线性代数英文ppt1

Chapter 3 Vector Spaces1.Suppose that(6204),(3157)αβ=-=-,find vector v , let βα43=-v2.Determine whether the vector β can be a linear combination of the vectors123,,ααα，if it can be ，write down the expressions 。

（1）123(83125),(1305),(2073),(4126)βααα=--=-=-=--3. Prove that any vector 1234()b b b b β=can be a linear combination of the vectors 1234(1000),(1100)(1110),(1111)αααα====。

4.Determine whether the following vectors are linearly independent 。

（1）123(1203),(2510),(3412).ααα=-=-=。

（2）1234(3425),(2503),(5012),(3335)αααα=-=--=-=-。

（3）232323123(1),(1),(1)aa ab b bc c c ααα===。

5. If 123,,ααα are linearly independent ，prove that whether 12233123,4,5αααααα+++ are linearly independent 。

6. Suppose that 1234,,,αααα are linearly independent ，then determine whether 12233441,,,αααααααα++++ are linearly independent, show the reason.7. Prove that in n R ，if 12,,,n ααα are linearly independent ，then any vector n R ∈β can be a linear combination of12,,,n ααα .8. If 1234,,,αααα are linearly dependent ，and any three of then are linearly independent, prove that there must have nonzero numbers 1234,,,k k k k such that112233440k k k k αααα+++=。

Chapter 1 Matrices and Systems of EquationsLinear systems arise in applications to such areas as engineering, physics, electronics, business, economics, sociology（社会学）, ecology （生态学）, demography(人口统计学), and genetics（遗传学）, etc. §1. Systems of Linear EquationsNew words and phrases in this section:Linear equation 线性方程Linear system，System of linear equations 线性方程组Unknown 未知量Consistent 相容的Consistence 相容性Inconsistent不相容的Inconsistence 不相容性Solution 解Solution set 解集Equivalent 等价的Equivalence 等价性Equivalent system 等价方程组Strict triangular system 严格上三角方程组Strict triangular form 严格上三角形式Back Substitution 回代法Matrix 矩阵Coefficient matrix 系数矩阵Augmented matrix 增广矩阵Pivot element 主元Pivotal row 主行Echelon form 阶梯形1.1 DefinitionsA linear equation (线性方程) in n unknowns（未知量）is1122...n na x a x a x b+++=A linear system of m equations in n unknowns is11112211211222221122...... .........n n n n m m m n n m a x a x a x b a x a x a x b a x a x a x b+++=⎧⎪+++=⎪⎨⎪⎪+++=⎩ This is called a m x n (read as m by n) system.A solution to an m x n system is an ordered n-tuple of numbers （n 元数组）12(,,...,)n x x x that satisfies all the equations.A system is said to be inconsistent （不相容的） if the system has no solutions.A system is said to be consistent （相容的）if the system has at least one solution.The set of all solutions to a linear system is called the solution set（解集）of the linear system.1.2 Geometric Interpretations of 2x2 Systems11112212112222a x a xb a x a x b +=⎧⎨+=⎩ Each equation can be represented graphically as a line in the plane. The ordered pair 12(,)x x will be a solution if and only if it lies on bothlines.In the plane, the possible relative positions are(1) two lines intersect at exactly a point; (The solution set has exactly one element)(2)two lines are parallel; (The solution set is empty)(3)two lines coincide. (The solution set has infinitely manyelements)The situation is the same for mxn systems. An mxn system may not be consistent. If it is consistent, it must either have exactly one solution or infinitely many solutions. These are only possibilities.Of more immediate concerns is the problem of finding all solutions to a given system.1.3 Equivalent systemsTwo systems of equations involving the same variables are said to be equivalent（等价的，同解的）if they have the same solution set.To find the solution set of a system, we usually use operations to reduce the original system to a simpler equivalent system.It is clear that the following three operations do not change the solution set of a system.(1)Interchange the order in which two equations of a system arewritten;(2)Multiply through one equation of a system by a nonzero realnumber;(3)Add a multiple of one equation to another equation. (subtracta multiple of one equation from another one)Remark: The three operations above are very important in dealing with linear systems. They coincide with the three row operations of matrices. Ask a student about the proof.1.4 n x n systemsIf an nxn system has exactly one solution, then operation 1 and 3 can be used to obtain an equivalent “strictly triangular system ”A system is said to be in strict triangular form (严格三角形) if in the k-th equation the coefficients of the first k-1 variables are all zero and the coefficient ofkx is nonzero. (k=1, 2, …,n)An example of a system in strict triangular form:123233331 2 24x x x x x x ++=⎧⎪-=⎨⎪=⎩Any nxn strictly triangular system can be solved by back substitution (回代法).(Note: A phrase: “substitute 3 for x ” == “replace x by 3”)In general, given a system of linear equations in n unknowns, we will use operation I and III to try to obtain an equivalent system that is strictly triangular.We can associate with a linear system an mxn array of numbers whose entries are coefficient of theix ’s. we will refer to this array as thecoefficient matrix (系数矩阵) of the system.111212122212.....................n nm m m n a a a a a a a a a ⎛⎫⎪ ⎪ ⎪ ⎪⎝⎭A matrix （矩阵） is a rectangular array of numbersIf we attach to the coefficient matrix an additional column whose entries are the numbers on the right-hand side of the system, we obtain the new matrix11121121222212n n s m m m na a ab a a a b b a a a ⎛⎫ ⎪ ⎪ ⎪⎝⎭We refer to this new matrix as the augmented matrix （增广矩阵） of a linear system.The system can be solved by performing operations on the augmented matrix. i x ’s are placeholders that can be omitted until the endof computation.Corresponding to the three operations used to obtain equivalent systems, the following row operation may be applied to the augmented matrix.1.5 Elementary row operationsThere are three elementary row operations:(1)Interchange two rows;(2)Multiply a row by a nonzero number;(3)Replace a row by its sum with a multiple of another row.Remark: The importance of these three operations is that they do not change the solution set of a linear system and may reduce a linear system to a simpler form.An example is given here to illustrate how to perform row operations on a matrix.★Example:The procedure for applying the three elementary row operations:Step 1: Choose a pivot element (主元)(nonzero) from among the entries in the first column. The row containing the pivotnumber is called a pivotal row（主行）. We interchange therows (if necessary) so that the pivotal row is the new firstrow.Multiples of the pivotal row are then subtracted form each of the remaining n-1 rows so as to obtain 0’s in the firstentries of rows 2 through n.Step2: Choose a pivot element from the nonzero entries in column 2, rows 2 through n of the matrix. The row containing thepivot element is then interchanged with the second row ( ifnecessary) of the matrix and is used as the new pivotal row.Multiples of the pivotal row are then subtracted form eachof the remaining n-2 rows so as to eliminate all entries belowthe pivot element in the second column.Step 3: The same procedure is repeated for columns 3 through n-1.Note that at the second step, row 1 and column 1 remain unchanged, at the third step, the first two rows and first two columns remain unchanged, and so on.At each step, the overall dimensions of the system are effectively reduced by 1. (The number of equations and the number of unknowns all decrease by 1.)If the elimination process can be carried out as described, we will arrive at an equivalent strictly triangular system after n-1 steps.However, the procedure will break down if all possible choices for a pivot element are all zero. When this happens, the alternative is to reduce the system to certain special echelon form（梯形矩阵）. AssignmentStudents should be able to do all problems.Hand-in problems are: # 7--#11§2. Row Echelon FormNew words and phrases:Row echelon form 行阶梯形Reduced echelon form 简化阶梯形 Lead variable 首变量 Free variable 自由变量Gaussian elimination 高斯消元Gaussian-Jordan reduction. 高斯-若当消元 Overdetermined system 超定方程组 Underdetermined systemHomogeneous system 齐次方程组 Trivial solution 平凡解2.1 Examples and DefinitionIn this section, we discuss how to use elementary row operations to solve mxn systems.Use an example to illustrate the idea.★ Example : Example 1 on page 13. Consider a system represented by the augmented matrix111111110011220031001131112241⎛⎫ ⎪--- ⎪ ⎪-- ⎪- ⎪ ⎪⎝⎭ 111111001120002253001131001130⎛⎫⎪ ⎪ ⎪ ⎪- ⎪ ⎪⎝⎭………..(The details will given in class)We see that at this stage the reduction to strict triangular form breaks down. Since our goal is to simplify the system as much as possible, we move over to the third column. From the example above, we see that the coefficient matrix that we end up with is not in strict triangular form,it is in staircase or echelon form (梯形矩阵).111111001120000013000004003⎛⎫ ⎪ ⎪ ⎪ ⎪- ⎪ ⎪-⎝⎭The equations represented by the last two rows are:12345345512=0 2=3 0=4 03x x x x x x x x x ++++=⎧⎪++⎪⎪⎨⎪-⎪=-⎪⎩Since there are no 5-tuples that could possibly satisfy these equations, the system is inconsistent.Change the system above to a consistent system.111111110011220031001133112244⎛⎫ ⎪--- ⎪ ⎪-- ⎪ ⎪ ⎪⎝⎭ 111111001120000013000000000⎛⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭The last two equations of the reduced system will be satisfied for any 5-tuple. Thus the solution set will be the set of all 5-tuples satisfying the first 3 equations.The variables corresponding to the first nonzero element in each row of the augment matrix will be referred to as lead variable .（首变量） The remaining variables corresponding to the columns skipped in the reduction process will be referred to as free variables （自由变量）.If we transfer the free variables over to the right-hand side in the above system, then we obtain the system:1352435451 2 3x x x x x x x x x ++=--⎧⎪+=-⎨⎪=⎩which is strictly triangular in the unknown 1x 3x 5x . Thus for each pairof values assigned to 2xand4x , there will be a unique solution.★Definition: A matrix is said to be in row echelon form (i) If the first nonzero entry in each nonzero row is 1.(ii)If row k does not consist entirely of zeros, the number of leading zero entries in row k+1 is greater than the number of leading zero entries in row k.(iii) If there are rows whose entries are all zero, they are below therows having nonzero entries.★Definition : The process of using row operations I, II and III to transform a linear system into one whose augmented matrix is in row echelon form is called Gaussian elimination （高斯消元法）.Note that row operation II is necessary in order to scale the rows so that the lead coefficients are all 1.It is clear that if the row echelon form of the augmented matrix contains a row of the form (), the system is inconsistent.000|1Otherwise, the system will be consistent.If the system is consistent and the nonzero rows of the row echelon form of the matrix form a strictly triangular system (the number of nonzero rows<the number of unknowns), the system will have a unique solution. If the number of nonzero rows<the number of unknowns, then the system has infinitely many solutions. (There must be at least one free variable. We can assign the free variables arbitrary values and solve for the lead variables.)2.2 Overdetermined SystemsA linear system is said to be overdetermined if there are more equations than unknowns.2.3 Underdetermined SystemsA system of m linear equations in n unknowns is said to be underdetermined if there are fewer equations than unknowns (m<n). It is impossible for an underdetermined system to have only one solution.In the case where the row echelon form of a consistent system has free variables, it is convenient to continue the elimination process until all the entries above each lead 1 have been eliminated. The resulting reduced matrix is said to be in reduced row echelon form. For instance,111111001120000013000000000⎛⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭ 110004001106000013000000000⎛⎫⎪- ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭Put the free variables on the right-hand side, it follows that12345463x x x x x =-=--=Thus for any real numbersαandβ, the 5-tuple()463ααββ---is a solution.Thus all ordered 5-tuple of the form ()463ααββ--- aresolutions to the system.2.4 Reduced Row Echelon Form★Definition : A matrix is said to be in reduced row echelon form if :(i)the matrix is in row echelon form.(ii) The first nonzero entry in each row is the only nonzero entry in its column.The process of using elementary row operations to transform a matrix into reduced echelon form is called Gaussian-Jordan reduction.The procedure for solving a linear system:(i) Write down the augmented matrix associated to the system; (ii) Perform elementary row operations to reduce the augmented matrix into a row echelon form;(iii) If the system if consistent, reduce the row echelon form into areduced row echelon form. (iv) Write the solution in an n-tuple formRemark: Make sure that the students know the difference between the row echelon form and the reduced echelon form.Example 6 on page 18: Use Gauss-Jordan reduction to solve the system:1234123412343030220x x x x x x x x x x x x -+-+=⎧⎪+--=⎨⎪---=⎩The details of the solution will be given in class.2.5 Homogeneous SystemsA system of linear equations is said to be homogeneous if theconstants on the right-hand side are all zero.Homogeneous systems are always consistent since it has a trivial solution. If a homogeneous system has a unique solution, it must be the trivial solution.In the case that m<n (an underdetermined system), there will always free variables and, consequently, additional nontrivial solution.Theorem 1.2.1 An mxn homogeneous system of linear equations has a nontrivial solution if m<n.Proof A homogeneous system is always consistent. The row echelon form of the augmented matrix can have at most m nonzero rows. Thus there are at most m lead variables. There must be some free variable. The free variables can be assigned arbitrary values. For each assignment of values to the free variables, there is a solution to the system.AssignmentStudents should be able to do all problems except 17, 18, 20.Hand-in problems are 9, 10, 16,Select one problem from 14 and 19.§3. Matrix AlgebraNew words and phrases:Algebra 代数Scalar 数量，标量Scalar multiplication 数乘 Real number 实数 Complex number 复数 V ector 向量Row vector 行向量 Column vector 列向量Euclidean n-space n 维欧氏空间 Linear combination 线性组合 Zero matrix 零矩阵Identity matrix 单位矩阵 Diagonal matrix 对角矩阵 Triangular matrix 三角矩阵Upper triangular matrix 上三角矩阵 Lower triangular matrix 下三角矩阵 Transpose of a matrix 矩阵的转置(Multiplicative ) Inverse of a matrix 矩阵的逆 Singular matrix 奇异矩阵 Singularity 奇异性Nonsingular matrix 非奇异矩阵 Nonsingularity 非奇异性The term scalar (标量，数量) is referred to as a real number (实数) or a complex number （复数）. Matrix notationAn mxn matrix, a rectangular array of mn numbers.111212122212.....................n nm m m n a a a a a a a a a ⎛⎫⎪ ⎪ ⎪ ⎪⎝⎭()ij A a =3.1 VectorsMatrices that have only one row or one column are of special interest since they are used to represent solutions to linear systems.We will refer to an ordered n-tuple of real numbers as a vector （向量）.If an n-tuple is represented in terms of a 1xn matrix, then we will refer to it as a row vector . Alternatively, if the n-tuple is represented by an nx1 matrix, then we will refer to it as a column vector . In this course, we represent a vector as a column vector.The set of all nx1 matrices of real number is called Euclidean n-space (n 维欧氏空间) and is usually denoted by nR.Given a mxn matrix A, it is often necessary to refer to a particular row or column. The matrix A can be represented in terms of either its column vectors or its row vectors.12(a ,a ,,a )n A = ora (1,:)a(2,:)a(,:)A m ⎛⎫ ⎪⎪= ⎪ ⎪⎝⎭3.2 EqualityFor two matrices to be equal, they must have the same dimensions and their corresponding entries must agree★Definition : Two mxn matrices A and B are said to be equal ifij ij a b =for each ordered pair (i, j)3.3 Scalar MultiplicationIf A is a matrix,αis a scalar, thenαA is the mxn matrix formed by multiplying each of the entries of A byα.★Definition : If A is an mxn matrix, αis a scalar, thenαA is themxn matrix whose (i, j) is ij a αfor each ordered pair (i, j) .3.4 Matrix AdditionTwo matrices with the same dimensions can be added by adding their corresponding entries.★Definition : If A and B are both mxn matrices, then the sum A+B is the mxn matrix whose (i,j) entry isij ija b + for each ordered pair (i, j).An mxn zero matrix (零矩阵) is a matrix whose entries are all zero. It acts as an additive identity on the set of all mxn matrices.A+O=O+A=AThe additive of A is (-1)A since A+(-1)A=O=(-1)A+A.A-B=A+(-1)B-A=(-1)A3.5 Matrix Multiplication and Linear Systems3.5.1 MotivationsRepresent a linear system as a matrix equationWe have yet to defined the most important operation, the multiplications of two matrices. A 1x1 system can be writtena xb =A scalar can be treated as a 1x1 matrix. Our goal is to generalize the equation above so that we can represent an mxn system by a single equation.A X B=Case 1: 1xn systems 1122... n n a x a x a x b +++=If we set()12n A a a a =and12n x x X x ⎛⎫ ⎪⎪= ⎪ ⎪⎝⎭, and define1122...n n AX a x a x a x =+++Then the equation can be written as A X b =。

Math1229A/BUnit3: Lines and Planes(text reference:Section1.3)c V.Olds201030Unit33Lines and PlanesLines inℜ2You are already familiar with equations of lines.In previous courses you will have written equations of lines in slope-point form,in slope-intercept form,and probably also in standard form for a line inℜ2.Recall that:y−y1=m(x−x1)is the slope-point form equation of the line through point(x1,y1)with slope m y=mx+b is the slope-intercept form equation of the line with slope m and y-intercept b ax+by=c is the standard form equation which either of the others can be rearranged to In this course,we don’t use the slope-point or slope-intercept forms of equations of lines.Instead, we use various other,vector-based,forms of equations.But we do still use the standard form.You already know that given any2distinct points,whether inℜ2or inℜ3,there is exactly one line which passes through both points.Suppose we have2points,P and Q.Letℓbe the line that passes through these two points.Both point P and point Q lie on lineℓ,and so do all the points between them.In fact,the directed line segment−−→P Q lies on lineℓ.When this directed line segment is translated to the origin,the resulting vector most likely doesn’t lie on lineℓ(unless the originhappens to lie on lineℓ),but if not,it does lie on a line which is parallel to lineℓ.It lies on the line parallel toℓwhich passes through the origin.So this vector does give us some information about the line.(Similar to the information given by knowing the slope of a line inℜ2,although it’s not quite the same information.)If a vector v lies on a particular line,or on a line parallel to that line,we say that v is parallel to, or is collinear with that line.And we call v a direction vector for the line.Not that the line actually has a direction associated with it.It doesn’t.It extends in both directions,but has no particular “forwards along the line”or“backwards along the line”associated with it.So don’t read too much meaning into the term direction vector.If v is a direction vector for lineℓ,then so is− v.And so is every other scalar multiple of v,except for0 v.Because of course0 v= 0which has no direction information.But every other scalar multiple of v starts at the origin,and goes either the same or the opposite direction as v and therefore also lies on the line parallel toℓwhich passes through the origin.So any such vector would be considered a direction vector for lineℓ.Definition:Any non-zero vector which is parallel to a lineℓis called a direction vectorfor lineℓ.For instance,consider the line x+y=2.The points(1,1),(2,0),(0,2),(−1,3),(3,−1),(−2,4), (4,−2),etc.,all lie on this line.So do(1/2,3/2)and(3/2,1/2)and infinitely many other points. Pick any2of these points,andfind the vector which is the translation to the origin of the directed line segment between them,and you have a direction vector for the line.And we know that for any points P and Q,letting p=−−→OP denote the vector from the origin to point P and q=−−→OQ denote the vector from the origin to point Q,the vector v= q− p is the translation of directed line segment −−→P Q to the origin.So for instance for points P(1,1)and Q(0,2),we have p=(1,1)and q=(0,2), and we see that v= q− p=(0,2)−(1,1)=(−1,1)is a direction vector for the line x+y=2.And other choices of P and Q give other direction vectors which are scalar multiples of this one.(Go ahead,pick some other points,and see what vectors you get.)Unit331 Point-Parallel Form32Unit3 Parametric EquationsUnit333 Example3.3.Write a point-parallel form equation for the line with parametric equationsx=1+5ty=2Solution:We use the x equation tofind thefirst components for our point-parallel equation,and the y equation tofind the second components.We need to recognize that in each equation,the number on the right hand side that isn’t multiplied by t is the coordinate of the known point,P,and that the number that is multiplied by t is the component of the direction vector, v.So from thefirst equation,i.e. the x-equation,we see that p1=1and v1=5.And from the second equation,since there’s no t multiplying it,the2must be p2.So where’s the t?It’s invisible,which means it must have a0 multiplier.That is,v2=0.So we have the point P(1,2)and the direction vector v=(5,0),which when we put it in the form x(t)= p+t v gives the point-parallel form equationx(t)=(1,2)+t(5,0)Two-Point Form-34Unit 3But nothing that we did here really required being in between P and Q .We could do something similar for any point on the line containing P and Q .The only difference is that t would no longer necessarily be between 0and 1.That is,for any point X on the line containing the points P and Q ,we could travel from the origin to point P ,and then travel some scalar multiple of the vector u to end up at the point X .For instance,consider the diagram shown below.As before,we have x = p +t u ,but now t is bigger than 1.Or if we needed to go the other direction along the line,from P ,then t would be negative.6X X X X X X X X X X X X X X X X X X X X ¡¡¡¡¡¡!P p Q q ¨¨¨¨¨¨¨¨¨B X x And so for any point X on the line,we havex = p +t u = p +t ( q − p )= p +t q −t p =(1−t ) p +t qfor some value t .That is,we can express the line containing two points P and Q as the line containing all points X (x,y )such that x =(1−t ) p +t q for some value t .And so we have another form of equation for the line.We call this the two-point form,and as before,we write x (t )instead of just x .Definition:The two-point form of equation for the line through points P and Q is:x (t )=(1−t ) p +t qExample 3.4.Write equations in two-point form for each of the lines in Example 3.1.Solution:(a)The line here is the line through the points P (3,1)and Q (0,6),so we have p =(3,1)andq =(0,6)and we get the two-point form x (t )=(1−t )(3,1)+t (0,6)(b)This time,we have the line through P (1,2)with direction vector v =(2,−1).We don’t know two points on the line,so we need to find a second point.We saw in Example 3.1(b)that a point-parallel form equation for this line is x (t )=(1,2)+t (2,−1).We can choose any value of t other than 0to get another point on the line.(Notice:We don’t want to use t =0,because that will just give us the point we already know.But any other value of t will do.)For instance,using t =1we have x (1)=(1,2)+1(2,−1)=(1,2)+(2,−1)=(3,1),so we see that Q (3,1)is another point on the same line.Now that we know two points on the line,we can find a two-point form equation.Notice,though,that since we have already been using t as the parameter for the point-parallel form equation,we should use a different name for the parameter in the two-point form equation.(Especially since we gave t a specific value.We wouldn’t want to get confused and think the parameter in the two-point form equation was supposed to have that value too.)Notice also that it doesn’t matter in the least what letter we use to represent the parameter (which is just a scalar multiplier).So we can use s instead.We get:x (s )=(1−s )(1,2)+s (3,1)Unit335 (c)This time,we have the line through the origin with direction vector(0,1).We know that the point(0,0)is on the line,and clearly the point(0,1)is also on the line(because the vector(0,1)is on the line,since the line does pass through the origin).So a two-point form equation for this line isx(t)=(1−t)(0,0)+t(0,1)Point-Normal Form36Unit3 Example3.6.(a)Find an equation in point-normal form for the line x(t)=(0,1)+t(2,−1).(b)Write an equation in point-parallel form for the line from Example3.5.(c)Write a point-normal form equation for the line with parametric equationsx=3+t and y=2t−4Solution:(a)We have x(t)=(0,1)+t(2,−1),which we recognize as a point-parallel form equation for the line through point(0,1)parallel to the vector(2,−1).Since the vector(2,−1)is parallel to the line,then the vector(1,2),obtained by switching the components and changing one of the signs, is perpendicular to the line.That is, n=(1,2)is a normal for this line.So a point-normal form equation for the line is(1,2)•( x−(0,1))=0(b)In Example3.5we found the point-normal form equation(−1,1)•( x−(1,2))=0for a particular line.Since(−1,1)is a normal for this line,and the vector(1,1)is orthogonal to(−1,1),then the vector(1,1)is parallel to the line,i.e.is a direction vector for the line.And of course(1,2)is a point on the line.So a point-parallel form equation for this line isx(t)=(1,2)+t(1,1)(c)From the parametric equations of the line we can identify both a point on the line and a direction vector for the line.Remember,the multiplier on t is the component of the direction vector,while the number without a t is the coordinate of the known point.Keeping this in mind allows us to correctly identify both the known point and the direction vector from the parametric equations, even when they look a bit different than we expect.Here,the parametric equations are given asx=3+ty=2t−4We’re more accustomed to seeing the form we have in the x equation.The form in the y equation, with the t term coming before the non-t term,is different.This is just done to avoid having a “leading negative”.Equations look less tidy when thefirst thing on one side of the equation is a negative sign,so mathematicians often avoid writing things that way.That is,the given parametric equations are just a tidier form ofx=3+ty=−4+2tIn this form we see that the corresponding point-parallel form equation is x(t)=(3,−4)+t(1,2). So(1,2)is a direction vector for the line and therefore(2,−1)is a normal for the line.Thus we can write a point-normal form equation as(2,−1)•( x−(3,−4))=0Unit337 Standard Form38Unit3 Example3.8.Write a point-normal form equation for the line x−2y=5.Solution:We use the coefficients of x and y as the components of a normal vector for the line.Of course, x−2y=1x+(−2)y,so the coefficients are1and−2.That is,we get n=(1,−2)as a normal vector for the line.Now,we just need tofind any point on the line.We plug in any convenient x-value and solve for y.Or we plug in any convenient y-value and solve for x.For instance,when y=0we have x−2(0)=5,so x−0=5.That is,we see that when y=0we must have x=5.So P(5,0)is a point on the line.Now we can write the point-normal form equation:(1,−2)•( x−(5,0))=0Example3.9.Write a standard form equation of the line x(t)=(3,2)+t(2,7).Solution:From the given point-parallel form equation,we see that P(3,2)is a point on the line and v=(2,7) is a direction vector for the line,i.e.is parallel to the line.And so n=(7,−2)is a normal vector to the line,so the standard form equation has7x−2y=c for some value c.And we canfind c usingc=(7,−2)•(3,2)=7(3)+(−2)(2)=21−4=17Therefore the standard form equation is7x−2y=17.Lines inℜ3Of course,we can have lines in3-space,as well as in the plane.And there’s a lot that’s the same inℜ3as it was inℜ2,so we use the same terminology and notation.For instance,when we move from2dimensions to3,it’s still true that given any2points,there is exactly one line that passes through both those points.And the vector equivalent to the directed line segment between those points is parallel to that line,so we still call it a direction vector for the line.That is,we define the term direction vector the same way inℜ3as we did inℜ2.Definition:If v∈ℜ3is parallel to some lineℓinℜ3,we say that v is a directionvector forℓ.As inℜ2,we can use a direction vector for a line(i.e.a vector parallel to the line)and any one point on the line to write a point-parallel equation for the line.And from that we can write parametric equations.Or we could write a2-point form equation,instead.The only difference is that now the points have3coordinates and the vectors have3components. Of course,for parametric equations this means that we have a third equation,corresponding to the z components of the vectors.These observations are summarized in the following definitions.Unit339 Definition:Let P(p1,p2,p3)and Q(q1,q2,q3)be any points inℜ3and let v=(v1,v2,v3)be any vector inℜ3.Then:1.Ifℓis the line which passes through P parallel to v(so that v is a direction vectorforℓ),thenx(t)=(p1,p2,p3)+t(v1,v2,v3)is an equation for lineℓin point-parallel form.2.If lineℓpasses through point P and v is a direction vector forℓ,then parametricequations of lineℓare:x=p1+tv1y=p2+tv2z=p3+tv33.If points P and Q are both on lineℓthen a two-point form equation forℓisx(t)=(1−t)(p1,p2,p3)+t(q1,q2,q3)Example3.10.Letℓbe the line which passes through the points P(1,2,3)and Q(1,−1,1).Write equations of lineℓin two-point form and in point-parallel form.Solution:In two-point form,we get the equation forℓ:x(t)=(1−t)(1,2,3)+t(1,−1,1)For a point-parallel form equation of lineℓwefirst need tofind a direction vector forℓ.The directed line segment−−→P Q is equivalent tov= q− p=(1,−1,1)−(1,2,3)=(0,−3,−2)which is parallel to(and hence is a direction vector for)ℓ.Using this direction vector and the point P which we know is on the line,we getx(t)=(1,2,3)+t(0,−3,−2)(Of course,we could have used point Q instead of point P to write the point-parallel form equation. Likewise,we could have used p− q=(0,3,2)as the direction vector.And in the two-point form equation,we could have switched the roles of P and Q.)Example3.11.Write parametric equations for the line through the point(0,1,−1)which is parallel to v=(2,1,0).Solution:An equation of the line in point-parallel form is x(t)=(0,1,−1)+t(2,1,0).This tells us that a point(x,y,z)is on this line if there is some value of t for which(x,y,z)=(0,1,−1)+t(2,1,0).So it must be true that,for the same value of t,we havex=0+2ty=1+1tz=−1+0t40Unit3 That is,we can write parametric equations of the line asx=2ty=1+tz=−1Example3.12.ℓ1is the line x(t)=(1−t)(2,1,−1)+t(0,1,2).ℓ2is the line with parametric equa-tions x=2t−2,y=1,z=5−3t.Areℓ1andℓ2the same line?Solution:Hmm.That’s different.Let’s see.Forℓ1we recognize that what we’ve been given is a two-point form equation.(We can tell because of the(1−t)multiplier.)From it we can see that P(2,1,−1) and Q(0,1,2)are two points on lineℓ1.This also tells us that the vectorv=−−→P Q= q− p=(0,1,2)−(2,1,−1)=(−2,0,3)is parallel to lineℓ1.Forℓ2we’re given parametric equations.It may be helpful to write these equations all the same way,with“constant+multiple of t”on the right hand side.We have:x=2t−2x=−2+2ty=1⇒y=1+0tz=5−3t y=5+(−3)tFrom the rearranged set of equations,using our knowledge of the form of parametric equations,we see that the point onℓ2used to write these parametric equations is R(−2,1,5).Also,the direction vector used for these equations is u=(2,0,−3).Since u=(2,0,−3)=−(−2,0,3)=− v,we see that these vectors are scalar multiples of one an-other,so they are collinear.That is,the direction vector u used to write the equation ofℓ2is parallel to the vector which we know is parallel toℓ1.Therefore u is also parallel toℓ1,and thus linesℓ1 andℓ2are parallel to one another.It’s possible that they could be the same line.How can we tell whether they are?Sinceℓ1andℓ2are parallel,then either they have no points in common or else they are the same line and have all points in common.So all we need to do is determine whether any point which is known to be on one line is also on the other.If it is,then they are actually the same line.But if it isn’t,then they must be different,but parallel,lines.We know that the point P(0,1,2)is on lineℓ1.Is it also on lineℓ2?If it is,then(x,y,z)=(0,1,2) must satisfy the parametric equations forℓ2,using the same value of t for each component(equation). Since the second coordinate of P is1,the equation y=1is satisfied.For thefirst coordinate,we see that we need to have x=2t−2satisfied for x=0.This gives0=2t−2⇒0+2=2t⇒2t=2⇒t=1Now,if we substitute t=1into the third of the parametric equations,we getz=5−3(1)=5−3=2Since z=2is the third coordinate of point P,we see that the point(x,y,z)=(0,1,2)does satisfy the parametric equations ofℓ2.That is,we have(0,1,2)=(−2,1,5)+1(2,0,−3)Unit341 so(x,y,z)=(0,1,2)satisfies x=−2+2t,y=1+0t and z=5−3t with t=1.Because the point (0,1,2)does satisfy the equations forℓ2,it is a point on lineℓ2.So now we know thatℓ1andℓ2are parallel lines,with a point in common,which means that they must have all points in common and be the same line.That is,sinceℓ1andℓ2are parallel and intersect at point P(0,1,2),they must intersect at all other points as well,and actually be the same line.Planes inℜ3We know that inℜ2,something of the form ax+by=c is the standard form of an equation of a line.What about the3-dimensional equivalent,ax+by+cz=d.Is that an equation of a line? Well,let’s see.Let’s think about a specific,uncomplicated,example.Consider the equation x+y+z=0. This equation is satisfied by the point P(2,−2,0)(because2+(−2)+0=0)and also by the point Q(3,−3,0).And we know that there’s a unique line that passes through those2points.Let’s call that lineℓing v= q− p=(3,−3,0)−(2,−2,0)=(1,−1,0)as a vector which is parallel toℓ1, we can write an equation ofℓ1asx(t)=(2,−2,0)+t(1,−1,0)Notice that for any point(x,y,z)on lineℓ1we havex=2+ty=−2−tz=0and so x+y+z=(2+t)+(−2−t)+0=2−2+t−t=0.That is,every point onℓ1satisfies the equation x+y+z=0.But those aren’t the only points which satisfy that equation.For instance,the point R(1,0,−1) also satisfies this equation.And this point is not onℓ1.The easiest way to tell is because from the parametric equations ofℓ1we can see that every point onℓ1has z=0,but the third coordinate of point R isn’t0,so it is not a point onℓ1.Hmm.Every point on lineℓ1satisfies x+y+z=0,but it’s not true that every point that satisfies x+y+z=0is on lineℓ1.So x+y+z=0cannot be an equation of lineℓ1.Then what is it?Well,actually,it’s the equation of a plane.ℜ2(i.e.2-space)is just a single plane.Butℜ3, which is to say3-space,contains infinitely many planes.(For instance,think of the walls,ceilings andfloors of the building you’re in.And also every other building you ever have been or ever could be in.And all the ramps you’ve ever seen.And what those ramps would look like if they were knocked offkilter.And...Each of those things lies in some particular plane inℜ3,and there are many other planes besides those.)So x+y+z=0is an equation of a plane.Let’s call it the planeΠ.(Planes are often named Π,which is just the Greek letter P,just like lines are namedℓ.ℓfor line,Πfor plane.Same idea.) Any plane contains infinitely many lines.One of the lines that lies in the particular planeΠwe’ve been talking about is the lineℓ1.But there are many others.For instance,we saw that P(2,−2,0) and R(1,0,−1)both lie on this plane,so the line on which those points lie is another line in plane Π.We can call that oneℓ2.And the vector r− p=(1,0,−1)−(2,−2,0)=(−1,2,−1)is parallel to ℓ2so we can expressℓ2as x(t)=(1,0,−1)+t(−1,2,−1).42Unit3 Notice that x+y+z=(1,1,1)•(x,y,z).Let’s think about the vector(1,1,1)whose components are the coefficients in the equation of the planeΠ.We know that(1,−1,0)is parallel to lineℓ1, which lies in planeΠ.Notice that(1,1,1)•(1,−1,0)=1(1)+1(−1)+1(0)=1−1+0=0, so the vector(1,1,1)is a normal for(i.e.is perpendicular to)lineℓ1.Likewise,we know that (−1,2,−1)is parallel to lineℓ2,which also lies in planeΠ.Notice that(1,1,1)•(−1,2,−1)= 1(−1)+1(2)+1(−1)=−1+2−1=0,so the vector(1,1,1)is also a normal for(perpendicular to) lineℓ2.However(1,−1,0)is not a scalar multiple of(−1,2,−1),so those vectors aren’t orthogonal (i.e.parallel)and therefore linesℓ1andℓ2aren’t parallel to one another.How can the same vector be perpendicular to both?Well,by being perpendicular to the whole plane in which both lines lie. This vector(1,1,1)is actually perpendicular to,i.e.a normal for,the planeΠ.Definition:A vector which is perpendicular to a particular plane inℜ3is said to benormal to the plane,and is called a normal for that plane,or a normal vector forthe plane.Point-Normal Form of an Equation of a PlaneUnit343 How do wefind a normal for the plane?Well,we know from the equation ofℓ1that the vector u=(1,−1,0)is parallel toℓ1,and likewise from the equation ofℓ2that the vector v=(−1,2,−1) is parallel toℓ2.Of course any vector n which is a normal forΠ(i.e.is perpendicular to this plane) must be perpendicular to any line that lies withinΠ.So if n is a normal forΠ,then n is perpen-dicular to bothℓ1andℓ2and therefore must be orthogonal to both u and v.(That is,any vector which is perpendicular toℓ1is also perpendicular to(orthogonal to)every vector that is parallel to ℓ1.And similarly forℓ2.)So how do wefind a vector which is perpendicular to both u and v?Well that’s easy.We know that the vector u× v is perpendicular to both u and v.So we can usen= u× v=(1,−1,0)×(−1,2,−1)=((−1)(−1),(0)(−1),(1)(2))−((2)(0),(−1)(1),(−1)(−1))=(1,0,2)−(0,−1,1)=(1,1,1)(Recall that we discussed previously that the vector(1,1,1)was a normal for the plane containing these linesℓ1andℓ2.)Now we know both a normal vector forΠand a point in planeΠso we can write the point-normal form equation.We get(1,1,1)•( x−(2,−2,0))=0Standard Form Equation of a Plane44Unit3 Example3.15.Write an equation in standard form for the plane with normal vector n=(1,2,3) which contains the point P(0,−1,2).Solution:Since n=(1,2,3)is a normal vector for the plane,then the standard form equation must have the form1x+2y+3z=d for some scalar d.But of course we would write that as x+2y+3z=d. How can wefind the value of d?Well,we know that the point P(0,−1,2)lies on the plane,so (x,y,z)=(0,−1,2)must satisfy this equation.That is,we plug in x=0,y=−2and z=2tofind the value of d.We get:x+2y+3z=d⇒0+2(−1)+3(2)=d⇒−2+6=d⇒d=6−2=4So a standard form equation of the plane is x+2y+3z=4.Notice:We could have used n• p=d,from rearranging the point-normal equation for the plane. What we did here is just another explanation of the exact same arithmetic.(Look back at the ex-amples in which we found point-normal equations of lines.We could have described the arithmetic we did there as“let x=p1and y=p2”instead of“find x• p”.)The Plane Determined by Three PointsUnit345 Example3.16.Find both a point-normal form equation and a standard form equation of the plane determined by the points P(−1,0,1),Q(1,2,3)and R(2,−1,5).Solution:The line passing through points P and Q lies in this plane,and so any vector parallel to that line is also parallel to the plane.And if we let u= q− p,then u is such a vector.Similarly,the vector v= r− p is parallel to the line which passes through both P and R,and since that line also lies in the plane, v is another vector which is parallel to the plane we need to describe.Also,we haveu= q− p=(1,2,3)−(−1,0,1)=(1−(−1),2−0,3−1)=(2,2,2)v= r− p=(2,−1,5)−(−1,0,1)=(2−(−1),−1−0,5−1)=(3,−1,4)and we can see that since u and v are not scalar multiples of one another then they are not collinear. We use these two non-collinear vectors which are both parallel to the plane tofind a normal for the plane:n= u× v=(8−(−2),6−8,−2−6)=(10,−2,−8)Now we use this normal vector and any one of the three points to write a point-normal equation of the plane.For instance,using point P,the form n•( x− p)=0gives:(10,−2,−8)•( x−(−1,0,1))=0Finally,we can also rearrange this equation to standard form.Letting x=(x,y,z),we get:(10,−2,−8)•((x,y,z)−(−1,0,1))=0⇒(10,−2,−8)•(x,y,z)−(10,−2,−8)•(−1,0,1)=0⇒10x−2y−8z=(10,−2,−8)•(−1,0,1)⇒10x−2y−8z=−10+0−8⇒10x−2y−8z=−18(Note:We might prefer to divide through the equation by2.That is,this plane would often be expressed as5x−y−4z=−9.)Determining the Distance between a Point and a Plane|| n||46Unit3That is,we simply need tofind the dot product of any normal vector to the plane with the vector equivalent to the directed line segment between the point P and any known point on the plane,discard the negative sign(if there is one),and divide by the magnitude of the normal vector used.(Notice:We have not explained why this gives −−→P′P ,so you should not be trying to understand that from the above.If you’re interested,look at the explanation given in the text.All we’ve done here is to assert that it can be shown that this is true.)Theorem3.3.Consider any planeΠ.Let n be any normal vector for planeΠand let Q be any point on planeΠ.Consider any other point P which is not on the planeΠ.Then the distance between point P and planeΠis given by:distance=| n•( q− p)|√612+22+12=and so the distance from P to the plane isdistance=| n•( q− p)|√√|(1,1,−1)•((5,0,0)−(0,0,0))| √|| n||=Unit347 Finding the Distance Between a Point and a Line|| n||Example3.19.Find the distance between the point P(1,2)and the lineℓdescribed by2x+y=1. Solution:Lineℓhas normal n=(2,1).We need tofind some point Q on lineℓ.Letting x=0we get 2(0)+y=1,so y=1.That is,the point on lineℓwhich has x-coordinate0has y-coordinate1,so the point Q(0,1)is a point on lineℓ.(Notice that for(x,y)=(1,2)we have2x+y=2(1)+2=4=1, so P(1,2)is not on lineℓ.)The distance between P andℓis| n•( q− p)|||(2,1)||=|(2,1)•(−1,−1)|22+12=|−2−1|4+1=35Finding the Intersection of Two Lines48Unit3 Example3.20.Find the point of intersection of the lineℓ1: x(t)=(1,0)+t(2,1)with the lineℓ2: x(s)=(1,1)+s(−1,0).Solution:Forℓ1we have parametric equations x=1+2ty=tand forℓ2we havex=1−sy=1.If some point P(x,y)is on both these lines,then it must be true that there are some values of t and s which give the same values of x and y.So we must have1+2t=1−s and t=1.Since t=1, then1+2t=3,so1−s=3and we see that s=1−3=−2.Notice that we’ve found values of the parameters,s and t,but we have not yet found the point on the line which corresponds to these values.That is,we know the value of t that gives the point on lineℓ1at which the two lines intersect,and likewise we know the value of s that gives that same point on lineℓ2.But we were asked tofind the actual point at which the two lines intersect.We’re notfinished until we’ve done that.And we have more information than we need tofind the point, since we know two ways to get it.So we can use the value of t we found,in the equation forℓ1,to get the point P.And then we can use the value of s we found,in the equation ofℓ2,to check our work.We get:t=1⇒(x,y)=(1,0)+t(2,1)=(1,0)+1(2,1)=(3,1)as the point onℓ1which we were looking for.We check that the point onℓ2is the same point: s=−2⇒(x,y)=(1,1)+s(−1,0)=(1,1)+(−2)(−1,0)=(1,1)+(2,0)=(3,1) Since we didfind the same point on each line,this is the point we were looking for.We see thatℓ1 andℓ2intersect at the point P(3,1).Note:As we observed above,we found values of both parameters,but really we only need one. As we have seen,the other allows us to check our work.We’re just checking that we didn’t make an arithmetic error.If we got a different point onℓ2than the one onℓ1that would tell us that somewhere in our calculations we made an arithmetic mistake.Either infinding the points,or(more likely)infinding the values of the parameters.We would need to re-do our calculations until we find the mistake,and thenfinish the problem(including the check)again.Example3.21.Find the point of intersection of the lineℓ1: x(t)=(1,1,2)+t(2,1,−1)with the line ℓ2: x(s)=(0,1,2)+s(1,−1,1).Solution:Forℓ1we have x=1+2ty=1+tz=2−tand forℓ2we havex=sy=1−sz=2+s.The point of intersection ofℓ1andℓ2is a point P(x,y,z)which satisfies both sets of equations at the same time,so we must have:1+2t=s(1)1+t=1−s(2)2−t=2+s(3) Equation(1)says that s=1+2t,so that1−s=1−(1+2t)=0−2t=−2t.Therefore equation (2)gives1+t=−2t,so1=−3t and thus t=−13into s=1+2t。