GeometriExtra material till kursen Algebra och Geometri

Med utvalda tentamensuppgifter

Augusti 2012, Roy Skjelnes

Institutionen for matmatik, KTH.

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In the first section we will recall the notion of vectors in the plane.The purpose is to go through these concepts in detail, before we con-tinue with the similar concepts for the 3-space. In addition to thetheory we provide some examples for the concepts in the plane. Thereare many exercises given in the end, exercises that earlier have oc-curred in the examination of the course “Algebra och Geometri”. Bydoing these exercises the student will in fact construct examples for thetheory in 3-space. We hope and believe that by actively doing theseexercises the theory will become transparent and the ideas will becomeaccessible.

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1. Euclidean Plane

1.1. The plane. The Euclidean plane R2 is the set consisting of all

ordered pairs of real numbers. An ordered pair we denote as

[ab

], and

the Euclidean plane is

R2 = { all ordered pairs

[ab

]with real numbers a, b}.

For typographic reasons we will often use a transposed notation, writing

(a, b) instead of

[ab

]. If (a, b) is an ordered pair, then the number a is

the first component, and the number b is the second component.

1.2. Vectors and points. An ordered pair (a, b) is an element in theplane. We will refer to elements in the plane as points, and also asvectors. These two names mean exactly the same thing, but are usuallypictured differently. When we refer to an ordered pair of numbers asa point P = (a, b) we usually draw a small cross in our picture of theplane at the position (a, b). Another natural way to visualise the sameelement (a, b) is to draw an arrow that starts in the zero vector (0, 0)and that ends in (a, b). We will refer to the arrow visualisation as avector, and then also usually use the symbol ~v = (a, b).

Example 1.3. In the example below we have pictured two elementsin the plane. One of these elements is the ordered pair (1, 2) whichwe have labelled as P , and pictured as a point. The other element isthe ordered pair (4, 1) which we have labelled as ~v, and pictured as anarrow.

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1.3.1. One should not get confused by these two different visualisa-tions, and arrow and a cross, of elements in the Euclidean plane. Thedefinition of the plane R2 is given in 1.1, and does not change. Theonly thing that changes is how we visualise elements in the plane; thisis helpful for our intuition.

1.4. Length. With the length of a vector ~v = (a, b) we simply meanthe distance from (0, 0) to (a, b). By the Pythagorean Theorem wehave that the length of ~v equals

√a2 + b2. The notation we use for the

length of a vector ~v is

|~v| = |[ab

]| =√a2 + b2.

Example 1.5. In the Example 1.3 we have the two elements P = (1, 2)and ~v = (4, 1) in the plane. The length of the vector ~v is |~v| =

√17.

We do not usually talk about the length of a point, but the length ofP considered as a vector is |P | =

√5.

1.6. Addition. We can add points P and Q in the plane, by addingtheir respective components. If P = (p1, p2), and Q = (q1, q2), then wewrite P +Q for the point

P +Q =

[p1 + q1p2 + q2

].

If we want to picture the addition with arrows we observe the following.Let ~v denote the arrow that starts in (0, 0) and ends at P = (p1, p2),and let ~w denote the arrow that starts in (0, 0) and ends at Q = (q1, q2).Then translate the arrow ~w, without changing direction or length, so itstarts at the point where the other arrow ~v ends. Then the compositionof the two arrows is an arrow that begins in (0, 0), and ends at the pointP +Q.

1.6.1. Before continue reading, pick your favourite two points in theplane. Add these points together. Therafter, represent the points asarrows, and see that by adding the arrows correctly, you do indeed getthe same additon. If you do not have favourite points, take P = (1, 2)and Q = (−2, 1).

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Example 1.7. In the example below we have taken two rather arbi-trary elements in the plane, denoted as P and Q. These two elementswe have pictured with arrows. Then we imagine translating the arrowrepresenting Q so that it starts in P . This translated arrow is denotedwith the rightmost dotted arrow in the picture. This dotted arrow nowpoints at the element P + Q. If we interchange the role of P and Q,and instead translate the arrow representing P so it starts in Q, thenwe get the leftmost dotted arrow. This arrow also points at the pointP +Q, picturing the fact that P +Q = Q+ P .

1.8. Subtraction. Having defined addition, we also get subtraction ofpoints in the plane. If P = (p1, p2) and Q = (q1, q2), then

Q− P =

[q1 − p1q2 − p2

].

We can also visualise the subtraction with our arrows. Again, let ~v bethe arrow representing P , and let ~w be the arrow representing Q, andlet ~u be the arrow representing Q− P . Then ~u is an arrow that startsin (0, 0) and ends at (q1−p1, q2−p2). Translate the arrow ~u such that itstarts at P . Then the composition of v followed with u gives an arrowthat starts in (0, 0), and ends at Q. We then have that ~v + ~u = ~w, orin terms of points P + (Q− P ) = Q.

1.8.1. Subtraction is more confusing than addition. Again, before con-tinuing reading, take your favourite points and subtract one from theother. Thereafter, do the same with the arrow representation.

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Example 1.9. In the picture below we imagine two elements in theplane, labelled as P and Q, that we have pictured with arrows. Theelement Q−P is also pictured with an arrow. If we translate the arrowrepresenting Q−P so that it starts at P , then we get the dotted arrow.The dotted arrow starts at P and ends at Q.

Example 1.10. Returning to the Example 1.3, and the two elementsP = (1, 2) and ~v = (4, 1) in the plane. We can add these two elements,and we have that P + ~v = (5, 3). We can subtract, and this givesP − ~v = (−3, 1) and ~v − P = (3,−1).

1.11. Notation. Let P and Q be two points in the plane. It is some-times convenient to write ~PQ for the vector Q− P . Remember that ifP and Q are given points, then we can draw the arrow that starts atP and ends in Q. When we translate that arrow so it starts in (0, 0),

then we have the vector ~PQ.

Example 1.12. Let P , Q and R be three arbitrary points in the plane.Then we have that

~PQ+ ~QR = ~PR.

Indeed, if we write what is hiding behind the notation, we are simplysaying that

~PQ+ ~QR = (Q− P ) + (R−Q) = R− P = ~PR.

1.13. Scalar multiplication. If P = (a, b) is a point, and r an num-ber, we multiply the point P with the number r by multiplying eachcomponent of P with r. We then get the point (ar, br), which we willdenote with rP . This operation is called scalar multiplication, and wewill in the sequel refer to the numbers r as scalars.

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If we visualise the vector P with an arrow ~v, then the geometricmeaning of r~v is to stretch the vector length with the factor r. Notethat if r < 0, then we turn the arrow around.

Example 1.14. In the picture below we have taken the point P =(2, 3) in the plane. We have then marked the points P , 2P , 0 · P ,12P , 3

2P and −P . The dotted line is a line through these points, and

indicates the scalar multiples tP when varying the scalars t.

Example 1.15. By now you should be able to do Exercise 4, andExercise 21 a.

2. Lines in the plane

We will in a sense continue the Example 1.14. The description ofthe line with vectors and a parameter is most probably new for you,so please pay attention and stay focused. Later on we will return tothe more familiar description of the lines using equations. When wefurther down the road take off to higher dimensions, the description oflines with vectors and parameters is the one that we will use.

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2.1. Lines and directional vectors. Let P = (a, b) be a point in theplane, and let ~v = (v1, v2) be a non-zero vector. We can add a pointwith a vector, and in particular we can add P + t~v, for any scalar t.The set L consisting of all points of the form P + t~v, with scalars tform a line. We have that

L = {[a+ tv1b+ tv2

]for all scalars t},

form a straight line in the plane. The line L goes through the point P ,and in the direction predicted by the vector ~v. We call ~v a directionalvector for the line L.

Example 2.2. In the picture below we have taken a point P and a(non-zero) vector ~v in the plane. Both P and ~v are pictured witharrows. Then we have marked the points P + 0 · ~v, P + ~v and P + 2~v.The dotted arrow is the line passing through these points, and describesthe line L consisting of points of the form P + t · ~v, with scalars t.

2.3. Parametric representation. Let L be a line in the plane, andlet P and Q be two different points on the line. Then we have that theline L can be described as

L = {P + t~v},with scalars t, and where ~v = Q − P . In particular a line L can bedescribed in many different ways. What one needs is a point P on theline and a (non-zero) directional vector ~v for the line L. A descriptionof the line L as the set {P + t~v for all scalars t} is called a parametricrepresentation of the line L.

Example 2.4. Consider the equation 2x − y = 1. The solution setto the equation is all ordered pairs (a, b) of real numbers such that2a − b = 1. We take for granted that you know that the solution set

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for the equation 2x− y = 1 form a line L in the plane. The line passesthrough the points P = (0,−1) and Q = (1, 1). A directional vector isthen ~v = Q − P = (1, 2). A parametric presentation of the line L isthen P + t~v, which reads

{ all ordered pairs

[t

−1 + 2t

]with scalars t}.

3. Orthogonal vectors

Material in this section many will encounter as abstract and difficult.The reader should keep in mind that we are still in the plane, andtherefore all concepts can be drawn and understood in very concreteterms.

3.1. Orthogonal vectors. We need first to establish what we meanwith vectors being orthogonal. Let us first consider an example. Con-sider the vectors (2, 3) and (−3, 2). If we draw these vectors as arrows,then we see that the two arrows are perpendicular. In fact, the vec-tor (2, 3) is perpendicular to (−3t, 2t) for all scalars t. We make thefollowing definition.

Definition 3.2. Let ~u = (a, b) and ~v = (c, d) be two, arbitrary, vectorsin the plane R2. Their scalar product is defined as the number

~u · ~v =

[ab

]·[cd

]= ac+ bd.

If the scalar product ~u · ~v = 0, then we say that the vectors ~u and ~vare orthogonal.

3.2.1. Note. We have now introduced both scalar multiplication andscalar product. These two concepts are only similar in names. Scalarmultiplication produces a vector from a scalar and a vector. Scalarproduct produces a number from two given vectors.

3.2.2. Note. With our definition we have that the zero vector ~0 = (0, 0)is orthogonal to any vector. Note also that the only vector ~v that isorthogonal to itself is the zero vector.

3.3. Orthogonal lines. Let L be a line in the plane, and ~n a vector.If a directional vector ~v of L is orthogonal to ~n, then any directionalvector of L is orthogonal to ~n. Therefore we say that two lines N andL in the plane are orthogonal if their respective directional vectors areorthogonal.

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3.4. Equations for lines in the plane. Let P = (p, q) be a point inthe plane, and let ~v = (a, b) be a non-zero vector. Then we can in anatural way consider two different lines. We can consider the line

L = {P + t~v for all scalars t},that goes through P with direction ~v. And we can consider the line Nthat goes through the point P , and is orthogonal to the line L. Thatis, we can consider the line N that goes through P and is orthogonalto ~v.

Example 3.5. In the picture below we have taken a point P and a(non-zero) vector ~v. Then we have drawn the line L passing throughthe point P , and having directional vector ~v. The line N we havedrawn is a picture of the line passing through P , but being orthogonalto L.

3.6. Normal vector. Let L be a line, and let ~n be a non-zero vector.If the vector ~n is orthogonal to the line L, then we say that ~n is anormal vector of L.

Example 3.7. In the Example 2.4 we looked at the line L given as thesolution set to the equation 2x − y = 1. We found that a parametricrepresentation for the line L was P + t~v, where P = (0,−1) and ~v =(1, 2), and where t is any scalar. A normal vector ~n for the line L isthen ~n = (2,−1). This is true since ~v · ~n = (1, 2) · (2,−1) = 0. Note

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also that the normal vector ~n = (2,−1) that we chose is in fact thecoefficients in front of the variables in the equation that defines theline.

Proposition 3.8. Let ~n = (a, b) be a non-zero vector. Then the lineL passing through a given point P = (p, q) and having ~n as a normalvector, is given as the solutions to the equation

ax+ by + c = 0 where c = −ap− bq.Proof. If X = (x, y) is a point, different than P , on the line L, thenX − P is a directional vector for L. We then have that the line L isthe set of points X = (x, y) such that the vector X − P is orthogonalto ~n. Using Definition 3.2 we can rewrite L as the set of point (x, y)such that

a(x− p) + b(y − q) = 0.

Recall now that the four numbers a, b, p, q were fixed, and therefore wehave now obtained an equation describing the line H. Having the pointP and the normal vector ~n = (a, b) fixed, we get that the solutions tothe equation

ax+ by + c = 0 where c = −ap− bq,describes the line through L and with normal vector ~n. Thus L is theline through P and orthogonal to ~n. �

Example 3.9. Let us return to the line L given as P + t~v, whereP = (0,−1) and where ~v = (1, 2). We want to find an equation whosesolution set is that line L. For that we need a point that the line passesthrough, and one such point we can find by letting t = 1. We then getthe point P1 = P + ~v = (1, 1). Then we need a normal vector to theline L. Since ~v = (1, 2) is a directional vector, we get a normal vector~n by letting ~n = (−4, 2). Note that the normal vector is not unique,and perhaps a more natural choice for the normal vector would havebeen (2,−1). Anyhow, the proposition above tells us that the solutionset to the equation

(3.9.1) −4x+ 2y + 2 = 0

is the line L. We have used that c = 2 = −(−4 ·1)−2 ·1. Note that theline L we are considering is precisely the same line that we consideredin Example 2.4. In that example the line was given as the solution setto the equation 2x − y = 1. If we multiply the equation 2x − y = 1with −2, and arrange the terms slightly we get the equation 3.9.1.

4. Projection and orthogonal decomposition

We will elaborate even further these new and slightly complicatedideas concerning orthogonal vectors. The ideas in this section will playan important role later on in the course. So please read, and readagain.

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4.1. Decomposition. Let L and M be two different lines in the plane,both passing through (0, 0). Let ~w be an arbitrary, but fixed vector. Itis clear that we can write

(4.1.1) ~w = ~wL + ~wM ,

where ~wL is a point on L, and ~wM is a point on M . This decompo-sition is furthermore unique, and we want to derive a formula for ~wL,and ~wM .

Example 4.2. In the picture below we have drawn two line passingthrough (0, 0) in the plane. These lines we have named L and M .Then we have taken a vector ~w, and represented that element as anarrow. We have thereafter drawn two elements ~wL and ~wM in theplane. The point ~wL is on the line L, and the point ~wM is on theline M . Furthermore, we have placed these points ~wL and ~wM suchthat their sum ~wL + ~wM = ~w. This fact is indicated with the arrowsrepresenting the points, and the dotted lines in the picture.

4.2.1. Note that if we know one of them, say ~wL, then we also knowthe other as

~wM = ~w − ~wL.

We are therefore satisfied with determining ~wL.

4.3. Orthogonal decomposition. We have our line L in the plane,passing through (0, 0). Let ~u be a directional vector for L. We thenhave that ~u = (a, b) is a non-zero vector, and that L = {t~u | scalars t}.

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Then the vector ~n = (−b, a) is also non-zero, and furthermore or-thogonal to L. The the line N = {t~n | scalars t} is normal to L. Anarbitrary vector ~w can then be written

~w = ~wL + ~wM ,

with orthogonal vectors ~wL and ~wN , and on the lines L and N , respec-tively. Such a decomposition is called an orthogonal decomposition.

4.4. Notation. Given a line L through (0, 0), and a vector ~w. Thenthe vector ~wL is called the projection of the vector ~w onto the line L.We denote this vector as

projL(~w).

Proposition 4.5. Let L denote a line through (0, 0), and with direc-tional vector ~u. Then, for any vector ~w, we have that

projL(~w) =~w · ~u|~u|2

~u.

Proof. Let N be the line through (0, 0), and orthogonal to L. We candecompose the vector ~w = projL(~w) + ~wN , where ~wN is a point on lineN . Since the point projL(~w) will be on the line L = {t~u | scalars t},we know that there is a scalar t0 such that

projL(~w) = t0~u.

We need only to determine the scalar t0. The trick is now to computethe scalar product ~w · ~u, which reads

~w · ~u = (t0~u+ ~wN) · ~u = t0(~v · ~v) + (~wN · ~u).

However, as N is orthogonal to L we have that ~wN · ~u = 0. Then theresult follows from the resulting equation

~w · ~u = t0~u · ~u.Indeed, we have that t0~u · ~u = t0(~u · ~u) is a product of two numbers t0and ~u · ~u. Since ~u was assumed to be a directional vector we have that~u is non-zero. Then the number ~u · ~u = |~u|2 is non-zero. Dividing theequation above with this non-zero number proves the statement. �

Corollary 4.6. Let L be a line through (0, 0), and let ~u = (a, b) be adirectional vector of L. For any vector ~w = (w1, w2) we have that

projL(~w) =aw1 + bw2

a2 + b2·[ab

].

Example 4.7. Let L denote the line given by the equation y−2x = 0.This is a line passing through (0, 0). A directional vector ~v for the lineL is ~v = (1, 2). Consider now the vector ~w = (4, 2). When we computethe projection of ~w onto the line L we find

projL(~w) =(1 · 4 + 2 · 2)

12 + 22·[12

]=

[85165

].

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A directional vector for the normal line N passing through (0, 0) is~n = (−2, 1). One gets that

projN(~w) =

[125−6

5

].

We have now a orthogonal decomposition w = projL(~w) + projN(~w) as

~w =

[42

]=

[85165

]+

[125−6

5

].

Corollary 4.8. Let L be a line given by the equation ax+ by + c = 0.Let P = (p, q) be a point. Then the distance from the point P to theline L is

|ap+ bq + c|√a2 + b2

.

Proof. By Proposition 3.8 we have that the vector ~n = (a, b) is orthog-onal to the line L. Take one point Q = (q1, q2) to be a point on theline L, that is aq1 + bq2 + c = 0. Then the line N = {t · ~n | scalars t}is a line through (0, 0), and orthogonal to L. We then have that thelength of the vector

~w = projN(P −Q)

is the distance from P to the line L. Now, using the projection formulaof Proposition 4.5, we obtain that

~w =(P −Q) · ~n|~n|2

~n.

Before computing the length of ~w we note that the length of r ·~n, wherer is a scalar, is |r||~n|. Thus, the length of ~w is

|~w| = |(P −Q) · ~n)||~n|2

|~n| = |(P −Q) · ~n||n|

.

Let us now compute the expression in the denominator. We have thatP −Q = (p− q1, q − q2). Therefore,

|(P −Q) · ~n| = |a(p− q1) + b(q − q2)| = |ap+ bq − aq1 − bq2|.

Then finally, using that the point Q is on the line, we have that thescalar c = −aq1 − bq2, proving the claim. �

4.9. Area of parallelograms. Two vectors ~u and ~v will naturallyform a s skew rectangle, a parallelogram. The corners will be the(0, 0), ~u,~v and ~u + ~v. One could argue that if the two vectors ~u and~v lie on the same line, these do not form a skew rectangle. We willhowever say that such a situation is a degenerate parallelogram, andin particular not exclude those possibilities.

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Proposition 4.10. Let ~u = (u1, u2) and ~v = (v1, v2) be two arbitraryvectors in R2. Then the area of the parallelogram determined by ~u and~v is the given by the scalar

|u1v2 − u2v1|.

Proof. The area of a parallelogram is given as the base times the height.If we let ~u form the base side in the parallelogram, then the length ~uwill be the lenght of the base. If ~u is the zero vector, then the areais zero and there is nothing to prove. We may therefore assume that~u is non-zero. Then we have the line N given as t(u2,−u1), scalars t,is orthogonal to the base of the parallelogram. In particular we havethat the length of the vector

projN(~v)

is the height of the parallelogram. The vector ~n = (u2,−u1) is a,non-zero, directional vector for N . Using the projection formula inProposition 4.5, we obtain

| projN(~v)| = 1

|~n|2|v1u2 − v2u1||~n|.

Then using that |~n| = |~u| proves the claim. �

4.11. Angles. If we are given two non-zero vectors ~u and ~v we candefine angle θ between these two vectors. We simply draw the twoarrows representing the vectors ~u and ~v, and then we let θ denote thesmallest, positive, angle between these two arrows, measured counterclockwise. The two arrows start both from (0, 0), and we want θ todenote the angle between these two arrows, not the angle around. Forcompleteness we say that the angle between the zero vector, and anyvector ~u is zero. Thus, for any two vectors ~u and ~v we have assignedan angle 0 ≤ θ ≤ 180.

Example 4.12. In the picture below we have drawn two elements ~vand ~w in the plane as two arrows. Their angle is denoted by θ. Eventhough you have seen this or similar picture many times before, pauseto recall what that the angle θ is: If we imagine a circle drawn withcentre in (0, 0), then we get an arc lying between the two vectors. Thelength of that arc divided by the length of our circle multiplied with360 is by definition the number θ. That number represents the arcdrawn in the picture below.

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Proposition 4.13. Let ~u = (u1, u2) and ~v = (v1, v2) be two vectorsin the plane, and let θ denote the angle between the vectors. Then wehave the identity

|~u| |~v| sin(θ) = |u1v2 − u2v1|.

Proof. By Proposition 4.10 that |u1v2 − u2v1| is the area of the paral-lelogram determined by the vectors ~u and ~v. The area is still the basetimes the height of the parallelogram. The lenght of the base is |~u|,and almost by the definition of the sinus function, we have that theheight is |~v| sin(θ). �

5. Three space

Having recalled the situation for the Euclidean plane, we show howthese concepts can be generalised to the Euclidean three space. We willsoup up these concepts quite quickly, and the reader (you) are assumedto familiarise with these concepts by doing the exercises given at theend.

5.1. The Euclidean three space R3 is the set of all ordered triples ofreal numbers,

R3 = {all ordered triples

xyz

with real numbers x, y and z}.

Again we will refer to the objects in R3 as vectors, or points, and wewill use the transpose notation (x, y, z) for elements in R3.

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5.2. Length. The length of a vector ~v = (a, b, c) is

|~v| =√a2 + b2 + c2,

and gives the distance from (0, 0, 0) to the point (a, b, c). We will from

now on denote the zero vector as ~0 = (0, 0, 0)

5.3. Lines in space. Let P = (p, q, r) be a point, and ~v = (a, b, c) bea non-zero vector. Then the set of points of the form

P + t~v =

p+ taq + tbr + tc

,with scalars t, form a line L in three space. This line is written asL = {P + t~v | scalars t}. The line L passes through the point P andhaving direction ~v.

5.4. Orthogonality. The right notion of orthogonality of vectors inthree space, turns out to be the following.

Definition 5.5. Let ~u = (a1, b1, c1) and ~v = (a2, b2, c2) be two arbitraryvectors in R3. Their scalar product is the number

~u · ~v = a1a2 + b1b2 + c1c2.

If their scalar product is zero, then we say that the vectors ~u and ~v areorthogonal.

5.5.1. Note. The length of a vector squared is the scalar product ofthe vector with itself,

|~u|2 = (~u · ~u).

Lemma 5.6. For any vectors ~u,~v in R3, and any scalars r and s wehave that

|r~u+ s~v|2 = r2|~u|2 + 2rs(~u · ~v)2 + s2|~v|2.

Proof. When proving this you need only to be patient and careful whenwriting out the expressions. �

5.7. Equations for planes. Let P = (p, q, r) be a point in threespace, and ~n = (a, b, c) be a non-zero vector. Then we have the lineN = {P + t~n | scalars t} passing through P and with directional vector~n. We can also consider the set H of vectors X = (x, y, z) in three spacesuch that X − P is orthogonal to ~n. The difference with the situationdiscussed in Section 3.4 is that in three space H is not a line, but aplane.

Proposition 5.8. Let ~n = (a, b, c) be a non-zero vector, and P =(p, q, r) be a point in R3. Then the plane H passing through P andhaving ~n as a normal vector is given by the equation

ax+ by + cz + d = 0 where d = −ap− bq − cr.

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Proof. If we write out what it means to be orthogonal, we get that His the set of points (x, y, z) such thatx− py − q

z − r

·abc

= 0.

In other words, the plane H that passes through P = (p, q, r) and thatis orthogonal to ~n = (a, b, c) is given as the solutions to the equation

a(x− p) + b(y − q) + c(z − r) = 0.

This means that ax+ by + cz + d = 0 where d = −ap− bq − cr. �

5.9. Projection and orthogonal decomposition. We will pose thesame problem in three space as we did for the Euclidean plane inSection 4. Let ~u be a non-zero vector in R3, and consider the lineL = {t~v | scalars t}. Any vector ~w in R3 can in a unique way bewritten as a sum

~w = ProjL(~w) + ~wH ,

with ProjL(~w) a point on L, and with the vector ~wH orthogonal to L.The vector ProjL(~w) is called the (orthogonal) projection of ~w onto theline L.

5.9.1. Uniqueness. When we looked at decomposition of a vector ~w inthe plane, the uniqueness of was clear. Uniqueness is not so immediatein three space, but here it is. Assume that we can write

~w = ~wL + ~wH = ~uL + ~uH

in two ways, with ~wL and ~uL on the line L, and with ~wH and ~uHorthogonal to L. It suffices to show that ~wL = ~uL. If both vectors arezero, there is nothing to prove. We can therefore assume that at leastone of them is non-zero, say ~wL. We take the fixed vector ~w and formits scalar product with ~wL. We then obtain

~w · ~wL = ~wL · ~wL + 0 = ~uL · ~wL + 0.

Thus ~wL · ~wL = ~uL · ~wL. Let ~v be directional vector for the line L. Sinceboth points are assumed to be on the line L we know that there arescalars s and t such that s~v = ~wL and t~v = ~vL. We need to see thats = t. The equality ~wL · ~wL = ~uL · ~wL becomes now s2|~v|2 = ts|~v|2. Thedirectional vector ~v is non-zero, thus the equality becomes s2 = ts. Wedid assume that ~wL was non-zero, hence s 6= 0. Then we get that s = tand we are done.

Proposition 5.10. Let L be a line through ~0, and let ~v be a directionvector for L. For any vector ~w in R3 we have that

ProjL(~w) =~v · ~w|~v|2

~v.

17

Proof. The proof is similar to the proof of Proposition 4.5, and is anexcellent exercise for the reader. �

Proposition 5.11. Let H be a plane in three space, given by the equa-tion ax + by + cz + d = 0, and let P = (p, q, r) be a point. Then thedistance from P to the plane is the number

|ap+ bq + cr + d|√a2 + b2 + c2

.

Proof. The proof is an application of the projection formula given inProposition 5.10, similar to the application given in Proposition 3.8.The reader is encouraged to write out this proof. �

5.12. Exercises. Relevant exercises are 1, 7, 10a, 13, 14

6. Vector product

In three space we will define the product of vectors. This construc-tion will be unique for R3.

Definition 6.1. Let ~u = (u1, u2, u3) and ~v = (v1, v2, v3) be two vectorsin R3. With ~u× ~v we mean the vectoru1u2

u3

×v1v2v3

:=

u2v3 − v2u3−u1v3 + v1u3u1v2 − v1u2

.The vector ~u× ~v is called the vector product of ~u with ~v.

Lemma 6.2. Let ~u and ~v be two vectors in R3. Then we have

(1) ~u× ~v = −~v × ~u.(2) The vectors ~u and ~v are both orthogonal to ~u× ~v.(3) We have |~u× ~v|2 = |~u|2|~v|2 − (~u · ~v)2.

Proof. These facts the reader can verify. The first and third statementone proves by simply writing out the expressions, and thereafter com-paring these. To prove the second, one will have to show that thenumber ~u · (~u× ~v) and the number ~v · (~u× ~v) both are zero. �

Example 6.3. There is a unique plane H passing through the threepoints P = (1, 2, 1), Q = (1, 3, 2) and R = (2,−2, 3). We want tohave an equation describing H. For this we need a point that H passesthrough, and a normal vector to H.

First we chose a point that the plane H passes through. We takeP = (1, 2, 1) as this point. Consider the vector ~u = Q − P = (0, 1, 1).Then the plane H contains the line P + t~u, for scalars t. And similarlywe have that the plane contains the line P + t~v, where t is scalars, andwhere ~v = R−P = (1,−4, 2). We then have two different lines passingthrough P , both contained in the plane H. It is then clear that a vector

18

~n that is non-zero, and orthogonal to both ~u and ~v is a normal vectorfor H. We therefore look at the vector

~u× ~v =

011

× 1−42

=

61−1

.We then have that an equation for H is 6x + y − z + d = 0. Theconstant d we determine by plugging in the coordinates for P . Thepoint P = (1, 2, 1) is a point on H, so

6 + 2− 1 + d = 0,

which means that d = −7. What happens if we in the last step use thepoint Q = (1, 3, 2), which also lies on H, instead of using the point P?

6.4. Exercises. Relevant exercises are 2, 3, 4, 9, 11, 15, 16, 17,18, 19, 20, 22, 23, 24, 25, 26, 28, 30, 31, 32, 34, 35, 39, 40.

6.5. Parallelogram. If we are given two vectors ~u and ~v in R3, thenthese two vectors form a parallelogram in a natural way. The fourcorners of the parallelogram are the points consisting of ~0, ~u, ~v and~u+ ~v.

Proposition 6.6. Let ~u and ~w be two vectors in R3. Then the area ofthe parallelogram they determine equals the length of the vector ~u× ~v.

Proof. We need to compute the area, which is the base |~u| times the

height of the parallelogram. Let L be the line through~0 with directionalvector ~u. Then the length of the vector

~h = ~v − projL(~v)

equals the height of the parallelogram. Using the projection formulagiven in Proposition 5.10, we obtain

~h =1

|~u|2(|~u|2~v − (~u · ~v)~u).

Instead of computing the length of ~h we will compute the square of the

length, that is |~h|2. Using Lemma 5.6 we get that

|~h|2 =1

|~u|4(|~u|4|~v|2 + (~u · ~v)2|~u2 − 2|~u|2(~u · ~v)(~u · ~v)

=1

|u|2(|~u2|~v|2 − 2(~u · ~v)2).

Using Lemma 6.2 we then get that |~h|2 = 1|~u|2 |~u×~v|

2. Thus the square

of the area of the parallelogram is |~u|2 |~h|2 = |~u × ~v|2, and we haveproved the claim. �

Corollary 6.7. The vector ~u × ~v is zero if and only if the vectors ~uand ~v lie on the same line L passing through ~0.

19

Proof. The only vector of length zero is the zero vector. Therefore,by the proposition, we have that ~u × ~v is zero if and only the area ofthe parallelogram determined by ~u and ~v is zero. The only possibil-ity for such a parallelogram is when the vectors ~u and ~v determine adegenerated parallelogram, that is they lie on the same line through~0. �

6.8. Cubes. Three vectors ~u,~v and ~w R3 determine a skew cube inR3. The corners are ~0, ~u,~v, ~w and ~u + ~v, ~u + ~w,~v + ~w and finally~u+ ~v + ~w.

Proposition 6.9. The volume of the skew cube determined by threevectors ~u,~v and ~w equals the absolute value of the number (~u× ~v) · ~w.

Proof. The volume of the skew cube is the area of its base times theheight. We know from Proposition 6.6 that the area of the base is|~u × ~v|. If the base has zero area, there is nothing to prove. We maytherefore assume that the vectors ~u and ~v determine a non-degenerateparallelogram. Then ~n = ~u × ~v is a non-zero vector. Let N = {t~n |scalars t} be the line through ~0 with directional vector ~n. Then Nis orthogonal to the base, and we have that the length of the vectorprojN(~w) is the height of the cube. Using the projection formula ofProposition 5.10 we obtain

| projN(~w)| = |(~u× ~v) · ~w|~u× ~v|2

||~u× ~v| |~u× ~v| = |(~u× ~v) · ~w|.

This proves our claim. �

6.10. Exercises. Relevant exercises are 5, 6, 10b, 21, 27, 33, 36,37, 38, 41