MAS111 Mathematics Core II
Spring 2021-22 (Version of 2022-02-02)
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© 2022 Frazer Jarvis, SoMaS
1 Two- and three-dimensional geometry
One of the main aims of this module is to begin a more systematic general study of some of the topics that will be familiar to you from school – algebra, geometry and calculus. One of the main things we shall do is to generalise ideas from calculus, such as the process of finding tangents, or of finding areas under curves, to functions of two variables. While this is already interesting, and has useful applications in physical applied mathematics, as you would see in mechanics, pure mathematicians like to generalise ideas to the most general extent possible, even if there are no obviously real-world applications, and understanding what happens for functions of two variables well is a very good start to understanding even more general situations.
This chapter will begin by studying some elementary geometry; we will first consider coordinate systems in two dimensions. Since differentiation is about gradients, we’ll start by recalling some facts about lines in two dimensions. We would expect tangents to functions of two variables to be planes in three dimensions, and so we will continue on to look at three dimensional geometry a little (although we will mostly only use the usual “Cartesian” coordinates), and then discuss planes and their geometry (but not in too much detail).
1.1 Two-dimensional coordinate systems
1.1.1 Cartesian coordinates
Recall that in a rectangular (or Cartesian) coordinate system we choose a pair of particular lines as the \(x\)-axis and \(y\)-axis of the system.
Figure 1.1: Cartesian coordinates
In general, a point \(P\) is represented by the ordered pair of numbers \((x,y)\). Remember that the set of ordered pairs is given by the Cartesian product of two sets; since \(x\) and \(y\) are real, the plane is given by \(\mathbb{R}\times\mathbb{R}\), which we usually abbreviate \(\mathbb{R}^2\). The name Cartesian derives from the French mathematician René Descartes (1596–1650).
1.1.2 Polar coordinates
The word coordinates simply means a way of describing the point in terms of some numbers. Given a point \((x,y)\in\mathbb{R}^2\), another way to describe it is in terms of the pair of the modulus, i.e., the distance of \((x,y)\) from the origin, and the argument, i.e., the angle made by the line joining \((x,y)\) to the origin makes with the \(x\)-axis.
Figure 1.2: Polar coordinates
By looking at the diagram, we see that \[x=r\cos\theta,\qquad y=r\sin\theta.\] To find \(r\) and \(\theta\) we have to solve these equations. For \(r\) this is easy; \(r=\sqrt{x^2+y^2}\). We always take \(r\ge0\). If \(r=0\), then \(\theta\) can have any value; it is not uniquely defined. If \(r>0\), then there is a unique \(-\pi<\theta\le\pi\) (or in any range of length \(2\pi\), such as \(0\le\theta<2\pi\)) such that \[\begin{aligned} \cos\theta&=\tfrac{x}{r},\\ \sin\theta&=\tfrac{y}{r}.\end{aligned}\] However any integer multiple of \(2\pi\) can be added or subtracted to \(\theta\); there are actually infinitely many solutions. This sometimes causes issues.
We can eliminate \(r\) from the above. If \(x\ne0\), we divide the second equation by the first to get \(\tan\theta=y/x\), i.e., \(\theta=\tan^{-1}(\tfrac{y}{x})\); if \(y\ne0\), we can write \(\theta=\cot^{-1}(\tfrac{x}{y})\). However, recall that \(\tan\) (and \(\cot\)) are only periodic with a period of \(\pi\), so these formulas may put \(\theta\) into the wrong quadrant, so some care should be taken.
1.2 Lines in two dimensions
Straight lines are particularly important examples of graphs in two dimensions. You will be familiar with the process of finding tangents by the differentiation process – what we are doing, in some sense, is finding the best line which approximates a function near a given point. Lines also have nice equations, and solving these leads to interesting algebra.
1.2.1 Straight lines
Let \(P=(x,y)\) be a point on a straight line \(\ell\) through two known points \(A=(x_1,y_1)\), \(B=(x_2,y_2)\).
Figure 1.3: Points on a straight line
Form similar triangles \(\dfrac{RP}{QB}=\dfrac{AR}{AQ}\). Thus, \[\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1},\] or \[\begin{equation} y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1). \tag{1.1} \end{equation}\] The quantity \(m=\dfrac{y_2-y_1}{x_2-x_1}(=\tan\theta)\) is called the slope (or gradient) of \(\ell\). Hence \[y=m(x-x_1)+y_1,\] or \[y=mx+(y_1-mx_1).\] The term in brackets is constant, thus the equation has the form \[\begin{equation} y=mx+c. \tag{1.2} \end{equation}\] Note that when \(x=0\), \(y=c\) is the intercept on the \(y\)-axis.
If \(m=0\) the line is horizontal;
if \(m>0\) the line rises from left to right;
if \(m<0\) the line falls from left to right;
A vertical line is represented by \(x=\mbox{const}\).
Example 1.1
Find the equations of the lines
passing through \((-2,-6)\), \((-5,2)\);
with slope \(-2\) and passing through \((3,-1)\);
passing through the points \((-2,-1)\), \((-2,2)\).
Let’s explain how to solve these:
Here equation (1.1) is the most convenient starting point: \[\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1},\] and substituting in the points gives \[\frac{y-(-6)}{2-(-6)}=\frac{x-(-2)}{-5-(-2)},\] and so \[\frac{y+6}{8}=\frac{x+2}{-3},\] which can be written as \(8x+3y=-34\) or \(y=-\frac{8}{3}x-\frac{34}{3}\).
An alternative is to start from \(y=mx+c\) (1.2), and to work out the gradient with \[m=\frac{2-(-6)}{-5-(-2)}=-\frac{8}{3},\] so the line must be of the form \[y=-\frac{8}{3}x+c.\] The line passes through \((-2,-6)\), so this point must satisfy this equation, so \(-6=-\frac{8}{3}(-2)+c\) and we can read off that \(c=-\frac{34}{3}\).
Here \(m=-2\), and so \(y=-2x+c\) in (1.2). We know a point on the line, so \(-1=-2(3)+c\), and can see that \(c=5\). Thus the line is \(y=-2x+5\). The \(y\)-intercept is 5, and the \(x\)-intercept is \(5/2\).
Using (1.1), we see: \[\frac{y-(-1)}{2-(-1)}=\frac{x-(-2)}{-2-(-2)},\] but the denominator \(-2-(-2)=0\), and we cannot proceed. Note that for both points, \(x=-2\), so the line is vertical, and the equation of the line is just \(x=-2\) (a line of infinite slope).
1.2.2 The intersection of straight lines
Consider the lines \[\begin{array}{rcrcl} 2x+5y&=&-1,&&(1)\\ 3x-2y&=&8.&&(2) \end{array}\] To find the point of intersection leads to simultaneous equations, where we need to solve both equations at the same time. Here, 3(1) \(-\) 2(2) gives \(15y+4y=-3-16\), and so \(19y=-19\), and \(y=-1\). Substituting back into either (1) or (2) gives \(x=2\), and the point of intersection is then \((2,-1)\).
Figure 1.4: The intersection of two straight lines
In general, two lines either
intersect at a unique point (as above) and the equations have a unique common solution, or
do not intersect and so are distinct and parallel (no solutions).
(A third possibility is that the two equations are identical, but then you just really have one equation, a line.)
For example, \[\begin{aligned} y&=2x+3\\ y&=2x-3\end{aligned}\] gives \(2x+3=2x-3\), which clearly has no solution. These lines are parallel, and one can see from the equations that they have the same gradient.
Figure 1.5: Two parallel lines
The key point here is that intersection of two lines corresponds to solving their equations simultaneously. Conversely, solving a pair of simultaneous equations in two variables means that, geometrically, we are find the point of intersection of the two lines.
1.2.3 Perpendicular lines
Consider a line \(\ell_1\) which has a slope \(m_1\) is not zero, and consider also a line \(\ell_2\) which is perpendicular to \(\ell_1\) with slope \(m_2\).
Figure 1.6: Perpendicular lines
Recall that the gradient \(m_1=\tan\theta_1\), where \(\theta_1\) is the angle of the line \(\ell_1\) from the \(x\)-axis, and similarly \(m_2=\tan\theta_2\). Since the lines are perpendicular, we have \[\theta_2=\theta_1+\pi/2,\] and so \[\tan\theta_2=\tan(\pi/2+\theta_1)=\frac{\sin(\pi/2+\theta_1)}{\cos(\pi/2+\theta_1)}=\frac{\cos\theta_1}{-\sin(\theta_1)}=-\cot\theta_1=-\frac{1}{\tan\theta_1}.\] Thus \(\tan\theta_2=-\frac{1}{\tan\theta_1}\), i.e., \(m_2=-\frac{1}{m_1}\), or \(m_1m_2=-1\).
Example 1.2 Find the equation of the line which passes through \(A(-1,4)\) and is perpendicular to the line which passes through \(A\) and \(B(5,2)\).
The slope of the line \(AB\) is \(\dfrac{2-4}{5-(-1)}=-\dfrac{1}{3}\). Thus the slope of the perpendicular line is \(\dfrac{-1}{-1/3}=3\). So we need a line of gradient \(3\) passing through \((-1,4)\), so this is \(y=3x+c\), where \(4=3(-1)+c\), and so \(c=7\). Thus the line is \(y=3x+7\).Before leaving this section, let’s look at the line \(ax+by=0\), and compare it with the vector \((a,b)\). It might seem natural to imagine that the line is in the direction of the vector \((a,b)\), but a moment’s reflection (perhaps thinking of lines like \(y=x\)) should persuade you that in fact the line \(ax+by=0\) is not in this direction, but is orthogonal to it. We can easily see this by rearranging the equation \(ax+by=0\) as \(y=-(a/b)x\), so we can see that the gradient of the line is \(-a/b\), whereas the line from \((0,0)\) to \((a,b)\) has gradient \(b/a\). Since these multiply to be \(-1\), the two lines are normal (i.e., orthogonal) to each other.
More generally, the line \(ax+by=c\) is just a translation of the line \(ax+by=0\), so its gradient is again \(-a/b\), and it is again orthogonal to the vector \((a~b)\).
1.2.4 Midpoints
Let \(P(x,y)\) be the midpoint of the line segment joining \(A(x_1,y_1)\) and \(B(x_2,y_2)\).
Figure 1.7: The midpoint of a line
By similar triangles \(AR=\frac{1}{2}AQ\), and \(RP=\frac{1}{2}QB\). Then \(x-x_1=\frac{1}{2}(x_2-x_1)\) and so \(x=\frac{1}{2}(x_1+x_2)\). Similarly (or by substituting this value of \(x\) back in), \(y-y_1=\frac{1}{2}(y_2-y_1)\) and \(y=\frac{1}{2}(y_1+y_2)\).
1.2.5 The distance between two points
Figure 1.8: The distance between two points
By Pythagoras’s Theorem, \[d=|AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.\]
Example 1.3 The distance between the points \((-1,4)\) and \((3,-2)\) is given by \[\sqrt{(3-(-1))^2+(-2-4)^2}=\sqrt{16+36}=\sqrt{52}\approx7.211.\]
1.2.6 Straight lines in parametric form
The position of a point in the plane is sometimes described by giving the coordinates \((x,y)\) as functions of a single variable \(t\), called a parameter. It is sometimes helpful to think of \((x,y)\) as the position of a point at a time \(t\).
Example 1.4 A particle moves in the plane with constant velocity which has component 2 in the \(x\)-direction and component 3 in the \(y\)-direction. Initially (at \(t=0\)) the particle is located at \((-1,4)\), then after time \(t\) its coordinates are \((x,y)=(-1+2t,4+3t)\). We can eliminate \(t\) to get \(y=4+3\frac{x+1}{2}\), or \(y=\frac{3}{2}x+\frac{11}{2}\).
Example 1.5 Two ships follow straight line courses at steady speeds. The position of Ship 1 is \((3t-10,t+4)\) and the position of Ship 2 is \((2t-3,-t+13)\), where \(t\) represents time (in hours) and distances are in km. What is the closest distance that the ships approach each other and at what time does this occur?
We can work out the distance at time \(t\) as above: \[d^2=(3t-10-(2t-3))^2+(t+4-(-t+13))^2=(t-7)^2+(2t-9)^2=5t^2-50 t+130,\] and completing the square gives \[d^2=5[(t-5)^2-25]+130=5(t-5)^2+5.\] So the closest approach is at time \(t=5\), when the distance is \(d=\sqrt{5}\) km.1.3 Vectors and scalars (MAS112)
1.3.1 Basic definition of vectors
Note that because vectors are often physical quantities, we tend to think of them as being three-dimensional objects, although this is not strictly necessary.
Figure 1.9: Displacement along a vector
We usually write a vector as a single lowercase letter, either in bold type (in books or these notes!) or with a line underneath (in handwriting). \[\mathbf{v}=\vec{AB}=\vec{PQ}.\] Vectors and scalars are fundamentally different types of quantity and both need to be used together. In handwriting, lines underneath vectors are strongly recommended to avoid confusion.
Definition 1.3
The magnitude of a vector \(\mathbf{a}\) is denoted \(|\mathbf{a}|\), and is equal to the length of any directed line segment specifying \(\mathbf{a}\). The magnitude of a vector \(\vec{AB}\) is equal to the length \(AB\). (You may also see the notation \(\|\mathbf{a}\|\) for this magnitude, but we will use \(|\mathbf{a}|\) in this module.)
Two vectors are equal if they have the same magnitude and direction.
Two vectors are parallel if they have the same direction. Two vectors are anti-parallel if their directions are opposite.
The zero vector \(\mathbf{0}\) is the vector with zero magnitude.
Choose one point in space as the origin \(O\). The position of any point \(A\) is completely specified by the displacement taking \(O\) to \(A\), which is \(\mathbf{a}=\vec{OA}\). In this case \(\mathbf{a}\) is the position vector of \(A\) (relative to \(O\)). (Note that we usually use the bold lower case letter (e.g., \(\mathbf{a}\), \(\mathbf{b}\)) to denote the position vector of the point denoted by the corresponding upper case letter (e.g., \(A\), \(B\)).
When considering a general point \(P\) in space, the position vector of \(P\) is usually written \(\mathbf{r}\) or \(\mathbf{x}\).
Let’s now explain how to add two vectors. Two displacements \(\mathbf{a}=\vec{OA}\) and \(\mathbf{b}=\vec{OB}\) are applied, first \(\mathbf{a}\) and then \(\mathbf{b}\). The net result is a single displacement \(\mathbf{c}=\mathbf{a}+\mathbf{b}\). \(O\) goes first to \(A\) and then to \(C\) (because \(\vec{AC}=\vec{OB}=\mathbf{b}\)).
Now apply the displacements in the reverse order, first \(\mathbf{b}\), then \(\mathbf{a}\). The net result is a single displacement \(\mathbf{b}+\mathbf{a}\). This time, \(O\) goes first to \(B\) but then to \(C\) because \(\vec{BC}=\vec{OA}=\mathbf{a}\).
Figure 1.10: Vector addition
We get the commutative law for vector addition:
\(\mathbf{a}+\mathbf{b}=\mathbf{b}+\mathbf{a}\).
Figure 1.11: Adding three vectors
With three displacements, \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\), we see that \[\begin{eqnarray*} (\mathbf{a}+\mathbf{b})+\mathbf{c}&=&(\vec{OA}+\vec{AD})+\vec{DE}=\vec{OD}+\vec{DE}=\vec{OE};\\ \mathbf{a}+(\mathbf{b}+\mathbf{c})&=&\vec{OA}+(\vec{AD}+\vec{DE})=\vec{OA}+\vec{AE}=\vec{OE}.\end{eqnarray*}\] We get the associative law for vector addition:
\((\mathbf{a}+\mathbf{b})+\mathbf{c}=\mathbf{a}+(\mathbf{b}+\mathbf{c})\).
This is written simply as \(\mathbf{a}+\mathbf{b}+\mathbf{c}\).
If \(\mathbf{a}\) is any vector and \(\mathbf{0}\) is the zero vector, \(\mathbf{a}+\mathbf{0}=\mathbf{0}+\mathbf{a}=\mathbf{a}\).
Let \(\lambda\) be a scalar. Then \(|\lambda|=\lambda\) if \(\lambda\ge0\) and \(|\lambda|=-\lambda\) if \(\lambda<0\). The quantity \(\lambda\mathbf{a}\) is a vector with magnitude \(|\lambda|.|\mathbf{a}|\). It has the same direction as \(\mathbf{a}\) if \(\lambda>0\) and the opposite direction to \(\mathbf{a}\) if \(\lambda<0\). It is parallel to \(\mathbf{a}\) if \(\lambda>0\) and anti-parallel to \(\mathbf{a}\) if \(\lambda<0\).
For \(\lambda=0\), we have \(\lambda\mathbf{a}=\mathbf{0}\) for all \(\mathbf{a}\), where \(\mathbf{0}\) is the zero vector. Also \(\lambda\mathbf{0}=\mathbf{0}\) for all \(\lambda\).
For vectors \(\mathbf{a}\) and \(\mathbf{b}\), and scalars \(\lambda\) and \(\mu\), we have the distributive laws:
\(\lambda(\mathbf{a}+\mathbf{b})=\lambda\mathbf{a}+\lambda\mathbf{b},\quad(\lambda+\mu)\mathbf{a}=\lambda\mathbf{a}+\mu\mathbf{a},\quad(\lambda\mu)\mathbf{a}=\lambda(\mu\mathbf{a})\).
The vector \(-\mathbf{a}\) has the same magnitude as \(\mathbf{a}\), but the opposite direction:
Figure 1.12: Scaling a vector
Define \(\mathbf{b}-\mathbf{a}=\mathbf{b}+(-\mathbf{a})\). If \(\mathbf{a}\) is the position vector of \(A\) and \(\mathbf{b}\) is the position vector of \(B\), then \(\vec{AB}=\mathbf{b}-\mathbf{a}\).
If \(\mathbf{a}\) is any non-zero vector with magnitude \(|\mathbf{a}|\), then \(\hat{\mathbf{a}}=\frac{\mathbf{a}}{|\mathbf{a}|}\) is a unit vector in the direction of \(\mathbf{a}\).
Figure 1.13: Diagram for example
Let \(\mathbf{a}=\vec{AB}=\vec{DC}\), \(\mathbf{b}=\vec{AD}=\vec{BC}\). Then \[\vec{AC}=\vec{AD}+\vec{DC}=\mathbf{b}+\mathbf{a}=\mathbf{a}+\mathbf{b}.\] To find \(\vec{AX}\), note that \(\vec{AX}=\vec{AD}+\vec{DX}\). As \(\mathbf{b}=\vec{AD}\), we need to find \(\vec{DX}\). Now \(\vec{DX}=\frac{2}{3}\vec{DE}\) from the information given in the question. Next, \(\vec{DE}=\vec{DA}+\vec{AE}\). Now \(\vec{DA}=-\vec{AD}=-\mathbf{b}\). From the question, \[\vec{AE}=\frac{1}{2}\vec{AB}=\frac{1}{2}\mathbf{a}.\] Then \[\vec{DE}=\vec{DA}+\vec{AE}=-\mathbf{b}+\frac{1}{2}\mathbf{a}.\] Working backwards, then \[\vec{DX}=\frac{2}{3}\vec{DE}=\frac{2}{3}\left(-\mathbf{b}+\frac{1}{2}\mathbf{a}\right)=\frac{1}{3}\mathbf{a}-\frac{2}{3}\mathbf{b}.\] This gives \[\vec{AX}=\vec{AD}+\vec{DX}=\mathbf{b}+\frac{1}{3}\mathbf{a}-\frac{2}{3}\mathbf{b}=\frac{1}{3}(\mathbf{a}+\mathbf{b})=\frac{1}{3}\vec{AC}.\]
Show that \(A'\) is the mid-point of the side \(BC\).
Find the position vector of the point \(G\) which lies on the line \(AA'\), where the distance \(AG\) is two-thirds of the distance from \(A\) to \(A'\).
Show that the point \(G\) lies on \(BB'\) and \(CC'\), where \(B'\) is the mid-point of \(CA\) and \(C'\) is the mid-point of \(AB\).
Solution. First draw a diagram:
Figure 1.14: Diagram for example
\[\vec{BC}=\vec{BO}+\vec{OC}=-\mathbf{b}+\mathbf{c}.\] Consider \[\vec{BA'}=-\mathbf{b}+\mathbf{a}'=-\mathbf{b}+\frac{1}{2}(\mathbf{b}+\mathbf{c})=\frac{1}{2}(-\mathbf{b}+\mathbf{c})=\frac{1}{2}\vec{BC}.\] Therefore \(A'\) is the mid-point of the side \(BC\).
Now \(\vec{OG}=\vec{OA}+\vec{AG}\) and \(\vec{AG}=\frac{2}{3}\vec{AA'}\). Therefore \[\vec{OG}=\vec{OA}+\frac{2}{3}\vec{AA'}.\] Now \[\vec{AA'}=-\mathbf{a}+\mathbf{a}'=-\mathbf{a}+\frac{1}{2}(\mathbf{b}+\mathbf{c}).\] This gives \[\vec{OG}=\mathbf{a}+\frac{2}{3}\left[-\mathbf{a}+\frac{1}{2}(\mathbf{b}+\mathbf{c})\right]=\mathbf{a}-\frac{2}{3}\mathbf{a}+\frac{1}{3}\mathbf{b}+\frac{1}{3}\mathbf{c}=\frac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c}).\]
Note that the result for the position vector of \(G\) is symmetric in \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\). Therefore, if we found the position vector of the point on \(BB'\) which is two-thirds of the distance from \(B\) to \(B'\), we would get the same answer. Similarly for \(CC'\). Therefore \(G\) lies on \(BB'\) and \(G\) lies on \(CC'\).
The medians of a triangle are the three lines joining the three vertices to the mid-points of the opposite sides. Therefore \(AA'\), \(BB'\) and \(CC'\) are the medians of the triangle \(ABC\). We have shown that the medians are concurrent, that is, they all pass through one point.
1.3.2 Lines
Consider a line passing through a fixed point \(A\) with position vector \(\mathbf{a}\) and parallel to a given vector \(\mathbf{c}\). Let \(\mathbf{r}\) be the position vector of an arbitrary point \(P\) on the line.
Figure 1.15: Vector equation of a line
Then \(\vec{AP}\) is parallel to \(\mathbf{c}\) and therefore \(\vec{AP}=\lambda\mathbf{c}\) for some real number \(\lambda\in\mathbb{R}\). Therefore \(\mathbf{r}=\vec{OP}=\vec{OA}+\vec{AP}=\mathbf{a}+\lambda\mathbf{c}\).
Any point on the line has a position vector of this form, and any value of \(\lambda\) gives a point on the line.
Solution. First draw a diagram.
Figure 1.16: Diagram for example
Let the point \(P\), with position vector \(\mathbf{r}\), be an arbitrary point on the line \(AB\). The line \(AB\) is parallel to the vector \(\vec{AB}=\mathbf{b}-\mathbf{a}\). Therefore the equation of the line is then \[\mathbf{r}=\mathbf{a}+\lambda(\mathbf{b}-\mathbf{a})=(1-\lambda)\mathbf{a}+\lambda\mathbf{b}.\]
Find the vector equations of the lines \(AC\) and \(BQ\).
Find the position vector of the point \(S\) where the lines \(BQ\) and \(AC\) meet.
Find \(AS/SC\).
Solution. First draw a diagram
Figure 1.17: Diagram for example
From Example 1.10, the vector equation of the line \(A\) is \[\begin{equation} \mathbf{r}=(1-\lambda)\mathbf{a}+\lambda\mathbf{c}. \tag{1.3} \end{equation}\] The direction of the line \(BQ\) is \[\vec{BQ}=-\mathbf{b}+\mathbf{q}=-\mathbf{b}+\frac{1}{2}\mathbf{a}+\frac{1}{4}\mathbf{b}+\frac{1}{4}\mathbf{c}=\frac{1}{2}\mathbf{a}-\frac{3}{4}\mathbf{b}+\frac{1}{4}\mathbf{c}.\] The vector equation of the line \(BQ\) is then \[\begin{equation} \mathbf{r}=\mathbf{b}+\mu\mathbf{q}=\mathbf{b}+\mu\left[\frac{1}{2}\mathbf{a}-\frac{3}{4}\mathbf{b}+\frac{1}{4}\mathbf{c}\right]=\frac{\mu}{2}\mathbf{a}+\left(1-\frac{3\mu}{4}\right)\mathbf{b}+\frac{\mu}{4}\mathbf{c}. \tag{1.4} \end{equation}\]
The point \(S\) lies on both \(AC\) and \(BQ\), so at \(S\) the position vectors (1.3) and (1.4) must be the same. This gives \[(1-\lambda)\mathbf{a}+\lambda\mathbf{c}=\frac{\mu}{2}\mathbf{a}+\left(1-\frac{3\mu}{4}\right)\mathbf{b}+\frac{\mu}{4}\mathbf{c}.\] This will be true if \[\begin{eqnarray*} 1-\lambda&=&\frac{\mu}{2}\\ 0&=&1-\frac{3\mu}{4}\\ \lambda&=&\frac{\mu}{4}.\end{eqnarray*}\] Solving these gives \(\mu=\frac{4}{3}\), \(\lambda=\frac{1}{3}\). Then the position vector of \(S\) is \[\mathbf{s}=\frac{2}{3}\mathbf{a}+\frac{1}{3}\mathbf{c}.\]
We have \[\begin{eqnarray*} \vec{AS}&=&-\mathbf{a}+\mathbf{s}=\frac{1}{3}(-\mathbf{a}+\mathbf{c}),\\ \vec{SC}&=&-\mathbf{s}+\mathbf{c}=\frac{2}{3}(-\mathbf{a}+\mathbf{c}).\end{eqnarray*}\] Therefore \[\frac{AS}{SC}=\frac{\frac{1}{3}|-\mathbf{a}+\mathbf{c}|}{\frac{2}{3}|-\mathbf{a}+\mathbf{c}|}=\frac{1}{2}.\]
Very important note: DO NOT DIVIDE VECTORS!
1.4 Three-dimensional coordinate systems
1.4.1 Cartesian coordinates
Again, the aim of a coordinate system in more than two dimensions is to specify the point by means of some numbers. The simplest way is again to use Cartesian coordinates.
In three dimensions, we have a third variable \(z\), representing the height, so that a point is specified by a triple \((x,y,z)\). For example, the triple \((1,2,3)\) represents the point which is \(1\) along in the \(x\)-direction and \(2\) along in the \(y\)-direction and at a height of \(3\). This means that it lies directly above the point \((1,2)\) in the 2-dimensional plane, and a distance of \(3\) from it.
Figure 1.18: Three-dimensional Cartesian coordinates
1.4.1.1 Interpretation with vectors
Consider a set of three unit vectors \(\{\mathbf{i},\mathbf{j},\mathbf{k}\}\):
Figure 1.19: Cartesian coordinates and vectors
Each pair of the three vectors are perpendicular (at right angles).
If \(\mathbf{j}\) and \(\mathbf{k}\) are in the plane of the paper in the directions shown, then \(\mathbf{i}\) is pointing out of the paper towards you. The set \(\{\mathbf{i},\mathbf{j},\mathbf{k}\}\) is a right-handed set.
Let \(O\) be the origin and \(P\) be any point, with position vector \(\mathbf{r}=\vec{OP}\). The point \(N\) is the foot of the perpendicular from \(P\) to the plane containing \(O\) and the vectors \(\mathbf{i}\) and \(\mathbf{j}\). Then \[\begin{eqnarray*} \mathbf{r}&=&\vec{OP}=\vec{ON}+\vec{NP}=(\vec{OX}+\vec{XN})+\vec{NP}\\ &=&(\vec{OX}+\vec{OY})+\vec{OZ}\\ &=&x\mathbf{i}+y\mathbf{j}+z\mathbf{k},\end{eqnarray*}\] where \(x=OX\), \(y=OY\), \(z=OZ\) and we have used the results \(\vec{XN}=\vec{OY}\), \(\vec{NP}=\vec{OZ}\).
We usually consider the special set of vectors \[\mathbf{i}=\begin{pmatrix}1\\0\\0\end{pmatrix},\qquad \mathbf{j}=\begin{pmatrix}0\\1\\0\end{pmatrix},\qquad \mathbf{k}=\begin{pmatrix}0\\0\\1\end{pmatrix},\] and then \(\mathbf{r}=\begin{pmatrix}x\\y\\z\end{pmatrix}\). Often you may see vectors written as row vectors, so that \[\mathbf{i}=\begin{pmatrix}1&0&0\end{pmatrix},\qquad \mathbf{j}=\begin{pmatrix}0&1&0\end{pmatrix},\qquad \mathbf{k}=\begin{pmatrix}0&0&1\end{pmatrix},\] and then \(\mathbf{r}=\begin{pmatrix}x&y&z\end{pmatrix}\). In this module, we will try to be consistent about using column vectors.
Using Pythagoras’ theorem, we find the magnitude of the vector \(\mathbf{r}\): \[|\mathbf{r}|^2=(OP)^2=(ON)^2+(NP)^2=\left[(OX)^2+(OY)^2\right]+(OZ)^2=x^2+y^2+z^2,\] so that \[|\mathbf{r}|=(x^2+y^2+z^2)^\frac{1}{2}=\sqrt{x^2+y^2+z^2}.\] The rules of vector algebra apply to vectors in component form. If \(\mathbf{r}_1=x_1\mathbf{i}+y_1\mathbf{j}+z_1\mathbf{k}\) and \(\mathbf{r}_2=x_2\mathbf{i}+y_2\mathbf{j}+z_2\mathbf{k}\) are \(\lambda\) is a scalar, then \[\mathbf{r}_1+\mathbf{r}_2=(x_1+x_2)\mathbf{i}+(y_1+y_2)\mathbf{j}+(z_1+z_2)\mathbf{k},\qquad\lambda\mathbf{r}_1=(\lambda x_1)\mathbf{i}+(\lambda y_1)\mathbf{j}+(\lambda z_1)\mathbf{k}.\]
1.4.2 Polar coordinates
There are now several alternative ways to specify points, and we will restrict ourselves to mentioning the main two, spherical polar coordinates and cylindrical polar coordinates:
Spherical polar coordinates
Here, a point \((x,y,z)\) in 3-dimensional space is specified by giving its distance \(r\) from the origin, along with the angle \(\theta\) that the pair \((x,y)\) makes with the \(x\)-axis (just like for the 2-dimensional polar coordinates), and also the angle \(\phi\) that the line joining the origin to the point makes with the vertical axis.
In the picture below, you can see how \(r\), \(\theta\) and \(\phi\) are related to a point \(P=(x,y,z)\). In fact, we can read off \[r=\sqrt{x^2+y^2+z^2},\qquad\theta=\tan^{-1}(y/x),\qquad\phi=\cos^{-1}(z/r),\] where \(r\ge0\), \(0\le\theta<2\pi\) as before, and also \(0\le\phi\le\pi\). Conversely, we have the formulae \[\begin{aligned} x&=r\cos\theta\sin\phi\\ y&=r\sin\theta\sin\phi\\ z&=r\cos\phi.\end{aligned}\] Notice that the surfaces with constant values of \(r\) are those whose points are at a fixed radius from the origin, so are spheres, centred at the origin.
Cylindrical polar coordinates
Cylindrical polar coordinates are a sort of halfway between Cartesian and spherical coordinates. Given a point \((x,y,z)\), we consider the pair \((x,y)\), and simply turn these into polar coordinates \((\rho,\theta)\), and specify our point as \((\rho,\theta,z)\). So here \(\rho\) is the distance of the point from the vertical axis, and \(\theta\) is the angle that the line joining the point to the vertical axis makes with the \(x\)-axis, and \(z\) is still the height of the point. Since these are just like polar coordinates, we have \[x=\rho\cos\theta,\qquad y=\rho\sin\theta,\] with \(\rho=\sqrt{x^2+y^2}\) and \(\theta=\tan^{-1}(y/x)\) (with the appropriate warnings on this value made above!). Notice that the surfaces with constant values of \(\rho\) are those whose points are at a fixed distance from the \(z\)-axis, so are cylinders, whose axis is vertical.
We can see the cylindrical coordinates \((\rho,\theta,z)\) on the picture below, as well as the spherical coordinates:
Figure 1.20: Polar coordinates in three dimensions
1.4.3 Distance between two points
In the usual Cartesian coordinates, the distance between \((0,0,0)\) and \((x,y,z)\) is given by \(\sqrt{x^2+y^2+z^2}\).
To see this, consider the distance from \((0,0,0)\) to \((x,y,0)\), which is \(\sqrt{x^2+y^2}\), since the points lie in the two-dimensional \(xy\)-plane, and note that the points \((0,0,0)\), \((x,y,0)\) and \((x,y,z)\) form a right-angled triangle. So the square of the distance between \((0,0,0)\) and \((x,y,z)\) is the sum of the squares of the distances between \((0,0,0)\) and \((x,y,0)\) and between \((x,y,0)\) and \((x,y,z)\), namely \[\left(\sqrt{x^2+y^2}\right)^2+z^2=x^2+y^2+z^2.\] In the same way, the distance between \((x_1,y_1,z_1)\) and \((x_2,y_2,z_2)\) is \[\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}.\] (This generalises to give distances in any number of dimensions, but we’ll simply use the 2- and 3-dimensional cases in this module.)
The distance between \((0,0,0)\) and \((r,\theta,\phi)\) in spherical polar coordinates is simply \(r\), of course. And the distance between \((0,0,0)\) and \((\rho,\theta,z)\) in cylindrical polar coordinates is \(\sqrt{\rho^2+z^2}\).
1.5 Planes in three dimensions
In three dimensions, we have a third variable, \(z\), and general loci (the plural of the Latin word locus, meaning “place”) are given by relations \(f(x,y,z)=0\) for a function \(f\) in three variables. For example, a sphere of radius \(1\) in three dimensions is given by the equation \(x^2+y^2+z^2-1=0\).
Later, when we look at calculus in three dimensions, we will really only look at loci of the form \(z-f(x,y)=0\), i.e., where we write \(z\) as a function of the two variables \(x\) and \(y\). So for each point \((x,y)\) in the plane, we get a height \(z\) for the function, and join them to give a graph. For example, we have a paraboloid \(z=x^2+y^2\) (which generalises the usual notion of parabola \(y=x^2\) in two dimensions), and which looks like:
Figure 1.21: The paraboloid
1.5.1 The equation of a plane
Just as lines are important in two dimensions, planes are important in three dimensions: we would hope that a “nice” surface \(z=f(x,y)\) will have a tangent plane at every point. Since one of our main goals is to develop calculus for functions of two variables, we should spend a little time thinking about planes.
It turns out that the equation of a plane is simply a linear equation (i.e., of degree 1): \[ax+by+cz=d.\] Earlier we remarked that the line \(ax+by=c\) was orthogonal to \((a~b)\). If we now look in 3 dimensions, something rather similar happens. Given an equation \(ax+by+cz=0\), we can view this as the set of all \((x~y~z)\) which are perpendicular to \((a~b~c)\). The easiest way to see this is using vectors, since we could rewrite the left-hand side of the equation as the dot product of \((a~b~c)\) and \((x~y~z)\) (if you don’t know what the dot product is, don’t worry – we’ll say a little about it later; the key thing is simply that two vectors are orthogonal when their dot product is zero). So \(ax+by+cz=0\) is a plane; it is perpendicular to \((a~b~c)\), and passes through the origin.
Figure 1.22: A plane in three dimensions
As before, \(ax+by+cz=d\) simply translates the plane in a parallel direction; it remains orthogonal to \((a~b~c)\), but doesn’t pass through the origin unless \(d=0\).
Figure 1.23: Displacing a plane
Given two vectors \(\mathbf{u}\) and \(\mathbf{v}\) lying in the plane and pointing in different directions, and a point \(\mathbf{x}\) in the plane, then every point in the plane can be written as \(\mathbf{x}+(s\mathbf{u}+t\mathbf{v})\); there are two parameters, and this reflects the fact that planes are two-dimensional.
Figure 1.24: Two vectors contained in a plane
Example 1.12 Find the plane containing the point \((1,2,3)\), and the vectors \((1,1,1)\) and \((0,1,-1)\).
This contains all the points of the form \[\begin{pmatrix}1\\2\\3\end{pmatrix}+s\begin{pmatrix}1\\1\\1\end{pmatrix}+t\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}1+s\\2+s+t\\3+s-t\end{pmatrix}.\] This is the parametric form of a plane, and its equation is of the form \(ax+by+cz=d\), where \((a,b,c)\) is perpendicular to \((1,1,1)\) and to \((0,1,-1)\). We could compute this using the cross product if we knew about vectors, or we could simply look for a vector \((a,b,c)\) such that \(a+b+c=0\) and \(b-c=0\) (these are the dot products of \((a,b,c)\) with the two vectors), and we can easily see that we could take \((a,b,c)=(-2,1,1)\), so the equation would be \(-2x+y+z=d\) – but we know that the plane contains the point \((1,2,3)\), so we find that \(d=3\), so the equation is \(-2x+y+z=3\).1.5.2 The vector equation of a plane
Two points \(A\), \(B\) have distinct position vectors \(\mathbf{a}\), \(\mathbf{b}\) respectively. Suppose that \(O\), \(A\), \(B\) are distinct and not collinear. Let \(P\) (with position vector \(\mathbf{r}\)) be any point in the plane through \(O\), \(A\) and \(B\).
Figure 1.25: The vector equation of a plane
Draw a line through \(P\) parallel to \(\mathbf{b}\) and a line through \(O\) parallel to \(\mathbf{a}\). These lines meet at the point \(S\), and \[\vec{OP}=\mathbf{r}=\vec{OS}+\vec{SP}=s\mathbf{a}+t\mathbf{b}.\]
Example 1.13
Find the vector equation of the line passing through the points \(A(1,1,1)\) and \(B(2,3,2)\).
Find the vector equation of the plane containing the origin and the points \(A\) and \(B\).
Find the points of intersection (if any) of the line \(L\) having equation \[\mathbf{r}=\begin{pmatrix}0\\0\\2\end{pmatrix}+\nu\begin{pmatrix}2\\1\\-4\end{pmatrix}\] with the plane \(OAB\) and the line \(AB\).
Solution. First draw a diagram.
Figure 1.26: Diagram for example
The position vector of \(A\) is \[\vec{OA}=\mathbf{a}=\begin{pmatrix}1\\1\\1\end{pmatrix}=\mathbf{i}+\mathbf{j}+\mathbf{k}.\] The position vector of \(B\) is \[\vec{OB}=\mathbf{b}=\begin{pmatrix}2\\3\\2\end{pmatrix}=2\mathbf{i}+3\mathbf{j}+2\mathbf{k}.\]
The vector equation of the straight line through \(A\) and \(B\) is \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{c}\) where \(\mathbf{c}\) is the direction vector of the line.
Now \[\mathbf{c}=\vec{AB}=-\mathbf{a}+\mathbf{b}=-\begin{pmatrix}1\\1\\1\end{pmatrix}+\begin{pmatrix}2\\3\\2\end{pmatrix}=\begin{pmatrix}1\\2\\1\end{pmatrix}.\] Therefore the vector equation of the line \(AB\) is \[\mathbf{r}=\begin{pmatrix}1\\1\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\1\end{pmatrix}=\begin{pmatrix}1+\lambda\\1+2\lambda\\1+\lambda\end{pmatrix}.\]
The equation of the plane through \(O\), \(A\) and \(B\) is \[\mathbf{r}=\sigma\mathbf{a}+\rho\mathbf{b}=\sigma\begin{pmatrix}1\\1\\1\end{pmatrix}+\rho\begin{pmatrix}2\\3\\2\end{pmatrix}=\begin{pmatrix}\sigma+2\rho\\\sigma+3\rho\\\sigma+2\rho\end{pmatrix}.\]
The plane \(OAB\) and the line \(L\) intersect when the position vectors \[\mathbf{r}=\begin{pmatrix}\sigma+2\rho\\\sigma+3\rho\\\sigma+2\rho\end{pmatrix},\qquad\mathbf{r}=\begin{pmatrix}0\\0\\2\end{pmatrix}+\nu\begin{pmatrix}2\\1\\-4\end{pmatrix}=\begin{pmatrix}2\nu\\\nu\\2-4\nu\end{pmatrix}\] are identical. This happens when \[\begin{eqnarray*} \sigma+2\rho&=&2\nu\\ \sigma+3\rho&=&\nu\\ \sigma+2\rho&=&2-4\nu.\end{eqnarray*}\] Comparing the first and third, we see that \(2\nu=2-4\nu\), so that \(\nu=1/3\). Taking the first from the second, we see that \(\rho=-\nu\), so \(\rho=-1/3\), and then substituting gives \(\sigma=4/3\).
Substitute for \(\nu\) in the equation of the line \(L\) to give the position vector of the point of intersection as \(\mathbf{r}=\begin{pmatrix}2/3\\1/3\\2/3\end{pmatrix}\).
Now we want to work out where \(L\) and the line \(AB\) intersect. We need the position vectors \[\mathbf{r}=\begin{pmatrix}1+\lambda\\1+2\lambda\\1+\lambda\end{pmatrix},\qquad\mathbf{r}=\begin{pmatrix}2\nu\\\nu\\2-4\nu\end{pmatrix}\] to be identical. This happens when \[\begin{eqnarray*} 1+\lambda&=&2\nu\\ 1+2\lambda&=&\nu\\ 1+\lambda&=&2-4\nu.\end{eqnarray*}\] As above, comparing the first and third gives \(\nu=1/3\), and then substituting gives \(\lambda=-1/3\). We check that this makes all three equations agree.
Substituting into the equation of the line \(AB\) gives the position vector of the point of intersection again as \(\mathbf{r}=\begin{pmatrix}2/3\\1/3\\2/3\end{pmatrix}\).
Note that the two answers to the third part agree; if we had considered the two lines first, we would have found that they intersected at the above point. That point must then be the point of intersection of the line \(L\) with the plane \(OAB\), because the line \(AB\) lies in the plane \(OAB\).
1.5.3 The normal to a plane
Given a plane \(ax+by+cz=d\), we have already indicated that it is perpendicular to the vector \((a~b~c)\). Indeed, this means that we can describe the “normal”, which again means the line perpendicular to the plane, which passes through the point \(P=(x_0,y_0,z_0)\) in a parametric way; the points are described by \[(x_0,y_0,z_0)+t(a,b,c)=(x_0+ta,y_0+tb,z_0+tc).\] This parametric way of describing a line is very common in three dimensions. Notice that when \(t=0\), we just get the point \(P\), so \(P\) is definitely in this set, and also as \(t\) varies, we get a line in the direction \((a,b,c)\).
Example 1.14 Find the normal to the plane \(x-y+z=1\) that passes through \((1,2,3)\).
We can simply read off this line as \[\begin{pmatrix}1\\2\\3\end{pmatrix}+t\begin{pmatrix}1\\-1\\1\end{pmatrix}=\begin{pmatrix}1+t\\2-t\\3+t\end{pmatrix}.\] If you want to go from a parametric form to a form simply involving \(x\), \(y\) and \(z\), we need to eliminate the parameter. So we see that \[t=x-1=2-y=z-3,\] so the line would be given by \[x-1=2-y=z-3.\] Notice that lines in three dimensions involve two equals signs. If we just had one equals sign, we’d have a plane. Having two equals signs means that the points have to satisfy two equalities; each one defines a plane, and so lines are viewed as the intersection of two planes.1.5.4 Intersections of planes
Given two planes, we can ask how they intersect. Generally, of course, it seems likely that they should intersect in a line, but you should be able to think of situations where two planes don’t intersect at all (since they are parallel) or are completely identical.
Two planes will be parallel if their left-hand sides are proportional. So if the first plane is given by \(ax+by+cz=d\), and the second by \(a'x+b'y+c'z=d'\), then the planes meet in a line unless \((a'~b'~c')=k(a~b~c)\) for some \(k\). If also \(d'=kd\), then the two equations are identical, apart from being scaled, and the two planes are the same. Otherwise, the two planes are parallel, and don’t have any intersection.
It may seem surprising that there is no single equation that defines a line in three dimensions. But when you think about an equation like \(z=f(x,y)\), you should see that the result is essentially two-dimensional – we can pick any pair \((x,y)\), and then there will be a corresponding \(z\)-value. This means that there should be a point on \(z=f(x,y)\) for every point in the plane, and since the plane is two-dimensional, so should \(z=f(x,y)\) be two-dimensional, and it should therefore be a surface. Instead, lines will correspond to pairs of planes (and there are many choices for any given line!), and are therefore the solutions to pairs of equations in three dimensions.
More generally, even if our equations aren’t linear, and perhaps have quadratic or other terms, we expect that in three dimensions, one equation will have essentially a two-dimensional solution (although we haven’t actually defined what this ought to mean!), and similarly, in \(n\) dimensions, one equation will have an \((n-1)\)-dimensional solution. Intersection of two equations ought generally to be \((n-2)\)-dimensional, and so on.
1.5.5 Intersection of three planes in three dimensions
Really, of course, it is better to treat intersection of planes as solutions of simultaneous equations, and we will do this beginning in the next chapter. But it is still worth thinking about the various configurations of three planes in three dimensions, and asking what their common points of intersection might look like. This will give us insight as to what to expect when we come to solve simultaneous equations in three variables.
Recall that the equations appearing in our systems of simultaneous equations are all of the form \(ax+by+cz=k\), and that these represent a plane in 3 dimensions orthogonal to \((a,b,c)^T\). Finding points lying on each of the planes is equivalent to solving a set of simultaneous equations.
Generally, three planes in three dimensions will intersect in a single point.
Figure 1.27: Intersection of three planes
In this case, we end up with a single solution to our simultaneous equations.
But there are a number of other things that might happen, where we don’t get a single solution, and there may be infinitely many solutions – or none at all!
- The three planes may intersect in a common line.
Figure 1.28: Three planes intersecting in a line
- The three planes may not intersect at all
Figure 1.29: Two parallel planes and another
Figure 1.30: Three planes intersecting in parallel lines
- Two or more planes may be the same.
Let’s think about this a little more carefully. If there are three planes, like in our system above, each one is orthogonal to the vector made up of the coefficients (as mentioned earlier). When you think about the situations where there are no common points, or a whole line (or more) of common points, you should see that this occurs when the three vectors involved all lie in a single plane.
Given three vectors \(\mathbf{u}\), \(\mathbf{v}\) and \(\mathbf{w}\) in \(\mathbb{R}^3\), there will usually not be any plane that contains all three vectors. This means that we can get from the origin to any point by travelling a certain (possibly negative) distance in the direction of \(\mathbf{u}\), then a certain distance in the direction of \(\mathbf{v}\), then a certain distance in the direction of \(\mathbf{w}\). The case where \(\mathbf{u}\), \(\mathbf{v}\) and \(\mathbf{w}\) all lie in a common plane will have special geometric significance in any purely mathematical problem, and will often have special physical significance in applied and engineering problems. Three vectors lie in the same plane occurs when one of the vectors, \(\mathbf{w}\) say, can be written in terms of the other two, so that \(\mathbf{w}=\alpha\mathbf{u}+\beta\mathbf{v}\) for some real numbers \(\alpha\) and \(\beta\). That is, there is a relationship between \(\mathbf{u}\), \(\mathbf{v}\) and \(\mathbf{w}\). Alternatively, we say that \(\mathbf{u}\), \(\mathbf{v}\) and \(\mathbf{w}\) are linearly dependent, and otherwise we say they are linearly independent. Along the same lines, we say that an expression \(\mathbf{w}=\alpha\mathbf{u}+\beta\mathbf{v}\) is a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\). The same terminology applies to anything where we can add two terms and multiply each by a number, so for example, \(2x+3x^2\) is a linear combination of \(x\) and \(x^2\), and \(2\sqrt{2}-\sqrt{3}\) is a linear combination of \(\sqrt{2}\) and \(\sqrt{3}\).
More formally:Let’s see this with some examples.
Example 1.15 Consider the three planes \[\begin{aligned} x+y+z&=2\\ 2x+3y-z&=1\\ x+4z&=5.\end{aligned}\] We claim that these intersect in a line, and that the normal vectors to the planes are linearly dependent.
To see that the three planes intersect in a line, we need to solve the equations simultaneously. We note that eliminating \(x\) from both the bottom two equations by using the first equation both give \(y-3z=-3\). (We’ll do this more systematically later!) So we have two equations, \[\begin{aligned} x+y+z&=2\\ y-3z&=-3.\end{aligned}\] We can put \(z=t\), an arbitrary parameter, and read off \(y=3t-3\), and then \(x=5-4t\), to get the solution \[\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}5-4t\\3t-3\\t\end{pmatrix}=\begin{pmatrix}5\\-3\\0\end{pmatrix}+t\begin{pmatrix}-4\\3\\1\end{pmatrix},\] which is a straight line passing through \(\begin{pmatrix}5\\-3\\0\end{pmatrix}\) in the direction \(\begin{pmatrix}-4\\3\\1\end{pmatrix}\).
The normal vectors to the planes are \(\begin{pmatrix}1\\1\\1\end{pmatrix}\), \(\begin{pmatrix}2\\3\\-1\end{pmatrix}\), \(\begin{pmatrix}1\\0\\4\end{pmatrix}\). We can see that there is a linear relationship: \[-3\begin{pmatrix}1\\1\\1\end{pmatrix}+\begin{pmatrix}2\\3\\-1\end{pmatrix}+\begin{pmatrix}1\\0\\4\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.\] So the normal vectors are linearly dependent.Example 1.16 Show that \(\begin{pmatrix}1\\0\\1\end{pmatrix}\), \(\begin{pmatrix}2\\1\\1\end{pmatrix}\) and \(\begin{pmatrix}1\\1\\1\end{pmatrix}\) are linearly independent.
We need to solve \[\alpha\begin{pmatrix}1\\0\\1\end{pmatrix}+\beta\begin{pmatrix}2\\1\\1\end{pmatrix}+\gamma\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.\] That is, \[\begin{aligned} \alpha+2\beta+\gamma&=0\\ \beta+\gamma&=0\\ \alpha+\beta+\gamma&=0\end{aligned}\] and it is easy to solve these to get \(\alpha=\beta=\gamma=0\).1.6 The scalar product of two vectors
Figure 1.31: The scalar product of two vectors
Note that
The scalar product of two vectors is a scalar.
The dot in the product \(\mathbf{a}.\mathbf{b}\) is essential.
Let’s think about some elementary properties of the scalar product.
Commutative property: \(\mathbf{a}.\mathbf{b}=\mathbf{b}.\mathbf{a}\).
Multiplication by scalars: \(\mathbf{a}.(\lambda\mathbf{b})=\lambda(\mathbf{a}.\mathbf{b})\) and more generally \((\lambda\mathbf{a}).(\mu\mathbf{b})=\lambda\mu(\mathbf{a}.\mathbf{b})\).
Distributive property: \(\mathbf{a}.(\mathbf{b}+\mathbf{c})=\mathbf{a}.\mathbf{b}+\mathbf{a}.\mathbf{c}\).
All these are easy to check. For example, let’s look at the commutative property. Take any vectors \(\mathbf{a}\) and \(\mathbf{b}\). Then \[\mathbf{a}.\mathbf{b}=|\mathbf{a}|\,|\mathbf{b}|\,\cos\theta=|\mathbf{b}|\,|\mathbf{a}|\,\cos\theta=\mathbf{b}.\mathbf{a}\] as required.
Note the following properties relating to the angles:
If \(\mathbf{a}\) and \(\mathbf{b}\) are parallel, them \(\theta=0\). So \(\cos\theta=1\), and \(\mathbf{a}.\mathbf{b}=|\mathbf{a}|\,|\mathbf{b}|\).
In particular, for any vector \(\mathbf{a}\), we have \(\mathbf{a}.\mathbf{a}=|\mathbf{a}|^2\).
If \(\mathbf{a}\) and \(\mathbf{b}\) are antiparallel, so that they point in opposite directions, \(\theta=\pi\), so that \(\cos\theta=-1\), and \(\mathbf{a}.\mathbf{b}=-|\mathbf{a}|\,|\mathbf{b}|\).
If \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular (i.e., orthogonal or at right-angles), then \(\theta=\pi/2\), so \(\cos\theta=0\) and \(\mathbf{a}.\mathbf{b}=0\).
Notice that this means that if \(\mathbf{a}.\mathbf{b}=0\), then either one of \(\mathbf{a}\) and \(\mathbf{b}\) is the zero vector \(\mathbf{0}\), or \(\mathbf{a}\) and \(\mathbf{b}\) are orthogonal.
We can use all this to compute the scalar product in Cartesian components. After all, the unit vectors \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\) have length \(1\), so that \(\mathbf{i}.\mathbf{i}=1\mathbf{j}.\mathbf{j}=\mathbf{k}.\mathbf{k}=1\), and are perpendicular to each other, so \(\mathbf{i}.\mathbf{j}=\mathbf{j}.\mathbf{k}=\mathbf{k}.\mathbf{i}=0\). Now consider two vectors \[\begin{eqnarray*} \mathbf{a}&=&\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k},\\ \mathbf{b}&=&\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}.\end{eqnarray*}\] Then \[\begin{eqnarray*} \mathbf{a}.\mathbf{b}&=&(a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}).(b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k})\\ &=&a_1b_1\mathbf{i}.\mathbf{i}+a_1b_2\mathbf{i}.\mathbf{j}+a_1b_3\mathbf{i}.\mathbf{k}+a_2b_1\mathbf{j}.\mathbf{i}+a_2b_2\mathbf{j}.\mathbf{j}+a_2b_3\mathbf{j}.\mathbf{k}+a_3b_1\mathbf{k}.\mathbf{i}+a_3b_2\mathbf{k}.\mathbf{j}+a_3b_3\mathbf{k}.\mathbf{k}\\ &=&a_1b_1+a_2b_2+a_3b_3.\end{eqnarray*}\] So \[\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}.\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}=a_1b_1+a_2b_2+a_3b_3.\]
Example 1.17 Find the angle between the vectors \(\mathbf{a}=\begin{pmatrix}7\\-4\\4\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}-2\\4\\4\end{pmatrix}\).
Solution. Use \(\mathbf{a}.\mathbf{b}=|\mathbf{a}|\,|\mathbf{b}|\,\cos\theta\) to find \(\theta\). We have \[\begin{eqnarray*} |\mathbf{a}|&=&\sqrt{7^2+(-4)^2+4^2}=\sqrt{49+16+16}=\sqrt{81}=9,\\ |\mathbf{b}|&=&\sqrt{(-2)^2+4^2+4^2}=\sqrt{4+16+16}=\sqrt{36}=6,\\ \mathbf{a}.\mathbf{b}&=&7\times(-2)+(-4)\times4+4\times4=-14-16+16=-14.\end{eqnarray*}\] Then \[\cos\theta=\frac{\mathbf{a}.\mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}=\frac{-14}{9\times6}=-\frac{7}{27},\] and we can compute that \(\theta=1.83\) radians, to 3 significant figures (remember that we always give angles in radians).Consider two non-zero vectors \(\mathbf{a}\) and \(\mathbf{b}\), as shown.
Figure 1.32: Projecting one vector along another
Define points \(A\) and \(B\) with position vectors \(\mathbf{a}\) and \(\mathbf{b}\) respectively. Let \(\hat{\mathbf{a}}\) be the unit vector in the direction of \(\mathbf{a}\).
Consider \[\mathbf{b}.\hat{\mathbf{a}}=|\mathbf{b}|\,|\hat{\mathbf{a}}|\,\cos\theta=|\mathbf{b}|\cos\theta=OB\,\cos\theta=ON.\] We say that ON is the projection of \(\mathbf{b}\) along OA. ON is the component of \(\mathbf{b}\) along OA. This process is also known as resolving \(\mathbf{b}\) along OA.
Example 1.18 Show that the lines with vector equations \[\begin{aligned} \mathbf{r}=\begin{pmatrix}2\\3\\2\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\1\end{pmatrix}\\ \mathbf{r}=\begin{pmatrix}0\\0\\2\end{pmatrix}+\mu\begin{pmatrix}2\\1\\-4\end{pmatrix}\end{aligned}\] are perpendicular.
Solution. The vector equation of a straight line has the form \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{c}\) where \(\mathbf{a}\) is the position vector of a point on the line and \(\mathbf{c}\) is the direction vector of the line.
So the direction vector of the first line is \(\begin{pmatrix}1\\2\\1\end{pmatrix}\ne\mathbf{0}\), and the direction vector of the second line is \(\begin{pmatrix}2\\1\\-4\end{pmatrix}\ne\mathbf{0}\). However \[\begin{pmatrix}1\\2\\1\end{pmatrix}.\begin{pmatrix}2\\1\\-4\end{pmatrix}=1\times2+2\times1+1\times(-4)=0,\] so the direction vectors of the two lines are perpendicular, and therefore the two lines are perpendicular.1.6.1 Another form for the vector equation of a plane
Consider the plane \(\Pi\) which passes through the point \(A\) (with position vector \(\mathbf{a}\)) and is normal (perpendicular) to the vector \(\mathbf{n}\).
Figure 1.33: Diagram for the vector equation of a plane
Let \(P\) be any point on the plane, with position vector \(\mathbf{r}\). Then \(\vec{AP}\) and \(\mathbf{n}\) are perpendicular, so \(\vec{AP}.\mathbf{n}=0\). But \(\vec{AP}=\mathbf{r}-\mathbf{a}\), so have \[(\mathbf{r}-\mathbf{a}).\mathbf{n}=0,\] and so \(\mathbf{r}.\mathbf{n}=\mathbf{a}.\mathbf{n}\). Since this right-hand side is a constant, \(d\), say, we get the equation \[\mathbf{r}.\mathbf{n}=d.\] (If you prefer, you can think of \(\mathbf{r}.\mathbf{n}=0\) as saying that \(\mathbf{r}\) is the plane through the origin consisting of all vectors perpendicular to \(\mathbf{n}\), and then \(\mathbf{r}.\mathbf{n}=d\) is just a shifting of the plane to a parallel one.)
If \(\mathbf{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\), and \(\mathbf{r}=\begin{pmatrix}x\\y\\z\end{pmatrix}\), we have \(\mathbf{r}.\mathbf{n}=ax+by+cz\), and we get the Cartesian equation of a plane as \[ax+by+cz=d.\] This is a plane perpendicular to \(\mathbf{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}\).
Example 1.19 Find the vector equation and Cartesian equation of the plane normal to the vector \(\mathbf{n}=\mathbf{i}+2\mathbf{j}+2\mathbf{k}\) and containing the point with position vector \(\mathbf{a}=2\mathbf{i}+\mathbf{j}+5\mathbf{k}\).
Solution. The vector equation of the plane is \(\mathbf{r}.\mathbf{n}=\mathbf{a}.\mathbf{n}\) where \[\mathbf{a}.\mathbf{n}=\begin{pmatrix}2\\1\\5\end{pmatrix}.\begin{pmatrix}1\\2\\2\end{pmatrix}=2\times1+1\times2+5\times2=14.\] So the vector equation of the plane is \[\mathbf{r}.\mathbf{n}=14,\] and if \(\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\), so that \(\mathbf{r}.\mathbf{n}=x+2y+2z\), we see that the Cartesian equation of the plane is \[x+2y+2z=14.\]Example 1.20 Find the Cartesian equation of the plane passing through the points with position vectors \[\mathbf{i},\qquad2\mathbf{i}+3\mathbf{j}+3\mathbf{k},\qquad-4\mathbf{j}-5\mathbf{k}.\] Hence find the direction of the normal to the plane.
Solution. The Cartesian equation of the plane is \(ax+by+cz=d\). For \(\mathbf{i}\) to lie on the plane, we need \(a\times1+b\times0+c\times0=d\), so \(a=d\). Similarly, using the other two points, we need \[\begin{eqnarray*} 2a+3b+3c&=&d\\ -4b-5c&=&d.\end{eqnarray*}\] Since \(a=d\), the first equation becomes \(3b+3c=-d\). Solving this simultaneously with the other equation gives \[b=-2d/3,\qquad c=d/3.\] The equation \(ax+by+cz=d\) becomes \[dx+(-2d/3)y+(d/3)z=d;\] dividing by \(d\) and multiplying by \(3\) gives the Cartesian equation as \[3x-2y+z=3.\] If \(\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\), as usual, and \(\mathbf{n}=3\mathbf{i}-2\mathbf{j}+\mathbf{k}\), then this can be rewritten as \(\mathbf{r}.\mathbf{n}=3\), which is the vector equation of the plane; we note that the normal vector is \(\mathbf{n}=\begin{pmatrix}3\\-2\\1\end{pmatrix}\).1.7 The vector product of two vectors
Note that
The vector product of two vectors is a vector.
The \(\times\) in \(\mathbf{a}\times\mathbf{b}\) is essential.
Note that for all vectors \(\mathbf{a}\), \(\mathbf{b}\) and all scalars \(\lambda\), \(\mu\):
Anti-commutative property: \(\mathbf{a}\times\mathbf{b}=-\mathbf{b}\times\mathbf{a}\).
Multiplication by scalars: \(\mathbf{a}\times(\lambda\mathbf{b})=\lambda(\mathbf{a}\times\mathbf{b})\), and more generally, \((\lambda\mathbf{a})\times(\mu\mathbf{b})=\lambda\mu(\mathbf{a}\times\mathbf{b})\).
Distributive property: \(\mathbf{a}\times(\mathbf{b}+\mathbf{c})=\mathbf{a}\times\mathbf{b}+\mathbf{a}\times\mathbf{c}\).
These are slightly less easy than the similar properties for the dot product. Let’s look at the anti-commutative property. Take vectors \(\mathbf{a}\) and \(\mathbf{b}\). If \((\mathbf{a},\mathbf{b},\hat{\mathbf{n}})\) is a right-handed triple, so is \((\mathbf{b},\mathbf{a},-\hat{\mathbf{n}})\).
Figure 1.34: Diagram for the vector product of two vectors
Therefore \[\mathbf{b}\times\mathbf{a}=|\mathbf{b}|\,|\mathbf{a}|\,\sin\theta\,(-\hat{\mathbf{n}})=-|\mathbf{a}|\,|\mathbf{b}|\hat{\mathbf{n}}\,\sin\theta=-\mathbf{a}\times\mathbf{b}.\] Notice that if \(\mathbf{a}\) and \(\mathbf{b}\) are parallel or antiparallel, so that \(\theta=0\) or \(\theta=\pi\), we have \(\mathbf{a}\times\mathbf{b}=\mathbf{0}\), since \(\sin\theta=0\). In particular, \(\mathbf{a}\times\mathbf{a}=\mathbf{0}\) for any vector \(\mathbf{a}\).
Conversely, if \(\mathbf{a}\times\mathbf{b}=\mathbf{0}\), either one of \(\mathbf{a}\) and \(\mathbf{b}\) is the zero vector, or \(\mathbf{a}\) and \(\mathbf{b}\) are parallel or antiparallel.
If \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular, so that \(\theta=\pi/2\), we get \(\mathbf{a}\times\mathbf{b}=|\mathbf{a}|\,|\mathbf{b}|\,\hat{\mathbf{n}}\).
Again, we can work out the vector product in Cartesian components, just by knowing that \[\mathbf{i}\times\mathbf{i}=\mathbf{0},\qquad\mathbf{j}\times\mathbf{j}=\mathbf{0},\qquad\mathbf{k}\times\mathbf{k}=\mathbf{0},\] and that \[\mathbf{i}\times\mathbf{j}=\mathbf{k},\qquad\mathbf{j}\times\mathbf{k}=\mathbf{i},\qquad\mathbf{k}\times\mathbf{i}=\mathbf{j};\] then anticommutativity gives \[\mathbf{j}\times\mathbf{i}=-\mathbf{k},\qquad\mathbf{k}\times\mathbf{j}=-\mathbf{i},\qquad\mathbf{i}\times\mathbf{k}=-\mathbf{j}.\] If \(\mathbf{a}=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}\), and \(\mathbf{b}=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}\), we see that \[\begin{eqnarray*} \mathbf{a}\times\mathbf{b}&=&(a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k})\times(b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k})\\ &=&a_1b_1\mathbf{i}\times\mathbf{i}+a_1b_2\mathbf{i}\times\mathbf{j}+a_1b_3\mathbf{i}\times\mathbf{k}\\ &~&+a_2b_1\mathbf{j}\times\mathbf{i}+a_2b_2\mathbf{j}\times\mathbf{j}+a_2b_3\mathbf{j}\times\mathbf{k}\\ &~&+a_3b_1\mathbf{k}\times\mathbf{i}+a_3b_2\mathbf{k}\times\mathbf{j}+a_3b_3\mathbf{k}\times\mathbf{k}\\ &=&a_1b_2\mathbf{k}-a_1b_3\mathbf{j}-a_2b_1\mathbf{k}+a_2b_3\mathbf{i}+a_3b_1\mathbf{j}-a_3b_2\mathbf{i}\\ &=&\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}.\end{eqnarray*}\]
Example 1.21 Find a vector perpendicular to the vectors \[\mathbf{p}=\mathbf{i}+3\mathbf{j}+3\mathbf{k},\qquad\mathbf{q}=-\mathbf{i}-4\mathbf{j}-5\mathbf{k}.\]
Solution. By the definition of the vector product, the vector \(\mathbf{p}\times\mathbf{q}\) is perpendicular to both \(\mathbf{p}\) and \(\mathbf{q}\). Using the formula: \[\mathbf{p}\times\mathbf{q}=\begin{pmatrix}p_2q_3-p_3q_2\\p_3q_1-p_1q_3\\p_1q_2-p_2q_1\end{pmatrix}=\begin{pmatrix}3\times(-5)-3\times(-4)\\3\times(-1)-1\times(-5)\\1\times(-4)-3\times(-1)\end{pmatrix}=\begin{pmatrix}-3\\2\\-1\end{pmatrix}.\] This final vector is perpendicular to both \(\mathbf{p}\) and \(\mathbf{q}\) (and one can check this by verifying that its dot products with \(\mathbf{p}\) and with \(\mathbf{q}\) are both \(0\)).Example 1.22 Let \(\mathbf{a}\) be the position vector of a point on the line \(L\) which has direction vector \(\mathbf{c}\). Show that the vector equation of the line \(L\) can be written in the form \(\mathbf{r}\times\mathbf{c}=\mathbf{d}\), where \(d\) is a constant vector which should be found in terms of \(\mathbf{a}\) and \(\mathbf{c}\).
Solution. The vector equation of the line is \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{c}\). Then \[\mathbf{r}\times\mathbf{c}=\mathbf{a}\times\mathbf{c}+\lambda\mathbf{c}\times\mathbf{c}=\mathbf{a}\times\mathbf{c}.\] So put \(\mathbf{d}=\mathbf{a}\times\mathbf{c}\), and then the vector equation of the line takes the form \(\mathbf{r}\times\mathbf{c}=\mathbf{d}\), as required.Example 1.23 Consider the point \(S\) and the line \(L\) with vector equation \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{c}\). Show that the shortest distance from \(S\) to the line \(L\) is \[\frac{|\vec{SA}\times\mathbf{c}|}{|\mathbf{c}|},\] where \(A\) is the point on the line with position vector \(\mathbf{a}\).
Solution. Let \(P\) be the point on the line such that \(\vec{SP}\) is perpendicular to \(L\). Let \(Q\) be any other point on the line \(L\). Since \(SPQ\) is a right-angled triangle, \(SQ>SP\). Then \(SP\) is the shortest distance of the point \(S\) from the line \(L\).
Consider \[|\vec{SQ}\times\mathbf{c}|=|\vec{SQ}|\,|\mathbf{c}|\,\sin\theta=|\mathbf{c}|.SQ\,\sin\theta=|\mathbf{c}|.SP.\] Then the distance of the point \(P\) to the line \(L\) is \[SP=|\vec{SQ}\times\mathbf{c}|/|\mathbf{c}|\] for any point on the line. Taking \(Q=A\), we get the required result.Solution. First draw a sketch:
Figure 1.35: Diagram for example
If the vector equation of the line is \(\mathbf{r}\times\mathbf{c}=\mathbf{d}\), then \(\mathbf{c}\) is the direction vector of the line. Here \(\mathbf{c}\) must be perpendicular to the plane.
Therefore \(\mathbf{c}\) is parallel to \(\mathbf{n}\), the normal to the plane. The vector equation of the plane has the form \(\mathbf{r}.\mathbf{n}=d\), where \(\mathbf{n}\) is the normal to the plane.
In our case, therefore, \(\mathbf{n}=2\mathbf{i}+\mathbf{j}+\mathbf{k}\). Therefore we can take \(\mathbf{c}=2\mathbf{i}+\mathbf{j}+\mathbf{k}\), and the equation of the line is \(\mathbf{r}\times(2\mathbf{i}+\mathbf{j}+\mathbf{k})=\mathbf{d}\).
To find \(\mathbf{d}\), let \(\mathbf{a}=-4\mathbf{j}+5\mathbf{k}\) be the position vector of a point on the line. Then \[\mathbf{d}=\mathbf{a}\times\mathbf{c}=\begin{pmatrix}0\\-4\\5\end{pmatrix}\times\begin{pmatrix}2\\1\\1\end{pmatrix}=\begin{pmatrix}-9\\10\\8\end{pmatrix}.\] Then the vector equation of the line is \[\mathbf{r}\times(2\mathbf{i}+\mathbf{j}+\mathbf{k})=-9\mathbf{i}+10\mathbf{j}+8\mathbf{k}.\]
Solution. First draw a sketch:
Figure 1.36: Diagram for example
The vector equation of the plane is \(\mathbf{r}.\mathbf{n}=d\), where \(\mathbf{n}\) is the normal to the plane.
Let’s find \(\mathbf{n}\). The vectors \(\vec{BA}\) and \(\vec{BC}\) lie in the plane. Consider the vector \(\mathbf{n}=\vec{BA}\times\vec{BC}\), which is perpendicular to both \(\vec{BA}\) and \(\vec{BC}\). Therefore \(\mathbf{n}\) is perpendicular (or normal) to the plane. We compute \[\begin{eqnarray*} \vec{BA}&=&\mathbf{a}-\mathbf{b}=-\mathbf{i}-3\mathbf{j}-3\mathbf{k};\\ \vec{BC}&=&\mathbf{c}-\mathbf{b}=-2\mathbf{i}-7\mathbf{j}-8\mathbf{k}.\end{eqnarray*}\] Then we can easily compute \(\mathbf{n}=\vec{BA}\times\vec{BC}=\begin{pmatrix}3\\-2\\1\end{pmatrix}\). So the vector equation will be \[\mathbf{r}.(3\mathbf{i}-2\mathbf{j}+\mathbf{k})=d.\]
To find \(d\), pick any point on the plane, \(A\) for example, and set \(\mathbf{r}=\mathbf{a}=\mathbf{i}\). Then \[d=\mathbf{a}.(3\mathbf{i}-2\mathbf{j}+\mathbf{k})=\mathbf{i}.(3\mathbf{i}-2\mathbf{j}+\mathbf{k})=3.\] Therefore the vector equation of the plane is \[\mathbf{r}.(3\mathbf{i}-2\mathbf{j}+\mathbf{k})=3.\]
Solution. First draw a sketch:
Figure 1.37: Diagram for example
The vector equation of a plane has the form \(\mathbf{r}.\mathbf{n}=d\) where \(\mathbf{n}\) is the normal to the plane. The normal \(\mathbf{n}_1\) to the plane \(\Pi_1\) is \(\mathbf{n}_1=2\mathbf{i}+\mathbf{j}+\mathbf{k}\); the normal \(\mathbf{n}_2\) to the plane \(\Pi_2\) is \(\mathbf{n}_2=\mathbf{i}+2\mathbf{j}+\mathbf{k}\).
The line of intersection \(L\) has parametric vector equation \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{c}\) where \(\mathbf{a}\) is the position vector of a point on the line and \(\mathbf{c}\) is the direction vector of the line.
The vector \(\mathbf{c}\) lies in the plane \(\Pi_1\) and \(\Pi_2\) so \(\mathbf{c}\) is perpendicular to \(\mathbf{n}_1\) and to \(\mathbf{n}_2\). Then \(\mathbf{c}\) is in the direction of \(\mathbf{n}_1\times\mathbf{n}_2=-\mathbf{i}-\mathbf{j}+3\mathbf{k}+(-1,-1,3)\).
To find \(\mathbf{a}\), we need to find the position vector of one point lying on both planes. Let this point be \(A\), with position vector \(\mathbf{a}=u\mathbf{i}+v\mathbf{j}+w\mathbf{k}\). \(A\) lies on the plane \(\Pi_1\), so \(\mathbf{a}.(2\mathbf{i}+\mathbf{j}+\mathbf{k})=1\). This gives \(2u+v+w=1\). In the same way, as \(A\) lies on the plane \(\Pi_2\), we need \(\mathbf{a}.(\mathbf{i}+2\mathbf{j}+\mathbf{k})=-3\). We only need one solution, so we might as well set \(v=0\). Solving these gives \(u=4\), \(w=-7\). Therefore the point \(A\) with position vector \(\mathbf{a}=4\mathbf{i}-7\mathbf{k}\) lies in both planes. The parametric vector equation of the line of intersection is then \[\mathbf{r}=4\mathbf{i}-7\mathbf{k}+\lambda(-\mathbf{i}-\mathbf{j}+3\mathbf{k}).\]
1.8 Products of three vectors
Three vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) can be combined using products in various ways. For example, \((\mathbf{b}.\mathbf{c})\mathbf{a}\) is the product of the scalar \(\mathbf{b}.\mathbf{c}\) and the vector \(\mathbf{a}\), so is just a scaled version of \(\mathbf{a}\). But there are two more interesting ways to take triple products.
1.8.1 The scalar triple product
We can first form the vector \(\mathbf{b}\times\mathbf{c}\) and take the dot prodct with \(\mathbf{a}\) to form the scalar triple product \(\mathbf{a}.(\mathbf{b}\times\mathbf{c})\). Notice that \(\mathbf{a}.(\mathbf{b}\times\mathbf{c})\) is a scalar
Suppose that \(\mathbf{a}=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}\), \(\mathbf{b}=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}\) and \(\mathbf{c}=\begin{pmatrix}c_1\\c_2\\c_3\end{pmatrix}\). From the earlier work expressing these in components: \[\mathbf{a}.(\mathbf{b}\times\mathbf{c})=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}.\begin{pmatrix}b_2c_3-b_3c_2\\b_3c_1-b_1c_3\\b_1c_2-b_2c_1\end{pmatrix}=a_1(b_2c_3-b_3c_2)+a_2(b_3c_1-b_1c_3)+a_3(b_1c_2-b_2c_1).\] We will see this quantity again in Chapter 4.
Properties of scalar triple product:
\(\mathbf{a}.(\mathbf{b}\times\mathbf{c})=(\mathbf{b}\times\mathbf{c}).\mathbf{a}\) (commutativity of the dot product)
\(\mathbf{a}.(\mathbf{b}\times\mathbf{c})=-\mathbf{a}.(\mathbf{c}\times\mathbf{b})\) (anticommutativity of the cross product)
As \(\mathbf{a}\) and \(\mathbf{a}\times\mathbf{b}\) are perpendicular, \(\mathbf{a}.(\mathbf{a}\times\mathbf{b})=0\). Similarly \(\mathbf{b}.(\mathbf{a}\times\mathbf{b})=0\), and then distributivity tells us that \((\lambda\mathbf{a}+\mu\mathbf{b}).(\mathbf{a}\times\mathbf{b})=0\) for any \(\lambda\) and \(\mu\). So if \(\mathbf{c}\) is any vector in the same plane as \(\mathbf{a}\) and \(\mathbf{b}\), we see that \(\mathbf{c}.(\mathbf{a}\times\mathbf{b})=0\). (We say that the three vectors are coplanar when they lie in the same plane.)
There is a cyclic property: \[\mathbf{a}.(\mathbf{b}\times\mathbf{c})=\mathbf{b}.(\mathbf{c}\times\mathbf{a})=\mathbf{c}.(\mathbf{a}\times\mathbf{b}).\]
We can interchange the products: \[\mathbf{a}.(\mathbf{b}\times\mathbf{c})=(\mathbf{a}\times\mathbf{b}).\mathbf{c}.\]
All these can be proven using components (or later results on determinants).
Let’s interpret the scalar triple product geometrically. Take three vectors, \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\), where we assume \((\mathbf{a},\mathbf{b},\mathbf{c})\) are a right-handed tripe, and consider the “parallelepiped” whose vertices are at \(\mathbf{0}\), \(\mathbf{a}\), \(\mathbf{b}\), \(\mathbf{c}\), \(\mathbf{a}+\mathbf{b}\), \(\mathbf{b}+\mathbf{c}\), \(\mathbf{c}+\mathbf{a}\) and \(\mathbf{a}+\mathbf{b}+\mathbf{c}\) (this is just a 3-dimensional analogue of a parallelogram in 2 dimensions).
Its volume \(V\) is the base area \(A\) multiplied by the perpendicular height \(h\). The base area is simply \(A=|\mathbf{b}|\,|\mathbf{c}|\,\sin\theta\), which is \(|\mathbf{b}\times\mathbf{c}|\), the length of \(\mathbf{b}\times\mathbf{c}\). The perpendicular height is \(h=|\mathbf{a}|\cos\phi\), where \(\phi\) is the angle between the perpendicular and \(\mathbf{a}\). So \[V=Ah=(|\mathbf{b}|\,|\mathbf{c}|\,\sin\theta)(|\mathbf{a}|\cos\phi)=|\mathbf{b}\times\mathbf{c}|\,|\mathbf{a}|\cos\phi=(\mathbf{b}\times\mathbf{c}).\mathbf{a}=\mathbf{a}.(\mathbf{b}\times\mathbf{c}).\]
If \((\mathbf{a},\mathbf{b},\mathbf{c})\) formed a left-handed triple, a similar argument would give \(V=-\mathbf{a}.(\mathbf{b}\times\mathbf{c})\); the argument would be the same, except that \(\phi\) would be obtuse, so that \(\cos\phi<0\). In general, then, the volume is \[V=|\mathbf{a}.(\mathbf{b}\times\mathbf{c})|.\]
If \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) are non-zero vectors, but have \(V=\mathbf{a}.(\mathbf{b}\times\mathbf{c})=0\), then the volume of the parallelepiped is \(0\), and the three vectors lie in the same plane, which shows the converse of the third property above.
Example 1.27 Show that the vectors \[\mathbf{a}=\begin{pmatrix}1\\2\\1\end{pmatrix},\qquad\mathbf{b}=\begin{pmatrix}-1\\-4\\-5\end{pmatrix},\qquad\mathbf{c}=\begin{pmatrix}0\\-1\\-2\end{pmatrix}\] are coplanar.
Find constants \(\lambda\) and \(\mu\) such that \(\mathbf{c}=\lambda\mathbf{a}+\mu\mathbf{b}\).
Solution. The three vectors are coplanar if \((\mathbf{a}\times\mathbf{b}).\mathbf{c}=0\). From the Cartesian components, we have \[\mathbf{a}\times\mathbf{b}=\begin{pmatrix}2\times(-5)-1\times(-4)\\1\times(-1)-1\times(-5)\\1\times(-4)-2\times(-1)\end{pmatrix}=\begin{pmatrix}-6\\4\\-2\end{pmatrix}.\] Then \[(\mathbf{a}\times\mathbf{b}).\mathbf{c}=(-6)\times0+4\times(-1)+(-2)\times(-2)=0.\] So the three vectors are coplanar.
If \(\mathbf{c}=\lambda\mathbf{a}+\mu\mathbf{b}\), we need \[\begin{pmatrix}0\\-1\\-2\end{pmatrix}=\lambda\begin{pmatrix}1\\2\\1\end{pmatrix}+\mu\begin{pmatrix}-1\\-4\\-5\end{pmatrix},\] so that \[\begin{eqnarray*} 0&=&\lambda-\mu\\ -1&=&2\lambda-4\mu\\ -2&=&\lambda-5\mu.\end{eqnarray*}\] The first equation gives \(\lambda=\mu\), and substituting in the second equation gives \(-1=-2\lambda\), so \(\lambda=1/2\) (we should check that \(\lambda=\mu=1/2\) works in the final equation too). So \[\mathbf{c}=\mathbf{a}/2+\mathbf{b}/2=\tfrac{1}{2}(\mathbf{a}+\mathbf{b}),\] which is actually the midpoint of \(\mathbf{a}\) and \(\mathbf{b}\).1.8.2 Vector triple product
Given three vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\), we can form \(\mathbf{a}\times(\mathbf{b}\times\mathbf{c})\). We can do this using components, and we find the result (exercise!): \[\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=(\mathbf{a}.\mathbf{c})\mathbf{b}-(\mathbf{a}.\mathbf{b})\mathbf{c}.\] Changing the roles of \(\mathbf{a}\) and \(\mathbf{c}\), we get: \[\mathbf{c}\times(\mathbf{b}\times\mathbf{a})=(\mathbf{c}.\mathbf{a})\mathbf{b}-(\mathbf{c}.\mathbf{b})\mathbf{a}.\] Using the anticommutativity of the cross product, we have \[\mathbf{c}\times(\mathbf{b}\times\mathbf{a})=-(\mathbf{b}\times\mathbf{a})\times\mathbf{c}=(\mathbf{a}\times\mathbf{b})\times\mathbf{c}.\] So we find that \[\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=(\mathbf{a}.\mathbf{c})\mathbf{b}-(\mathbf{a}.\mathbf{b})\mathbf{c}.\] Note that this is a different expression than \((\mathbf{a}\times\mathbf{b})\times\mathbf{c}\), so the brackets in the expression are essential.
Example 1.28 Given the vectors \(\mathbf{a}=\begin{pmatrix}1\\2\\1\end{pmatrix}\), \(\mathbf{b}=\begin{pmatrix}-1\\-4\\-5\end{pmatrix}\), \(\mathbf{c}=\begin{pmatrix}0\\-1\\-2\end{pmatrix}\), show that \[(\mathbf{a}\times\mathbf{b})\times\mathbf{c}\ne\mathbf{a}\times(\mathbf{b}\times\mathbf{c}).\]
Solution. In Example 1.27 we computed \(\mathbf{a}\times\mathbf{b}=\begin{pmatrix}-6\\4\\-2\end{pmatrix}\), and a similar computation gives \(\mathbf{b}\times\mathbf{c}=\begin{pmatrix}3\\-2\\1\end{pmatrix}\). Then \[\begin{eqnarray*} (\mathbf{a}\times\mathbf{b})\times\mathbf{c}&=&\begin{pmatrix}-6\\4\\-2\end{pmatrix}\times\begin{pmatrix}0\\-1\\-2\end{pmatrix}=\begin{pmatrix}-10\\-12\\6\end{pmatrix};\\ \mathbf{a}\times(\mathbf{b}\times\mathbf{c})&=&\begin{pmatrix}1\\2\\1\end{pmatrix}\times\begin{pmatrix}3\\-2\\1\end{pmatrix}=\begin{pmatrix}4\\2\\-8\end{pmatrix}.\end{eqnarray*}\] Therefore \((\mathbf{a}\times\mathbf{b})\times\mathbf{c}\ne\mathbf{a}\times(\mathbf{b}\times\mathbf{c})\).
Alternatively, we can compute \(\mathbf{a}.\mathbf{c}\) and \(\mathbf{a}.\mathbf{b}\): \[\begin{eqnarray*} \mathbf{a}.\mathbf{c}&=&1\times0+2\times(-1)+1\times(-2)=-4\\ \mathbf{a}.\mathbf{b}&=&1\times(-1)+2\times(-4)+1\times(-5)=-14\end{eqnarray*}\] and use \(\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=(\mathbf{a}.\mathbf{c})\mathbf{b}-(\mathbf{a}.\mathbf{b})\mathbf{c}\). Similarly, we can compute \(\mathbf{b}.\mathbf{c}=14\), and work out \((\mathbf{a}\times\mathbf{b})\times\mathbf{c}\) using \((\mathbf{a}.\mathbf{c})\mathbf{b}-(\mathbf{b}.\mathbf{c})\mathbf{a}\).