Title: Default for 9.2 Cartesian Components of Vectors
Course: HELM Workbooks 2023-2024

Two-dimensional coordinate frames

The figure below shows a two-dimensional coordinate frame. Any point P in the \(xy\) plane can be defined in terms of its \(x\) and \(y\) coordinates.

[Image: A set of x-y axes with a point P marked on it at position P]

A unit vector pointing in the positive direction of the \(x\)-axis is denoted by \(\underline{i}\). (Note that it is common practice to write this particular unit vector without the hat \(\hat{\ }\).) It follows that any vector in the direction of the \(x\)-axis will be a multiple of \(\underline{ i}\). The figure below shows vectors \(\underline{ i}\), \(2\underline{ i}\), \(5\underline{ i}\) and \(-3\underline{ i}\). In general a vector of length \(\ell\) in the direction of the \(x\)-axis can be written \(\ell\underline{ i}\).

[Image: 4 vectors of various lengths pointing in the x direction.]

Similarly, a unit vector pointing in the positive \(y\)-axis is denoted by \(\underline{ j}\). So any vector in the direction of the \(y\)-axis will be a multiple of \(\underline{ j}\). The figure below shows \(\underline{ j}\), \(4\underline{ j}\) and \(-2\underline{ j}\). In general a vector of length \(\ell\) in the direction of the \(y\)-axis can be written \(\ell\underline{ j}\).

[Image: Three vectors in the y-direction of different lengths.]

Key Point

\(\underline{i}\) represents a unit vector in the direction of the positive \(x\)-axis

\(\underline{j}\) represents a unit vector in the direction of the positive \(y\)-axis

Example

Draw the vectors \(5\underline{ i}\) and \(4\underline{ j}\). Use your diagram and the triangle law of addition to add these two vectors together. First draw the vectors \(5\underline{ i}\) and \(4\underline{ j}\). Then, by translating the vectors so that they lie head to tail, find the vector sum \(5\underline{ i} + 4\underline{ j}\).

Solution

[Image: On the left, the two vectors 4i and 5j are drawn, starting from the same point. On the right the are drawn with the base]

We now generalise the situation in the Example above. Consider the figure below.

[Image: A right angle triangle ABC with sides ai, bj and hypotenuse r.]

It shows a vector \(\underline{ r}=\overrightarrow{AB}\). We can regard \(\underline{ r}\) as being the resultant of the two vectors \(\overrightarrow{AC}=a\underline{ i}\) and \(\overrightarrow{CB}=b\underline{ j}\). From the triangle law of vector addition we have \[\overrightarrow{AB} = \overrightarrow{AC}+\overrightarrow{CB} = a\underline{i} + b\underline{j}\]

We conclude that any vector in the \(xy\) plane can be expressed in the form \(\underline{ r}= a\underline{ i} + b\underline{ j}\). The numbers \(a\) and \(b\) are called the components of \(\underline{r}\) in the \(x\) and \(y\) directions. Sometimes, for emphasis, we will use \(a_{x}\) and \(a_{y}\) instead of \(a\) and \(b\) to denote the components in the \(x\)- and \(y\)-directions respectively. In that case we would write \(\underline{ r}= a_x\underline{ i} + a_y\underline{ j}\).

Position vectors

Now consider the special case when \(\underline{ r}\) represents the vector from the origin \(O\) to the point \(P(a,b)\). This vector is known as the position vector of \(P\) and is shown in the figure below.

Key Point

[Image: A set of x-y with the point P(a,b) marked on it and the position vector r pointing from the origin to P]

The position vector of \(P\) with coordinates \((a,b)\) is \(\underline{r} = \overrightarrow{OP} = a\underline{i} + b\underline{j} \)

Unlike most vectors, position vectors cannot be freely translated. Because they indicate the position of a point they are fixed vectors in the sense that the tail of a position vector is always located at the origin.

Example

State the position vectors of the points with coordinates:

1. \(P(2,4)\)
2. \(Q(-1,5)\)
3. \(R(-1,-7)\)
4. \(S(8,-4)\)

Solution

1. \(2\underline{i} + 4\underline{j}\)
2. \(-\underline{i} + 5\underline{j}\)
3. \(-\underline{i} - 7\underline{j}\)
4. \(8\underline{i} - 4\underline{j}\)

Example

Sketch the position vectors:

1. \(\underline{r}_{1} = 3\underline{i}+4\underline{j}\)
2. \(\underline{r}_{2} = -2\underline{i}+5\underline{j}\)
3. \(\underline{r}_{3} = -3\underline{i}-2\underline{j}\)

Solution

The vectors are shown below. Note that all position vectors start at the origin.

[Image: Three vectors pointing from the origin to the three points specified in the question.]

The modulus of any vector \(\underline{r}\) is equal to its length. As we have noted earlier, the modulus of \(\underline{ r}\) is usually denoted by \(|\underline{ r}|\). When \(\underline{ r}=a\underline{i}+b\underline{ j}\) the modulus can be obtained using Pythagoras’ theorem. If \(\underline{r}\) is the position vector of point \(P\) then the modulus is, clearly, the distance of \(P\) from the origin.

Key Point

If \(\underline{ r}=a\underline{i}+b\underline{ j}\) then \(|\underline{r}| = \sqrt{a^2 + b^2}\)

Example

Find the modulus of each of the vectors shown in the previous Example.

Solution

1. \(|\underline{r}_{1}| = |3\underline{i}+4\underline{j}| = \sqrt{3^2+4^2} = \sqrt{25} = 5\)
2. \(|\underline{r}_{2}| = \sqrt{(-2)^2+5^2} = \sqrt{4+25} = \sqrt{29}\)
3. \(|\underline{r}_{3}| = \sqrt{(-3)^2+(-2)^2} = \sqrt{9+4} = \sqrt{13}\)

Three-dimensional coordinate frames

The real world is three-dimensional and in order to solve many engineering problems it is necessary to develop expertise in the mathematics of three-dimensional space. An important application of vectors is their use to locate points in three dimensions. When two distinct points are known we can draw a unique straight line between them. Three distinct points which do not lie on the same line form a unique plane. Vectors can be used to describe points, lines, and planes in three dimensions. These mathematical foundations underpin much of the technology associated with computer graphics and the control of robots. In this part of the quiz we shall introduce the vector methods which underlie these applications.

[Image: First panel shows a 3D set of coordinate axes with mutually perpendicular x,y,z axes. Middle panel shows a point P(a,b,c]

The figure above shows a three-dimensional coordinate frame. Note that the third dimension requires the addition of a third axis, the \(z\)-axis. Although these three axes are drawn in the plane of the paper you should remember that we are now thinking of three-dimensional situations. Just as in two-dimensions the \(x\) and \(y\) axes are perpendicular, in three dimensions the \(x\),\(y\) and \(z\) axes are all perpendicular to each other. We say they are mutually perpendicular. There is no reason why we could not have chosen the \(z\)-axis in the opposite direction. However, it is conventional to choose the directions shown in the figure above.

Any point in the three dimensional space can be defined in terms of its \(x\), \(y\) and \(z\) coordinates. Consider the point \(P\) with coordinates \((a,b,c)\) as shown in panel (b) of the figure above. The vector from the origin to the point \(P\) is known as the position vector of \(P\), denoted \(\overrightarrow{OP}\) or \(\underline{r}\). To arrive at \(P\) from \(O\) we can think of moving \(a\) units in the \(x\) direction, \(b\) units in the \(y\) direction and \(c\) units in the \(z\) direction.

A unit vector pointing in the positive direction of the \(z\)-axis is denoted by \(\underline{ k}\). Noting that \(\overrightarrow{OQ}=a\underline{ i}+b\underline{ j}\) and that \(\overrightarrow{QP}=c\underline{ k}\) we can state \[\underline{ r}=\overrightarrow{OP} = \overrightarrow{OQ}+\overrightarrow{QP} = a\underline{ i} +b\underline{ j} + c\underline{ k}\]

We conclude that the position vector of the point with coordinates \((a,b,c)\) is \(\underline{ r}=a\underline{ i}+b\underline{j}+c\underline{ k}\). (We might, for convenience, sometimes use a subscript notation. For example we might refer to the position vector \(\underline{ r}\) as \(\underline{ r}=r_x\underline{i}+r_y\underline{j}+r_z\underline{ k}\) in which \((r_{x}, r_{y}, r_{z})\) have taken the place of \((a,b,c)\).)

Key Point

If \(P\) has coordinates \((a,b,c)\) then its position vector is \[\underline{r} = \overrightarrow{OP} = a\underline{i} +b\underline{j} + c\underline{k}\]

The modulus of the vector \(\overrightarrow{OP}\) is equal to the distance \(OP\), which can be obtained by Pythagoras' theorem:

Key Point

If \(\underline{r} = a\underline{ i} +b\underline{ j} + c\underline{ k}\) then \[|\underline{r}| = \sqrt{a^2+b^2+c^2}\]

Example

Points \(A\), \(B\) and \(C\) have coordinates \((-1,1,4)\), \((8,0,2)\) and \((5,-2,11)\) respectively.

1. Find the position vectors of \(A\), \(B\) and \(C\).
2. Find \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\)
3. Find \(|\overrightarrow{AB}|\) and \(|\overrightarrow{BC}|\)

Solution

1. Denoting the position vectors of \(A\), \(B\) and \(C\) by \(\underline{a}\), \(\underline{b}\) and \(\underline{c}\) respectively, we find \[\underline{a} = -\underline{i} + \underline{j} + 4\underline{k},\] \[\underline{b} = 8\underline{i} + 2\underline{k},\] \[\underline{c} = 5\underline{i} - 2\underline{j} + 11\underline{k}\]
2. \(\overrightarrow{AB} = \underline{b} - \underline{a} = 9\underline{i} - \underline{j} - 2\underline{k}.\) \(\overrightarrow{BC} = \underline{c} - \underline{b} = -3\underline{i} - 2\underline{j} + 9\underline{k}.\)
3. \(|\overrightarrow{AB}| = \sqrt{9^2+(-1)^2+(-2)^2}=\sqrt{86}\). \(|\overrightarrow{BC}| = \sqrt{(-3)^2+(-2)^2+9^2} = \sqrt{94}\).

Exercise

Given that \(\underline{a} = [value]\) and \(\underline{b} = [value]\) find \(\underline{a}+\underline{b}.\)

\(\underline{a}+\underline{b}=\) \(\underline{i}\) \(+\) \(\underline{j}\)
Feedback: Given that \(\underline{a} = [value]\) and \(\underline{b} = [value]\) \[\begin{array}{rcl}\displaystyle \underline{a}+\underline{b}&\displaystyle =&\displaystyle \left([value] + ([value])\right)\underline{i} + \left([value] + ([value])\right)\underline{j},\\ &\displaystyle =&\displaystyle [value]. \end{array}\]

It is very important to note from the last task that vectors in Cartesian form can be added by simply adding their respective \(\underline{ i}\) and \(\underline{ j}\) components.

Thus, if \(\underline{ a}=a_x\underline{ i} + a_y\underline{ j}\) and \(\underline{ b}=b_x\underline{ i}+b_y\underline{ j}\) then \[\underline{ a}+\underline{ b}=(a_x+b_x)\underline{ i}+ (a_y+b_y)\underline{ j}.\] A similar, and obvious, rule applies when subtracting: \[\underline{ a}-\underline{ b}=(a_x-b_x)\underline{ i}+(a_y-b_y)\underline{ j}.\]

Exercise

Given \(\underline{p} = [value]\) and \(\underline{q} = [value]\), express \(\underline{p}\) and \(\underline{q}\) as column vectors.

\(\underline{p} =\)

\(\underline{q} =\)
Feedback: In column vector notation we arrange the elements with the \(x\)-component at the top and the \(y\)-component below. So, \[\underline{p} = [value] = [value]\] \[\underline{q} = [value] = [value]\]

Q: Exercise

Which of the below describes a free vector?
- A vector which may be translated to any point in space provided their direction and length remain unchanged.
- A vector which always starts at the origin of the coordinate system. (A free vector is not tied to any specific point in space.)
- A vector whose length may change as long as its direction remains the same. (A vector is defined by both its magnitude and direction therefore by changing either of these we change the vector we are defining.)
- A vector whose direction may change as long as its length stays the same. (A vector is defined by both its magnitude and direction therefore by changing either of these we change the vector we are defining.)
A: A vector which may be translated to any point in space provided their direction and length remain unchanged.

Q: Exercise

Which of the below describes a position vector?
- A vector which may be translated to any point in space provided their direction and length remain unchanged. (A position vector describes the position of a point relative to the origin, and so needs to be fixed to start at the origin.)
- A vector which always starts at the origin of the coordinate system.
- A vector whose length may change as long as its direction remains the same. (A vector is defined by both its magnitude and direction therefore by changing either of these we change the vector we are defining.)
- A vector whose direction may change as long as its length stays the same. (A vector is defined by both its magnitude and direction therefore by changing either of these we change the vector we are defining.)
A: A vector which always starts at the origin of the coordinate system.

Exercise

1. State the position vector of the point \(P([value],[value])\)
\(\underline{i}\) + \(\underline{j}\)
2. State the coordinates of the point \(Q\) with position vector \([value]\)
\((\) , \()\)
Feedback: 1. The \(x\) coordinate gives the \(\underline{i}\) component of the vector and the \(y\) coordinate gives the \(\underline{j}\) component of the vector. So, \(\vec{OP} = [value].\) 2. \(Q\) with position vector \(\vec{OQ} = [value]\) has coordinates \( ([value],[value]).\)

Exercise

Find the modulus of each of the following vectors (writing your answer in the form of a surd, i.e. \(\sqrt{x}\) ):

1. \(\underline{r} = [value]\), \(\quad |\underline{r}| = \)
2. \(\underline{s} = [value]\), \(\quad |\underline{s}| = \)
3. \(\underline{t} = [value]\underline{i} + [value]\underline{j}\), \(\quad |\underline{t}| = \)
Feedback: Considering the general case in part (c), for the vector, \(\underline{t} = [value]\underline{i} + [value]\underline{j}\) we calculate the modulus by applying Pythagoras' theorem: \(|\underline{t}| = \sqrt{[value]^2 + [value]^2}\). Applying this we find: For \(\underline{r} = [value], \quad |\underline{r}| = [value].\) For \(\underline{s} = [value] \quad |\underline{s}| = [value].\)

Exercise

Given that \(A\) is the point \(([value],\,[value],\,[value])\) and that \(B\) is the point \(([value],\,[value],\,[value])\) find:

1. \(\vec{OA} =\) \(\underline{i}\) \(+\) \(\underline{j}\) \(+\) \(\underline{k}\)

2. \(\vec{OB} =\) \(\underline{i}\) \(+\) \(\underline{j}\) \(+\) \(\underline{k}\)

3. \( \vec{AB} = \) \(\underline{i}\) \(+\) \(\underline{j}\) \(+\) \(\underline{k}\)

4. \( |\vec{AB}| = \)
Feedback: 1. The position vectors of the points \(A\) and \(B\) are [value] and [value] respectively. 2. Given that \(\vec{A}\) is \([value]\) and that \(\vec{B}\) is \([value]\) then the vector from A to B is \[\begin{array}{rcl}\displaystyle \underline{r}&\displaystyle =&\displaystyle \left([value] - ([value])\right)\underline{i} + \left([value] - ([value])\right) \underline{j} + \left([value] - ([value])\right)\underline{k},\\ &\displaystyle =&\displaystyle [value]. \end{array}\] 3. The modulus of \(\vec{AB}\) is calculated using Pythagoras' theorem as \[ |\vec{AB}| = \sqrt{ ([value])^2 + ([value])^2 + ([value])^2}\]

Exercise

Given that \(A\) is the point \(([value],\,[value])\) and that \(B\) is the point \(([value],\,[value])\) find:

1. \(\vec{OA} =\) \(\underline{i}\) \(+\) \(\underline{j}\)
2. \(\vec{OB}\) = \(\underline{i}\) \(+\) \(\underline{j}\)
3. \( \vec{AB} = \) \(\underline{i}\) \(+\) \(\underline{j}\)
4. \( |\vec{AB}| = \)
Feedback: 1. The position vectors of the points \(A\) and \(B\) are \([value]\) and \([value]\) respectively. 2. Given that \(\vec{A}\) is \([value]\) and that \(\vec{B}\) is \([value]\) then the vector from A to B is \[\begin{array}{rcl}\displaystyle \mathbf{r}&\displaystyle =&\displaystyle \left([value] - ([value])\right)\underline{i} + \left([value] - ([value])\right) \underline{j},\\ &\displaystyle =&\displaystyle [value]. \end{array}\] 3. The modulus of \(\vec{AB}\) is calculated using Pythagoras' theorem as \[ |\vec{AB}| = \sqrt{([value])^2 + ([value])^2} = [value]\]

Exercise

Find the modulus of each of the following vectors (writing your answer in the form of a surd, i.e. \(\sqrt{x}\)):

1. \(\underline{r} = [value], \quad |\underline{r}| = \)
2. \(\underline{s} = [value], \quad |\underline{s}| = \)
3. \(\underline{t} = [value]\underline{i} + [value]\underline{j} - [value]\underline{k}, \quad |\underline{t}| = \)
Feedback: Considering the general case in part (c), for the vector, \(\underline{t} = [value]\underline{i} + [value]\underline{j} - [value]\underline{k}\) we calculate the modulus by applying Pythagoras' theorem: \(|\underline{t}| = \sqrt{[value]^2 + [value]^2 + (-[value])^2}\). By applying this we get: For \(\underline{r} = [value], \quad |\underline{r}| = [value] .\) For \(\underline{s} = -[value], \quad |\underline{s}| = [value].\)

Column vector notation

An alternative, useful, and often briefer notation is to write the vector \(\underline{r}=a\underline{i}+b\underline{j}\) in notation as \[\underline{r}= \begin{bmatrix} a\\ b \end{bmatrix}\,.\]