^\circ=1.\) So \({\underline i}\times {\underline j}\) is a unit vector.The unit vector perpendicular to \(\underline{i}\) and \(\underline{j}\) in the sense defined by the right-handed screw rule is \(\underline{k}\) as shown in the first figure below (from left to right). Therefore \(\underline{i}\times \underline{j}=\underline{k}\) as required.

[Image]

Similarly you should verify that \(\underline{j}\times \underline{k}=\underline{i}\) (second figure above, from left to right) and \(\underline{k}\times \underline{i}=\underline{j}\) (third figure above from left to right).

Key Point

\[\underline{i}\times \underline{j}=\underline{k}, \; \underline{j}\times \underline{k}=\underline{i}, \; \underline{k}\times \underline{i}=\underline{j}\] \[\underline{j}\times \underline{i}=-\underline{k}, \; \underline{k}\times \underline{j}=-\underline{i}, \; \underline{i}\times \underline{k}=-\underline{j}\]

To help remember these results you might like to think of the vectors \(\underline{i}, \underline{j}\) and \(\underline{k}\) written in alphabetical order like this:

\[\underline{i}\quad \underline{j}\quad \underline{k}\quad \underline{i}\quad \underline{j}\quad\underline{k}\]

Moving left to right yields a positive result: e.g. \(\underline{k}\times \underline{i} = j\).

Moving right to left yields a negative result: e.g. \(\underline{j}\times \underline{i}=-\underline{k}.\)

A formula for finding the vector product

We can use the results in a previous Key point to develop a formula for finding the vector product of two vectors given in Cartesian form: Suppose \(\underline{a}=a_1\underline{i}+a_2\underline{j}+a_3\underline{k}\) and \(\underline{b}=b_1\underline{i}+b_2\underline{j}+b_3\underline{k}\) then \[\begin{aligned} \underline{a}\times\underline{b}&=(a_1\underline{i}+a_2\underline{j}+a_3\underline{k})\times (b_1\underline{i}+b_2\underline{j}+b_3\underline{k})\\ &= a_1\underline{i}\times(b_1\underline{i}+b_2\underline{j}+b_3\underline{k}) \\ & \quad + \quad a_2\underline{j}\times(b_1\underline{i}+b_2\underline{j}+b_3\underline{k}) \\ & \qquad\quad +\quad a_3\underline{k}\times(b_1\underline{i}+b_2\underline{j}+b_3\underline{k}) \\ &= a_1b_1(\underline{i}\times \underline{i}) + a_1b_2(\underline{i}\times \underline{j}) +a_1b_3(\underline{i}\times \underline{k}) \\ & \quad +\quad a_2b_1(\underline{j}\times \underline{i}) + a_2b_2(\underline{j}\times \underline{j}) +a_2b_3(\underline{j}\times \underline{k}) \\ & \qquad \quad +\quad a_3b_1(\underline{k}\times \underline{i}) + a_3b_2(\underline{k}\times\underline{j}) +a_3b_3(\underline{k}\times \underline{k})\end{aligned}\] Using something you learnt in a previous key point, this expression simplifies to \[\qquad\underline{a}\times \underline{b} = (a_2b_3-a_3b_2)\underline{i}-(a_1b_3-a_3b_1)\underline{j}+(a_1b_2-a_2b_1)\underline{k}\]

This gives us the following Key point.

Key Point

If \(\underline{a}=a_1\underline{i}+a_2\underline{j}+a_3\underline{k}\) and \(\underline{b}=b_1\underline{i}+b_2\underline{j}+b_3\underline{k}\), then

\[\underline{a}\times \underline{b} = (a_2b_3-a_3b_2)\underline{i}-(a_1b_3-a_3b_1)\underline{j}+(a_1b_2-a_2b_1)\underline{k}\]

Example

Evaluate the vector product \(\underline{a}\times \underline{b}\) if \(\underline{a} = 3\underline{i}-2\underline{j}+5\underline{k}\) and \(\underline{b}= 7\underline{i} +4\underline{j} -8\underline{k}\).

Solution

Identifying \(a_1=3\), \(a_2=-2\), \(a_3=5\), \(b_1=7\), \(b_2=4\), \(b_3=-8\) we find \[\begin{aligned} \underline{a}\times \underline{b}&= ((-2)(-8)-(5)(4))\underline{i}-((3)(-8)-(5)(7))\underline{j}+((3)(4)-(-2)(7))\underline{k}\\ &= -4\underline{i} +59\underline{j} +26\underline{k}\end{aligned}\]

Exercise

Use the key point of this section to find the vector product of
\(\underline{ p}=[value]\underline{ i}+[value]\underline{ j}\) and \(\underline{ q}=[value]\underline{ i}-[value]\underline{ j}\).

(Note that in this case there are no \(\underline{k}\) components so \(a_3\) and \(b_3\) are both zero).

\(\qquad \underline{ p}\times\underline{ q}=\) \(\underline{k}\)
Feedback: We are ginen the vectors: \[ \underline{ p}=[value]\underline{ i}+[value]\underline{ j} \, \text{ and } \, \underline{ q}=[value]\underline{ i}-[value]\underline{ j}.\] Identifying \(a_1=[value]\), \(a_2=[value]\), \(a_3=0\), \(b_1=[value]\), \(b_2=-[value]\), \(b_3=0\) we find \[\begin{aligned} \underline{a}\times \underline{b}&= (([value])(0)-(0)(-[value]))\underline{i}-(([value])(0)-(0)([value]))\underline{j}+(([value])(-[value])-([value])([value]))\underline{k}\\ &= [value]\underline{k}\end{aligned}.\]

Using determinants to evaluate a vector product

Evaluation of a vector product using the formula in a previous Key point is very cumbersome. A more convenient and easily remembered method is to use determinants. As we hope you will remember, for a \(3\times 3\) determinant,

\[ \left| \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} \right| = a\left| \begin{matrix} e & f\\ h & i\\ \end{matrix}\right| - b\left| \begin{matrix} d & f\\ g & i\\ \end{matrix}\right| +c\left| \begin{matrix} d & e\\ g & h\\ \end{matrix}\right|\]

The vector product of two vectors \(\underline{a}=a_1\underline{i}+a_2\underline{j}+a_3\underline{k}\) and \(\underline{b}=b_1\underline{i}+b_2\underline{j}+b_3\underline{k}\) can be found by evaluating the determinant:

\[ \underline{a}\times\underline{b}= \left| \begin{matrix} \underline{i} & \underline{j}& \underline{k}\\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{matrix}\right|\]

in which \(\underline{i}\), \(\underline{j}\) and \(\underline{k}\) are (temporarily) treated as if they were scalars.

Key Point

If \(\underline{a}=a_1\underline{i}+a_2\underline{j}+a_3\underline{k}\) and \(\underline{b}=b_1\underline{i}+b_2\underline{j}+b_3\underline{k}\) then \[\underline{a}\times \underline{b} = \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{matrix} \right|=\underline{i}(a_2b_3-a_3b_2) -\underline{j}(a_1b_3-a_3b_1)+\underline{k}(a_1b_2-a_2b_1)\]

Example

Find the vector product of \(\underline{a}=3\underline{i}-4\underline{j} +2\underline{k}\) and \(\underline{b}=9\underline{i}-6\underline{j}+2\underline{k}\).

Solution

We have \( \underline{a}\times\underline{b}=\left|\begin{matrix} \underline{i} & \underline{j} & \underline{k}\\ 3 & -4 & 2 \\ 9 & -6 & 2 \\\end{matrix}\right|\) which, when evaluated, gives

\(\qquad \underline{a}\times \underline{b}=\underline{i}(-8-(-12)) - \underline{j}(6-18) + \underline{k}(-18-(-36)) =4\underline{i} +12\underline{j} +18\underline{k}\)

Example

The area \(A_T\) of the triangle shown in the Figure below is given by the formula \(A_T=\dfrac{1}{2} bc\sin \alpha\). Show that an equivalent formula is \(A_T = \dfrac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|\).

[Image: triangle with vertices A,B and C and standard notation for sides and angles]

Solution

We use the definition of the vector product \(\quad |\overrightarrow{AB}\times \overrightarrow{AC}|= |\overrightarrow{AB}|\,| \overrightarrow{AC}| \sin \alpha.\)

Since \(\alpha\) is the angle between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\), and \(|\overrightarrow{AB}|=c\) and \(|\overrightarrow{AC}|=b\), the required result follows immediately: \(\qquad \dfrac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|=\dfrac{1}{2}c\cdot b\cdot\sin\alpha\).

Moments

The moment (or torque) of the force \(\underline{F}\) about a point \(O\) is defined as

\(\underline{M}_o=\underline{r}\times\underline{F}\)

where \(\underline{r}\) is a position vector from \(O\) to any point on the line of action of \(\underline{F}\) as shown in the following Figure.

[Image]

It may seem strange that any point on the line of action may be taken but it is easy to show that exactly the same vector \({\underline M}_o\) is always obtained.
By the properties of the cross product the direction of \(\underline{M}_o\) is perpendicular to the plane containing \(\underline{r}\) and \(\underline{F}\) (i.e. out of the screen). The magnitude of the moment is

\(\qquad |\underline{M}_0|=|\underline{r}||\underline{F}|\sin\theta\).

From the figure above, \(|\underline{r}|\sin\theta= OD\). Hence \(|\underline{M}_o|=OD|\underline{F}|\). This would be the same no matter which point on the line of action of \(\underline{F}\) was chosen.

Example

Find the moment of the force given by \(\underline{F}=3\underline{i}+4\underline{j}+5\underline{k}\) (N) acting at the point \((14,-3,6)\) about the point \(P(2,-2,1)\).

[Image]

Solution

The vector \(\underline{r}\) can be any vector from the point \(P\) to any point on the line of action of \(\underline{F}\). Choosing \({\underline r}\) to be the vector connecting \(P\) to \((14,-3,6)\) (and measuring distances in metres)

 we have: \(\qquad \underline{r}=(14-2)\underline{i}+(-3-(-2))\underline{j}+(6-1)\underline{k}=12\underline{i}-\underline{j}+5\underline{k}\).

The moment is \(\quad \underline{M}_\rho=\underline{r}\times \underline{F}=\left| \begin{matrix} \underline{i}&\underline{j}&\underline{k}\\ 12&-1&5\\ 3&4&5\end{matrix}\right|= -25\underline{i}-45\underline{j}+51\underline{k}\) (N m)

Exercise

If \(\underline{ a}\) and \(\underline{ b}\) are parallel vectors then their vector product is the zero vector.
Feedback: If \(\underline{a}\) and \(\underline{b}\) are parallel then the angle \(\theta\) between them is zero. Consequently \(\sin\theta=0\) from which it follows that \(\underline{a}\times \underline{b}= |\underline{a}|\,|\underline{b}|\,\sin \theta \,\hat{\underline{e}}=\underline{0}\).

Exercise

Find \(\underline{p}\times\underline{q}\) where \(\underline{p}=-[value]\underline{ i}-[value]\underline{ j}\) and \(\underline{ q}=[value]\underline{ i}+[value]\underline{ j}\).

\(\underline{k}\)
Feedback: We have \(\quad\underline{p}\times\underline{q}=\left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ -[value] & -[value] & 0 \\ [value] & [value] & 0 \\ \end{matrix} \right|\) which, when evaluated, gives \(\qquad\underline{p}\times \underline{q}=\underline{i}((-[value])(0)-(0)([value])) - \underline{j}((-[value])(0)-(0)([value])) + \underline{k}((-[value])([value])-(-[value])([value]))\) \(\qquad\qquad \; =[value]\underline{k}\).

Exercise

If \(\underline{ a}=[value]\underline{ i} +2\underline{ j} -[value]\underline{ k}\) and \(\underline{ b}=[value]\underline{ i} +3\underline{ j} +[value]\underline{ k}\), then:

1. \(\underline{ a}\times \underline{ b} = \)  \(\underline{i}\) \(+\)   \(\underline{j}\) \(+\) \(\underline{k}\)
2. \(\underline{ b}\times \underline{ a} = \)  \(\underline{i}\) \(+\)   \(\underline{j}\) \(+\) \(\underline{k}\)
3. In general, does \(\underline{ a}\times\underline{ b}= \underline{ b} \times \underline{ a}\)?
Feedback: 1. We have: \(\begin{aligned} \underline{ a}\times \underline{ b} &= \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ [value] & 2 & -[value] \\ [value] & 3 & [value] \\ \end{matrix} \right| \\ & = \underline{i}((2)([value])-(-[value])(3)) - \underline{j}(([value])([value])-(-[value])([value])) + \underline{k}(([value])(3)-(2)([value])) \\ & = [value] \underline{i} - [value] \underline{j} + [value] \underline{k} \end{aligned}\) 2. We have: \(\begin{aligned} \underline{ b}\times \underline{ a} &= \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ [value] & 3 & [value] \\ [value] & 2 & -[value] \\ \end{matrix} \right| \\ & = \underline{i}((3)(-[value])-([value])(2)) - \underline{j}(([value])(-[value])-([value])([value])) + \underline{k}(([value])(2)-(3)([value])) \\ & = -[value] \underline{i} + [value] \underline{j} - [value] \underline{k} \end{aligned}\) 3. In general, \(\underline{ a}\times\underline{ b} \ne \underline{ b} \times \underline{ a}\)

Exercise

Points \(A\), \(B\) and \(C\) have coordinates \(([value],1,-[value])\), \(([value],1,[value])\), and \((1,0,1)\) respectively. Find the vector product \(\overrightarrow{AB}\times \overrightarrow{AC}\).

  \(\underline{i}\) \(+\) \(\underline{j}\) \(+\) \(\underline{k}\)
Feedback: First, we have that: \(\overrightarrow{AB} = ([value]-[value])\underline{i} + (1-1)\underline{j} + ([value]-(-[value]))\underline{k} = [value]\underline{i} + [value] \underline{k}.\) \(\overrightarrow{AC} = (1-[value])\underline{i} + (0-1)\underline{j} + (1-(-[value]))\underline{k} = -[value]\underline{i} -1\underline{j} + [value] \underline{k}.\) Then: \(\begin{aligned} \overrightarrow{AB}\times \overrightarrow{AC} &= \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ [value] & 0 & [value] \\ -[value] & -1 & [value] \\ \end{matrix} \right| \\ & = \underline{i}((0)([value])-([value])(-1)) - \underline{j}(([value])([value])-([value])(-[value])) + \underline{k}(([value])(-1)-(0)(-[value])) \\ & = [value]\underline{i}-[value]\underline{j} -[value]\underline{k}. \end{aligned}\)

Exercise

Find a nonzero vector that is perpendicular to both of the vectors \(\underline{a}=-[value]\underline{ i} +[value]\underline{ j} -[value]\underline{ k}\) and \(\underline{b}=[value]\underline{ i} + [value]\underline{ j} + [value]\underline{ k}\).

  \(\underline{i}\) \(+\) \(\underline{j}\) \(+\)   \(\underline{k}\)
Feedback: As you could check in the definition of the vector product, \(\underline{a} \times \underline{b}\) is perpendicular to both vectors \(\underline{a}\) and \(\underline{b}\), so we proceed to compute \(\underline{a} \times \underline{b}\): \(\begin{aligned} \underline{ a}\times \underline{ b} &= \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ -[value] & [value] & -[value] \\ [value] & [value] & [value] \\ \end{matrix} \right| \\ & = \underline{i}(([value])([value])-(-[value])([value])) - \underline{j}((-[value])([value])-(-[value])([value])) + \underline{k}((-[value])([value])-([value])([value])) \\ & = [value]\underline{i}+[value]\underline{j} [value]\underline{k} \end{aligned}\). Any nonzero muliple of this vector is also a valid answer.

Exercise

Find a nonzero vector which is perpendicular to the plane containing \(-[value]\underline{ i} +[value]\underline{ k}\) and \([value]\underline{ i} -[value]\underline{ j}\)

\(\underline{i}\) \(+\) \(\underline{j}\) \(+\) \(\underline{k}\)
Feedback: As you checked in the definition of the vector product, \(-[value]\underline{ i} +[value]\underline{ k} \times [value]\underline{ i} -[value]\underline{ j}\) is perpendicular to the plane containing the vectors \(-[value]\underline{ i} +[value]\underline{ k} \) and \([value]\underline{ i} -[value]\underline{ j}\), so we proceed to compute \(-[value]\underline{ i} +[value]\underline{ k} \times [value]\underline{ i} -[value]\underline{ j}\): \(\begin{aligned} -[value]\underline{ i} +[value]\underline{ k} \times [value]\underline{ i} -[value]\underline{ j} =& \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ -[value] & 0 & [value] \\ [value] & -[value] & 0 \\ \end{matrix} \right| \\ =& \underline{i}((0)(0)-([value])(-[value])) - \underline{j}((-[value])(0)-([value])([value])) \\ & + \underline{k}((-[value])(-[value])-(0)([value])) \\ =& [value]\underline{i}+[value]\underline{j} +[value]\underline{k} \end{aligned}\). Any nonzero multiple of this would also be a valid answer.

Exercise

For the vectors \(\underline{ a}=[value]\underline{ i} +[value]\underline{ k}\), \(\underline{ b}=-3\underline{ j} +[value]\underline{ k}\), and \(\underline{ c}=[value]\underline{ i} -2\underline{ j}\):

1. Compute \(\underline{ a}\times(\underline{ b}\times\underline{ c}) = \) \(\underline{i}\) \(+\) \(\underline{j}\) \(+\) \(\underline{k}\)
2. Find \((\underline{ a}\times\underline{ b})\times\underline{ c} = \) \(\underline{i}\) \(+\) \(\underline{j}\) \(+\) \(\underline{k}\)
3. In general, is the vector product associative?
Feedback: 1. First we have: \(\begin{aligned} \underline{ b}\times \underline{ c} &= \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ 0 & -3 & [value] \\ [value] & -2 & 0 \\ \end{matrix} \right| \\ & = \underline{i}((-3)(0)-([value])(-2)) - \underline{j}((0)(0)-([value])([value])) + \underline{k}((0)(-2)-(-3)([value])) \\ & = [value]\underline{i}+[value]\underline{j} +[value]\underline{k} \end{aligned}\). Thus: \(\begin{aligned} \underline{ a}\times(\underline{ b}\times\underline{ c}) &= \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ [value] & 0 & [value] \\ [value] & [value] & [value] \\ \end{matrix} \right| \\ & = \underline{i}((0)([value])-([value])([value])) - \underline{j}(([value])([value])-([value])([value])) + \underline{k}(([value])([value])-(0)([value])) \\ & = -[value] \underline{i} - [value] \underline{j} + [value] \underline{k} \end{aligned}\) 2. First we have: \(\begin{aligned} \underline{ a}\times \underline{ b} &= \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ [value] & 0 & [value] \\ 0 & -3 & [value] \\ \end{matrix} \right| \\ & = \underline{i}((0)([value])-([value])(-3)) - \underline{j}(([value])([value])-([value])(0)) + \underline{k}(([value])(-3)-(0)(0)) \\ & = [value]\underline{i}-[value]\underline{j} -[value]\underline{k} \end{aligned}\). Thus: \(\begin{aligned} (\underline{ a}\times\underline{ b})\times\underline{ c} &= \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ [value] & -[value] & -[value] \\ [value] & -2 & 0 \\ \end{matrix} \right| \\ & = \underline{i}((-[value])(0)-(-[value])(-2)) - \underline{j}(([value])(0)-(-[value])([value])) + \underline{k}(([value])(-2)-(-[value])([value])) \\ & = -[value] \underline{i} - [value] \underline{j} + [value] \underline{k} \end{aligned}\) 3. In general, the vector product is not associative.

Exercise

The area of the triangle with vertices at the points with coordinates \(([value],2,[value])\), \(([value],-3,[value])\) and \(([value],1,[value])\) is:

units\(^2\) (please give your answer to one decimal place).
Feedback: As you checked in a previous example, the area of a triangle \(A_T\) defined by 3 points \(A\), \(B\) and \(C\) is given by: \(A_T = \dfrac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|\). If we define \(A=([value],2,[value])\), \(B=([value],-3,[value])\) and \(C=([value],1,[value])\), then, we have that \( \overrightarrow{AB} = [value] \underline{i} -5\underline{j} + [value]\underline{k} \) and \( \overrightarrow{AC} = [value] \underline{i} -\underline{j} + [value]\underline{k} \), thus: \(\begin{aligned} \overrightarrow{AB}\times \overrightarrow{AC} &= \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ [value] & -5 & [value] \\ [value] & -1 & [value] \\ \end{matrix} \right| \\ & = \underline{i}((-5)([value])-([value])(-1)) - \underline{j}(([value])([value])-([value])([value])) + \underline{k}(([value])(-1)-(-5)([value])) \\ & = [value]\underline{i}-([value])\underline{j} +([value])\underline{k} \end{aligned}\). Thus: \( A_T = \dfrac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}| = \dfrac{1}{2} \sqrt{([value])^2 + (-([value]))^2 + ([value])^2} = [value] \) In conclusion,the area of the triangle with vertices at the points with coordinates \(([value],2,[value])\), \(([value],-3,[value])\) and \(([value],1,[value])\) is [value] units\(^2\).

Exercise

For the vectors \(\underline{ r}=[value]\underline{ i} -9\underline{j} +[value]\underline{ k}\), \(\underline{ s}=[value]\underline{i} -3\underline{ j} -[value]\underline{ k}\), and \(\underline{ t}=[value]\underline{ i} -2\underline{ j} +[value]\underline{k}\):

1. Evaluate \((\underline{ r}\cdot\underline{ t})\underline{ s} - (\underline{s}\cdot\underline{ t})\underline{ r} = \)   \(\underline{i}\) \(+\)   \(\underline{j}\) \(+\)   \(\underline{k}\)
2. Find \((\underline{ r}\times \underline{s})\times \underline{ t} = \)   \(\underline{i}\) \(+\)   \(\underline{j}\) \(+\)   \(\underline{k}\)
3. \((\underline{ r}\cdot\underline{ t})\underline{s} - (\underline{ s}\cdot\underline{ t})\underline{ r} = (\underline{ r}\times \underline{s})\times\underline{ t} \)?
Feedback: 1. We have that \( \underline{r} \cdot \underline{t} = ([value])([value]) + (-9)(-2) + ([value])([value]) = [value] \) \( \underline{s} \cdot \underline{t} = ([value])([value]) + (-3)(-2) + (-[value])([value]) = [value] \) So: \(\begin{aligned} (\underline{ r}\cdot\underline{ t})\underline{ s} - (\underline{s}\cdot\underline{ t})\underline{ r} & = ([value])([value]\underline{i} -3\underline{ j} -[value]\underline{ k}) - ([value])([value]\underline{ i} -9\underline{j} +[value]\underline{ k}) \\ & = [value] \underline{i} + [value] \underline{j} - [value] \underline{k} \end{aligned}\) 2. First we have: \(\begin{aligned} \underline{ r}\times \underline{ s} &= \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ [value] & -9 & [value] \\ [value] & -3 & -[value] \\ \end{matrix} \right| \\ & = \underline{i}((-9)(-[value])-([value])(-3)) - \underline{j}(([value])(-[value])-([value])([value])) + \underline{k}(([value])(-3)-(-9)([value])) \\ & = [value]\underline{i}+[value]\underline{j} +[value]\underline{k} \end{aligned}\). Thus: \(\begin{aligned} (\underline{ r}\times\underline{ s})\times\underline{ t} &= \left| \begin{matrix} \underline{i} & \underline{j} & \underline{k} \\ [value] & [value] & [value] \\ [value] & -2 & [value] \\ \end{matrix} \right| \\ & = \underline{i}(([value])([value])-([value])(-2)) - \underline{j}(([value])([value])-([value])([value])) + \underline{k}(([value])(-2)-([value])([value])) \\ & = [value] \underline{i} + [value] \underline{j} - [value] \underline{k} \end{aligned}\) 3. This is always true.