4.1 Distance and Length

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1 Chapter Vector Geometry In this chapter we will look more closely at certain geometric aspects of vectors in R n. We will first develop an intuitive understanding of some basic concepts by looking at vectors in R and R where visualization is easy, then we will extend these geometric intuitions to R n for any n. The basic geometric concepts that we will look at involve measurable quantities such as length, angle, area and volume. We also take a closer look at the two main types of equations covered in this course: parametric-vector equations and linear equations. We begin with a reminder. We defined a vector in R n as an n-tuple, i.e., as an n matrix. This is an algebraic definition of a vector where a vector is just a list of numbers. The geometric objects we will look at in this chapter should be seen as geometric interpretations of this algebraic definition. One difficulty that students encounter at this stage is that there are many different geometric interpretations that can be given to a vector. For example, a vector in R n can be interpreted geometrically as an arrow starting at the origin. an arrow with a certain length and direction but no fixed location. a point (or more exactly, the coordinates of a point relative to some reference point). a directed line segment between two points. a displacement (i.e., a translation). This multiplicity of interpretations is a strength of the vector concept not a weakness. Vectors have many applications and depending on the application one geometric interpretation may be more relevant than another but no matter what geometric interpretation is chosen the underlying vector algebra remains the same. We will interpret a vector in R n as a position vector as described in section. of Lay s textbook. A position vector is just a pointer to a certain location in R n. When using position vectors it is not necessary to make a firm distinction between a vector and its endpoint. For example, when we say that a line is a set of vectors we mean that the endpoints of the vectors lie on the line. If we want to stress the direction of the vector we will usually represent it as an arrow. If we want to stress the particular location that the vector is pointing to we will usually represent it by a point. EXAMPLE.. If A = (x, x,..., x n) and B = (y, y,..., y n) are two points then the vector from A to B (represented by AB ) is defined as follows y x y x AB = 6. 5 y n x n You can think of this as letting A be the origin of a new coordinate system and then the entries in AB give the location of B relative to A. Or you can imagine translating both A and B by subtracting A from both points so that A is translated to the origin. Finally, you can think of AB as an arrow from A to B. So, for example, if we have P(,5, ) and Q(,,) then PQ = 55 = 6 5. The entries in this vector indicate that when you travel from P to Q you move 6 units in the x direction, units in the x direction and units in the negative x direction. These entries express the location of Q relative to P. If PQ is drawn with the intial point at the origin then the terminal point would be (6,, -). We will usually represent a vector as an n matrix but there is another standard way of representing vectors that is frequently used. In R we define» i =» j = It then follows that any vector in R can be written as»»» a a = + = ai + bj b b

2 Chapter. Vector Geometry Similarly in R we define i = 5 j = 5 k = 5 and then any vector in R can be written a b5 = ai + bj + ck c You should realize that in R the vectors i and j are just the vectors which we have called e and e, the standard basis of R. Similarly in R the vectors i, j and k are the standard basis of R.. Distance and Length The first geometric concept we want to look at is the the length of a vector. We define this to be the usual Euclidean distance from the intial point (the origin) to the end point of the vector. The length any vector v in R n will be represented by v. This quantity is also referred» to as the magnitude or norm of v. u Let u = be a vector in R u. The length of this vector would be the distance from the origin (, ) to the point (u, u ) and this is given by the Pythagorean Theorem as q u = u + u» 5 EXAMPLE.. Let u =. Figure. shows u and by the Pythagorean Theorem we can find the norm of u as q u = 5 + ( ) = In R a similar argument based on the Pythagorean Theorem gives q u = u + u + u for any vector u = u u 5. u We can extend the above formulas to R n by defining q u = u + u + + u n u u Notice that if u = 6. 5 is any vector in Rn then u n u u T u = ˆu u u u n 6. 5 = u + u + + u n u n We then have the following concise formula which is valid for vectors in R n for all n u = u T u EXAMPLE.. Let u be any vector in R n and k be a scalar then ku = ku T (ku) Taking square roots then gives = k u T u = k u ku = k u This shows that multiplying any vector in R n by a scalar k scales the length of the vector by k. We will sometimes make a distinction between the sense of a vector and the direction of a vector. When a vector is multiplied by a negative scalar the reversal of the arrow is described by saying the sense has been reversed but the direction has stayed the same. 5 6 Definition.. The distance between two vectors u and v in R n is defined as u v. EXAMPLE.. The distance between u = i + k and v = j k is q u v = i j + k = + ( ) + = 6 Unit Vectors Figure.. (5, ) A unit vector is a vector whose length is. If u is any non-zero vector in R n then u is a unit vector. u This can be seen by applying the formula v = v T v to the vector u. This gives: u

3 .. Distance and Length «T «u u u u = u ut u = u u = The process of multiplying a vector by the reciprocal of its length to obtain a unit vector is called normalization. Notice that this procedure doesn t alter the direction or sense of the vector. EXAMPLE.5. Normalize the vector v = 6 5. We have v = = 9 = so / 6 5 = 6 / 5 / is a unit vector parallel to v. Note: Just to avoid any possible confusion, when we say that two non-zero vectors, u and v, are parallel we mean that they have the same direction. Each one is a scalar multiple of the other.

4 Chapter. Vector Geometry Problems. If A = (, ) and» AB = what is B?. If B = (5,,) and AB = 6 5 what is A?. Find the length of the following vectors:» a. b. 5 c. 6 5» cos θ d. sinθ e. cos(s)sin(t) cos(s) cos(t) 5 sin(s) f. i + j + k g. i j k h. t i + tj + tk. Let v = 6. 5 be the vector in Rn all of whose entries are. What is v? 5. Find the lengths of the sides of triangle ABC where the vertices are given by 9. If AB = 5 and BC = what are the possible values for AC?»» cos(s) cos(t). Let u = and v =. These are two unit sin(s) sin(t) vectors p in R. Show that the distance from u to v is cos(s t). Prove that in R the length of v v 5 is given by v q v + v + v.. True or False: a. u = uu T b. u = u T u c. u = u T u d. u+v = u T u+v T v e. If u = v then u+ v = u + v. f. Au = u T A T Au. Under what conditions will u + v = u + v?. Suppose A is an n n matrix such that A T A = I. Let v be any vector in R n. Show Av = v» cos θ sinθ 5. Let A =. Show that if v is any vector in R sinθ cos θ then Av = v. a. A(, ), B(, ), C(5, ) b. A(, ), B(, 5), C(, ) c. A(,, ), B(,, ), 6. a. If» AB =», BC = 5 C(,,) d. A(,, ), B(,, ), C(,,), and A is the point (, ) what is C? Draw a diagram illustating this problem. b. If PQ = 5, QR = 5, and R is the point (, 5, ) what is P? k. Let u = 5 and v = k + 5. Use calculus to find the k + value of k for which the distance from u to v is a minimum. 8. Find a unit vector parallel to each of the following vectors:» a. b. 5 5 c. 6 5 d. i 5j + k» e. t f.» cos t + sint cos t sint

5 .. The Dot Product 5. The Dot Product The Dot Product in R Suppose we have two vectors in R, u = illustrated in Figure. v u - v» u u» v and v = as v (This formula uses the standard convention of interpreting the matrix u T v as a scalar.)»» EXAMPLE.6. If u = and v = then and u v = ()() + ()() = 6 u = v = It then follows that if θ is the angle between u and v we have cos θ = 6 = 5 u θ and the angle between u and v is arccos(/5) 5.. The dot product is often useful in geometric problems involving angles. If the problem is stated in terms of points then it should be translated into vector terminology before using the dot product. Look at the following example: Figure.. EXAMPLE.. Draw the parabola y = x and let P be the point (, ) on this parabola. If O is the origin find another point Q on the parabola such that the angle between OP and OQ is. These two vectors determine a triangle whose third side would be u v translated. If we let θ be the angle between u and v then we can apply the law of cosines to the triangle. This gives u + v u v cos θ = u v = (u v ) + (u v ) = u + v u u v v Cancelling out common factors and terms we get Q u v cos θ = u v + u v The expression on the right hand side of the last line is given a special name. It is called the dot product of u and v and is written u v. Thus we have the following two formulas x.5. P and u v = u v + u v u v = u v cos θ where θ is the angle between u and v. Since θ is one angle of a triangle we have θ 8. This means that θ is the smallest positive angle between u and v.» u There is another way of indicating the dot product. If u = u» v and v = then v u T v = ˆu»v u = u v v + u v and so we have the nice formula u v = u T v Figure.. In terms of vectors we have» OP = and» x OQ =. x We also have OP OP = x + x, OP =, and OQ = x + x. If the angle between these vectors is the dot product formula gives x + x = p x + x cos Substituting cos = and squaring both sides gives x + x + x = `x + x We omit the algebra but you should be able to solve this equation and find two values of x that work x = + and x =.

6 6 Chapter. Vector Geometry The Dot Product in R n The dot product can be generalized to vectors in R n. Definition.. Let u and v be vectors in R n then their dot product is defined by u v = u T v u v u v The above definition implies that if u = 6. 5 and v = u n v n then u v = u v + u v + + u nv n The fundamental properties of the dot product are summarized by the following theorem. Theorem.. If x, y, and z are vectors in R n and c is a scalar then a. x y = y x b. x (y + z) = x y + x z c. (cx) y = c(x y) d. x x and x x = if and only if x =. Proof. The proof of part (a) is left as an exercise. Since u v = u T v parts (b) and (c) follow from the corresponding properties of matrix multiplication. The proof of part (d) is more complicated because it makes several claims. First it claims that the inner product of a vector with x x itself can never be negative. To see this let x = Then x n x x = x +x + +x n and this value can never be negative since it is the sum of squares. Next it says that the inner product of the zero vector with itself is. This is easy: if x = then x x = =. Finally part (d) claims that if the dot product of a vector with itself is then the vector must be the zero vector. To see this suppose x x =. Then x + x + +x n =. Since the left hand side of this equation has no negative terms the only way the terms can add up to is if each term is. So x =, x =,..., x n = and we have x =. This last property says that in R n the only vector with length is the zero vector. Orthogonal Vectors The formula u v = u v cos θ relates the dot product to the angle between vectors. If u and v are non-zero vectors then the right hand side of this expression is positive if θ < 9 and negative if 9 < θ 8. More importantly the dot product is if θ = 9. This means that two non-zero vectors are perpendicular if and only if their dot product is. This leads to the following definition. Definition.. Two vectors in R n are said to be orthogonal if their dot product is. This definition implies that the zero vector in R n is orthogonal to every vector in R n. EXAMPLE.8. Show that the triangle with vertices A(, ), B(5, 5), and C(, ) is a right triangle. We want to interpret the sides of this triangle as vectors. If we treat A as the origin then the vector from A to B would be AB =» 5 ( ) 5 =». Similarly AC =» ( ) =» Now notice that AB AC = () + ( ) =. Since AB AC = we have AB AC cos θ =. We can then conclude that cos θ = and so θ = 9.» k EXAMPLE.9. For what value(s) of k are the vectors u = k» k + and v = orthogonal? k We just have to determine any values of k which make u v =. u v = k(k + ) + k(k ) = k + k + k k = k k = k(k ) It should then be clear that the vectors are orthogonal for k = and k = /. We end this section with three theorems which state some important properties of vectors. Theorem.5. (Pythagorean Theorem) Let u and v be vectors in R n then u + v = u + v if and only if u and v are orthogonal. Proof. Note that the statement of this theorem is another if and only if statement. This means that the theorem is making two claims. These will be proved separately below. First we must show that if u and v are orthogonal then u + v = u + v. The argument is as follows u + v = (u + v) (u + v) (.). = u u + u v + v v (.) = u u + + v v (.) = u + v (.) Although this formula was proved only for vectors in R it is applicable in all R n. Justification for this will be given shortly.

7 .. The Dot Product Equation. is a consequence of the distributive and commutative properties of the dot product. Equation. is a consequence of u and v being orthogonal (their dot product is ). Next we must show that if u + v = u + v then u and v are orthogonal. The argument goes as follows u + v = u + v (.5) u u + u v + v v = u u + v v (.6) u v = (.) u v = (.8) The last line tells us that u and v are orthogonal. Theorem.6. ( Cauchy-Schwarz Theorem) If u and v are vectors in R n then u v u v Proof. If v is the zero vector then both sides of the inequality would be zero and the theorem would be true so we will assume that v is not the zero vector. Part d of Theorem. tells us that u u v v v v u u v v v v since the dot product of any vector with itself is always greater than or equal to. Simplifying the left hand side of the above gives u u v v v v u u v v v v = u u u v (u v) u v + v v (v v) v v = u u (u v) v v Replacing this last expression in the original inequality gives u Rearranging these terms gives (u v) v u v (u v) Taking the square root of both sides gives u v u v The Cauchy-Schwarz Theorem guarantees that u v u v for any non-zero vectors u and v in R n. This allows us to define the angle θ between any non-zero vectors u and v in R n by the formula cos θ = u v u v (where θ 8 ). EXAMPLE.. Find the angle between u = 6 5 and v = 6 5. We have u v = + =, u = = =, and v = =. Thus cos θ = u v u v =.86 which gives θ.5 Theorem.. ( Triangle Inequality) If u and v are vectors in R n then u + v u + v Proof. u + v = u u + u v + v v (.9) u u + u v + v v (.) u u + u v + v v (.) = ( u + v ) (.) Equation. follows from the Cauchy-Schwarz theorem The above shows that u + v ( u + v ). The theorem then follows by taking the square root of both sides. Direction Angles The dot product gives us a new way of looking at unit vectors. Any particular entry in a unit vector cannot be larger than or less than -. The entries in a unit vector turn out to have a very simple geometric interpretation. In R the angles between any vector v = v v 5 and the x, x v and x axes are called the direction angles of v and are represented by α, β, γ respectively. These are just the angles between v and the unit vectors i, j, and k. So we have cos α = v i v i = v v cos β = v j v j = v v cos γ = v k v k = v v We can now write v = v v cos α v 5 = v cos β5 = v cos α cos β5 v v cos γ cos γ The values cos α, cos β, and cos γ are called the direction cosines of v.

8 8 Chapter. Vector Geometry The last example can be generalized in the following way (we leave the details as an exercise). Let v be any non-zero vector in R n. If we normalize v then the entries in the normalized vector are just the cosines of the angles between v and the vectors in the standard basis of R n. That is, the entries in any unit vector are the direction cosines of that vector. EXAMPLE.. Let u = 6 5. If we normalize u we get 6 / /5. Now cos () = π/, cos (/) = π/, cos ( /) = / π/, and cos ( /) = π/. So u lies at an angle of 9 from e, at an angle of 6 from e, at an angle of from e, at an angle of 5 from e Note that for any non-zero vector v there are two unit vectors in the same direction as v. One has the same sense as v, the other has the opposite sanse of v.

9 Problems 9 Problems. For the following pairs of vectors u and v calculate u, v, u v, and the angle between u and v.»» a. u =, v = d. u = 6 5, v = 6 5»» b. u =, v = 5»» a b e. u =, v = b a c. u = 5, v = 5 f. u = i j + k, v = i + j + k. Use the dot product to find the angles of the triangles with the following vertices a. A(, ), B(, ), C(6, ) b. A(,,), B(,,), C(,, ) c. A(,,, ), B(,,, ), C(,,, ). Given the points A(, ), B(, ), and C(, k) find the values of k for which triangle ABC is a right triangle.. Let u = 5 and u = 5. Show that any vector in Span {u,u } is orthogonal to v = 5.»» a cos(θ) sin(θ) 5. Let x = and R =. Use the dot product to find the angle between x and Rx. b sin(θ) cos(θ)»» 6. Let u = and v =. For what value(s) of k k a. do u and v have the same length? b. are u and v orthogonal? c. are u and v parallel? d. is the distance from u to v one unit?. Let u = 5 and v = + k5. For what value(s) of k k a. do u and v have the same length? b. are u and v orthogonal? c. are u and v parallel? d. is the distance from u to v units?»» cos θ sinθ 8. Let u = and v =. For what value(s) of θ sinθ cos θ a. do u and v have the same length? b. are u and v orthogonal? c. are u and v parallel? 9. Normalize the following vectors and find the direction angles in each case: a. v = 5 b. v = 5 c. v = a. Suppose v is a vector in R and you know that this vector forms an angle of π/ with i. Is this enough information to determine vector v? What are the possible values for the angle between v and j? b. Suppose v is a vector in R which forms an angle of π/6 with i. What are the possible values for the angle between v and k? c. Suppose v is a vector in R which forms an angle of π/6 with i and an angle of π/6 with j. What are the possible values for the angle between v and k?. Use the dot product to prove that x + y + x y = ` x + y. Show that if u +v and u v have the same magnitude then u and v are orthogonal.. It was pointed out in this section that u T v = v T u for any two vectors u and v in R n. Is it also true that uv T = vu T? Prove this or give a counter-example.. Suppose x and y are vectors in R n that have the same length. Show that x + y bisects the angle between x and y. 5. Prove part a of Theorem.. 6. Let u and v be two vectors in R n. a. Justify the following steps u T v = (u T v) T = v T u b. The following chain of equalities resembles the above example but is not valid. What is wrong with the following? uv T = (uv T ) T = vu T

10 Chapter. Vector Geometry. Lines Lines in R We have already seen in Chapter of Lay s textbook that a line in R can be represented by an equation of the form x = x + tv. Such an equation represents the line through x parallel to v. This is called a parametric-vector equation of the line. We now want to look at another way of representing a line in R. How can we specify the direction of a line in R? One standard way is to give the slope of the line and another is to give a vector parallel to the line. There is a third way of specifying the direction of a line in R which is not so obvious and which at first might seem unnecessarily complicated. We can specify the direction by giving a vector perpendicular to the line. Such a vector is said to be a normal vector to the line. Figure.. normal vector (xo, yo)»» x x Suppose x = lies on the straight line through x y = y» a with normal n = then x x b is parallel to the line and is therefore perpendicular to n. We must then have n (x x ) =»» a x x = b y y a(x x ) + b(y y ) = ax + by ax by = The equation of the line can then be written as or more concisely ax + by = ax + by ax+by=c where c = ax + by. We will call this the normal equation of a straight line. Notice that this equation could be written in vector notation as n T x = n T x This is not the usual term for this equation. Most books would call it the standard equation or the Cartesian equation of a line. We are calling it the normal equation just to stress the fact that the coefficients give a normal vector to the line. Now you already knew that any straight line in R could be written in the form ax + by = c but what is new here is that in an equation of this form the coefficients of x and y give a normal vector to the line. EXAMPLE»».. Given the parametric-vector equation of a line x = + t find the normal equation of this line.» a We will use the following observation: If u = is any vector b» in R b then is orthogonal to u. This can be easily confirmed a by a simple dot product.»» a b = ab ab = b a Now it should be clear» that the direction of the given line is determined by the vector. We want a normal vector to this» line and the above comments show that n = would be such a vector. We now know that the normal equation of the line is x y = c with the value of» c yet to be determined. Now we use the fact that lies on the line and so these values must satisfy the equation. We then have () () = c. So c = 9 = 5 and the equation we are looking for is x y = 5 EXAMPLE.. Find an equation for the line through A(, 5) and B(6, ). The vector»» 6 AB = = is a direction vector for this 5» line. The line x = t would be a line through the origin parallel to the desired line. We only have to add a translation. A parametric-vector equation for this line would be»» x = t + 5 This is illustrated in Figure.5. But we can also find a normal vector to this line, n =». The line therefore has an equation of the form x+y = c. Substituting the coordinates of A gives Lines in R n x + y = In general a line in R n can be written in parametric-vector form x = u + tv where v determines the direction of the line and u lies on the line.

11 .. Lines. represented by normal or Cartesian equations only in R Figure.5. The line through A and B. EXAMPLE.. Find the equation of the line containing both p = 6 5 and q = 6 5. If we translate this line to pass through the origin by subtracting p from all points on the line then a direction vector for the line would be q p = = 6 5 The line x = t 6 5 would therefore be a line through the origin parallel to the line we are looking for. We just have to translate the line so that it passes through p and q and we do this by adding either one of these to our equation. So one possible answer would be x = t 6 5 You should understand that points on the line are generated by different values of the parameter t. In particular when t = in the above equation we have x = p and when t = we have x = q. In R n with n a vector that is orthogonal to a line can be pointing in infinitely many different directions. For example in R any vector in the x, y plane will be orthogonal to the z axis. This means that it is no longer possible to determine the direction of a line just by specifying a normal vector in these spaces. So parametric equations are more versatile since they can be used to represent lines in spaces of any dimension. Lines can be A. x B 6. EXAMPLE.5. Let L be the line defined by x = s 6 5 and L be the line defined by x = t 6 5 It should be obvious that these lines intersect at 6 5. The angle between the direction vectors of these lines is cos = cos 5.radians. This 5 is the angle formed by the two intersecting lines. Even though these are lines in R there is still a plane (a flat -dimensional surface) containing these lines. It should also be obvious that this plane is defined by the equation x = s t 6 5 Does 6 5 lie in this plane? This is equivalent to asking if the equation s t 6 5 = 6 5 has a solution. This vector equation gives us the augmented matrix 6 5 This augmented matrix reduces to So this point does lie in the plane. It corresponds to the values s = and t = 9 in the equation of the plane.

12 Chapter. Vector Geometry Problems. Write the following lines in parametric-vector form. a. x + x = 5 b. x 5x = c. x = d. x =. Write the following lines in normal form»»» a. x = + t c. x = t»» 5 b. x = + t. Given the line L : x = d. x = ti + ( t)j»» + t a. Find the value of x that corresponds to t =. Find the value of x that corresponds to t =.» b. Find the value of t that corresponds to x =.» c. Find all x on L that lie units from. d. Find all x on L that lie 5 units from the origin. e. Illustrate all the above with a picture.. Given the line x = 5 + t 5 a. Find the value of x that correspond to t =. Find the values of x that correspond to t =. b. Find the value of t that corresponds to x = / /5. / c. Is 5 on this line? For what value(s) of k is 5 on k this line? d. Find all x on the line that lie units from 5. e. Find all x on the line that lie units from the origin. 5. Find all values of a and b such that the following two equations represent the same line»» x = + t»» a b x = + t. Find all values of a and b such that the following two equations represent the same line x + x = x + ax = b 8. Let n and x be vectors in R. Show that the line through x normal to n can be written n T x = n T x. 9. Find a parametric vector equation of the line containing p and q where a. p = 5 q = 5 b. p = 6 5 q = In R find three different lines through the origin orthogonal to the vector k.. For what values of a and b do the following equations give the same line? x = a 5 + t 5 and x = s b 5 a + b +. a. Show that the line in R n which contains p and q can be represented by the equation x = ( t)p + tq.» b. Find the normal equation of the line x = ( t) + 5» t.. Let p and q be vectors in R n. Describe the difference between the following expressions (where the parameters s and t are assumed to range over all real numbers). a. sp + tq b. sp + ( s)q c. sp + q 6. Find all values of a and b such that the following two equations represent the same line»» x = + t»» a b x = + t

13 .. Planes and Hyperplanes. Planes and Hyperplanes Planes in R. We have already seen in Chapter of Lay s textbook that a plane in R can be expressed in parametric-vector form as x = x + su + tv It is the equation of a plane with normal vector 5 which contains the point 5. A plot of the plane and normal vector is shown in Figure.6. The expression su + tv corresponds to a plane through the origin generated by vectors u and v (also called the span of u and v). The addition of x to this expression translates this plane in R. Now a plane in R doesn t have a slope and a plane doesn t point in any particular direction but a plane does have an orientation and parallel planes have the same orientation. The orientation of a plane in R can be specified by giving two independent vectors parallel to the plane or by specifying one vector that is normal to the plane. Suppose we want to determine a condition on x = x y5 z that tells us when it is in the plane containing x = x y 5 with z y z x normal vector n = a b5. If both x and x lie in the plane then the c vector x x must be parallel to the plane and so we have n (x x ) = a b5 x x y y 5 = c z z a(x x ) + b(y y ) + c(z z ) = ax + by + cz ax by cz = This equation can be written ax + by + cz = ax + by + cz or just ax + by + cz = d where d = ax +by +cz. We will call this the normal equation of a plane in R. Note that this equation can be expressed in terms of the relevant vectors as n T x = n T x EXAMPLE.6. The equation x + y + z = can be expressed in the form n T x = n T x as ˆ x y5 = ˆ 5 z It is more common to call this the standard equation or the Cartesian equation of a plane. Figure.6. The plane x + y + z = and a normal vector. EXAMPLE.. Find a parametric-vector equation of the plane with normal equation x y + z = We can look at this equation as a very simple system of equations. It is just a system of one equation with three unknowns. The general solution will therefore require two free variables. The general solution will be x + s t y5 = s 5 = 5 + s 5 + t 5 z t and this is a parametric-vector equation of the plane. Notice that the equation x y+z = tell us that n = 5 is a normal vector to the plane. The parametric-vector equation tells us that the plane is a translation of the plane generated by the vectors 5 and 5. A couple of simple dot product calculations shows us that n is perpendicular to both of these vectors as we should expect.

14 Chapter. Vector Geometry EXAMPLE.8. Find the normal equation of the plane containing p = 5, q = 5 and r = 5. The equation that we are looking for has the form ax+by+cz = d. Since the plane contains p we know that x =, y = and z = must satisfy this equation. So we have a() + b() + c() = d. Repeating this procedure for q and r results in the system a + b d = a + c d = b + c d = This gives the the coefficient matrix 5 which reduces to /5 / a / The general solution of this system is 6b c5 = t 6/ /5. Choosing d t = gives the particular solution a = b = c =, d = and so one possible equation for the plane would be x + y + z = (There are infinitely many solutions to this system which means there are infinitely many equations for this plane, but they are all just scalar multiples of the above equation.) 6 x Figure.. Intersecting planes. EXAMPLE.9. Find the line of intersection of the planes x + y + z = and x + y z = 5. y It should be clear that these planes are not parallel since their normals are not parallel. Since they are not parallel they should intersect along a straight line. The planes and the line of intersection are illustrated in Figure.. The points of intersection are the points that satisfy both equations. So we want the solution of the system x + y + z = x + y z = 5 But this system can be written in the form Ax = b where the solutions are now represented as vectors. The augmented matrix and its reduce form would be» 5» 5 It should now be easy to write down the solution x = 5t + t5 = 5 + t 5 5 t This is a parametric vector representation of the line of intersection. The direction vector of this line is 5 5. This vector must be parallel to both planes. Why? It then follows that this vector should be orthogonal to the normals for both planes. Verify that this is the case. EXAMPLE.. Find the point of intersection between the plane x + y z = and the line x = 55 + t 5. The equation of the line can be written as x y5 = 55 + t 5 = + t 5 + t5 z + t This means that any point on the line must satisfy the equations x = + t, y = 5 + t, and z = + t for some value of the parameter t. If the point also lies on the plane it must satisfy the equation x + y z =. Substitution then gives x + y z = ( + t) + ( 5 + t) ( + t) = t 6 = t = The point of intersection occurs when t = so the point of intersection is x = + =, y = =, z = + 6 =, or in vector form x = 5.

15 .. Planes and Hyperplanes 5 Hyperplanes We have seen that in R the equation n T x = n T x defines a line through x with normal n and in R this equation defines a plane through x with normal n. In general, in R n the equation n T x = n T x defines what is called a hyperplane through x with normal n. This type of equation is a linear equation with n variables. There are then n free variables and this means that in R n a hyperplane will be an n dimensonal subspace that might or might not have been translated away from the origin. So in R a hyperplane is a (possibly translated) dimensional subspace, a line. In R a hyperplane is a dimensional subspace that has possibly been translated. EXAMPLE.. Let n = 6 5 and x = 6 55 then the normal equation of the hyperplane through x with normal n would be of R n. If we represent such a subspace by V then we can translate the line or plane to a new position by adding a vector to each vector in V. The translated line or plane can be represented by V + u. In other words V + u = {x : x = v + u,u V }. But there is no need to restrict ourselves to lines and planes. If V is any subspace of R n then V + u represents V translated by u to a new position. If V is -dimensional then V + u is a line. If V is -dimensional then V + u is a plane. If V is n dimensional then V + u is a hyperplane. In general if V is k dimensional then V + u is called a k-flat. n T x = n T x x ˆ 6x x 5 = ˆ 6 55 x x + x + x + x = This equation contains four variables so the general solution would involve three free variables. If we represent the free variables by parameters x = r, x = s, and x = t then the general solution would be x r s t 6x x 5 = 6 r s 5 x t = r s t 6 5 This would be a parametric-vector representation of the same hyperplane. This representation makes it clear that the hyperplane is a three dimensional subspace of R with basis 8 9 >< 6 5 >:, 6 5, 6 >= 5 >; that has been translated by the vector 6 5. Another notation for lines and planes A line through the origin in R n is a one-dimensional subspace of R n. A plane through the origin in R n is a two-dimensional subspace

16 6 Chapter. Vector Geometry Problems. Find a normal equation of the plane or hyperplane containing x with normal n. a. n = 5, x = 5 b. n = 5, x = 5 5 c. n = i + j + k, x = i + 5j k d. n = 6 5, x = 6 5. Find a parametric-vector equation of each of the following planes or hyperplanes. a. x + x x = b. x + x + x = 5 c. x + x + x + x = d. x x + x x + 5x 5 =. The plane x + y z = is in fact a subspace of R. Find a basis for this subspace and find an equation for the line through the origin normal to this plane.. The hyperplane x +x x +x = is in fact a subspace of R. Find a basis for this subspace and find an equation for the line through the origin normal to this hyperplane. 5. Find an equation for the plane containing 5 parallel to the plane x y + z =. 6. Find an equation for the line which containins 5 5 and is also perpendicular to the plane x + y z =.. The angle between two hyperplanes is defined to be the smallest possible angle between normals of the hyperplanes. This angle must be between and π/ radians (why?). This means that you should choose normals such that their dot product is positive. Find the angle between the following hyperplanes: a. x + y + z =, x y z = b. x + x + x + x =, x + x + x + x = 8. In R n let n = 6. 5 and x = 6. Find the normal equation and the parametric-vector equation of the hyperplane. 5 containing x with normal n for a. R b. R c. R d. R 5 9. Find the point of intersection of the following planes and lines a. The plane x + y + z = 8 and the line x = 5 + t 5. b. The plane x y z = 5 and the line x = 5+t c. The plane x+y+z = and the line x = 5+t 5. d. The plane x+y+z = and the line x = 5+t 5 e. The hyperplane x + x x + x = and the line 8 x = t Let P be the plane x + y + z = 9 and let L be the line x = 5 + t k k + 5. For what value(s) of k a. are P and L orthogonal? b. are P and L parallel? 8 c. do P and L intersect at 5.. In R let P be the hyperplane with equation x +x +x + x =, P be the hyperplane with equation x x +x =, and P be the hyperplane with equation x x =. Find the parametric-vector equation for a. The intersection of P and P. b. The intersection of P and P. c. The intersection of P, P and P.. Let u = 5, v = 5, and w = 5. Let V = Span {u,v}. a. Find a normal equation for the plane V + w. b. Find a parametric vector equation for V + w.. Let p, q, and r be vectors in R n. Show that the plane which contains p, q and r can be represented by the equation x = ( s t)p + sq + tr.. Suppose you are told that the equations x + y z = 8 and x = 5 + s 5 + t 5 represent the same plane. Which equation would be easiest to use to answer each of the following problems?

17 Problems a. Does 65 lie in this plane? 5 b. Give 5 points which lie in this plane. 5. Suppose V + v and W + w represent the same plane then which of the following are true a. V = W. That is, V and W must be the same plane. b. v = w. c. If v then V + v does not contain the origin. d. V + v + w must also be the same plane as V + v and W + w.

18 8 Chapter. Vector Geometry.5 Projections Suppose u and v are vectors in R n with v. If we drop a perpendicular line from u onto the line determined by v as shown in Figure.8 we obtain a vector called the projection of u onto v which we will represent by Proj v u. proj u perp 5 Figure.8. The projection of u onto v. The projection of u onto v must be in the same direction as v so we have Proj v u = kv for some scalar k. We also know that u Proj v u is perpendicular to v so we can write (u Proj v u) v = (u kv) v = u v kv v = We then have kv v = u v and so k = u v v v and Proj v u = u v v v v As mentioned above the vector u Proj v u is perpendicular to v and is called the orthogonal component of the projection. We will denote this orthogonal component by Perp v u»» EXAMPLE.. Let u = and v = then and Proj v u = u v v v v =» =» Perp v u = u Proj v u = v» / /» / = /» / / Step. Translate the line and the point so that the line goes through the origin. The simplest way to do this is to subtract 5 from the line and the point. The translated line is just x = t 5 and the translated point would be u = 5 5 = 5. Step. Let v = 5 be the direction vector of the line and find Proj v u. This gives Proj v u = 5 5 = / 5 / Step. The distance is just the length of Perp v u. We have Perp v u = 5 / 5 = / 5 / / Thus the distance is q (/) + ( /) = p 5/ = Distance from a Point to a Plane The distance from a point to a plane (or hyperplane) can be found as illustrated in Figure.9 idea is that we project a vector onto a normal to the plane and the distance we want is just the length of this projection. distance n (normal) x Distance from a Point to a Line in R n The method we will use for finding the distance from a point to a line is illustrated by Figure.8. The distance will just be Perp v u. Figure.9. Distance to a Plane. EXAMPLE.. Find the distance from t 5. 5 to x = 5 + EXAMPLE.. Find the distance from v = 55 to the plane x + x + x = 6.

19 .5. Projections 9 We will need any point that lies on the given plane to use as a reference point. We can use the x intercept of the plane, x = 5. We will also need a normal to the plane, n = 5. Step. Translate the point and the plane so that the plane passes through the origin. We can do this by subtracting x from each. The normal to the plane would not change. The point v would be be translated to v x = 5 5. Step. Project v x onto n. Step. The distance we want is just the length of this projection. There is a simple formula for finding the distance from a point to a hyperplane. Suppose we have the hyperplane n T x = n T u and the point x. The point u lies on the hyperplane and so we can translate the hyperplane and point by subtracting u. The hyperplane then becomes n T x = and the point becomes x u. Next we project the point onto the normal to the hyperplane. This projection would be n T (x u) n T n n The distance is just the length of this projection and this length is n T (x u) nt n T n n = (x u) n n n T x n T u = n Now if we apply the above to a hyperplane» in R (that is a line) a with equation ax + by = c then n = and n b T u = c. If we let» x x = then the above formula gives us the distance y but to find the point on the plane that lies this distance from x requires that we go back to basics. A normal to the plane is given by n = 5. The x intercept of the plane is 5. If we treat this intercept as the origin the vector v = 5 5 = 5 is a vector from the plane to x. Then ( ) + () + () Proj nv = 5 = 8 5 = / 8/ /. It then follows that the point closest to x is / / x Proj nv = 5 8/ 5 = /5 / 9/ We will check this answer. Does / /5 lie on the plane? Substitution into the left hand side of the equation of the plane 9/ gives / + ( /) + (9/) = / / + / = 8/ = Further computation will confirm that the distance from / /5 9/ to x is 8/. A General Method for Finding Distances Every problem that involves finding a distance can be reduced to a situation as depicted in Figure.. ax + by c a + b If we apply the formula to a hyperplane in R with equation ax + by + cz = d then n = a b5 and n T u = d. If we let x = x y 5 c z then the above formula gives us the distance ax + by + cz d a + b + c (a)? (b)? EXAMPLE.5. Find the point on the plane x + y + z = closest to x = 5. It is easy to find the distance from x to the plane. The formula derived above tells us that this distance will be + () + () = 8 Figure.. Distance (a) to a line (b) to a plane. In Figure. (a) the distance you are looking for is the distance from one vector to the line generated by another vector. In Figure. (b) the distance is the distance from one vector to the plane generated by two other vectors. You should be able to see how this pattern could extend to higher dimensional situations with hyperplanes.

20 Chapter. Vector Geometry Such problems can always be seen as finding the distance to a subspace from some vector not in the subspace. So now suppose we want to find the distance from u to the subspace spanned by v,v,...,v k (where these vectors are assumed to be linearly independent). We will illustrate a method for obtaining this information (and more) that seems almost magical. If you want an explanation of why this method works you will have to take a more advanced course in linear algebra. Here s the method: Let A = ˆv v v k u. Compute A T A (this is sometimes called the Gram matrix). Put the Gram matrix A T A in row echelon form using only addrow operations (no row swaps or multiplying rows by constants). This amounts to finding matrix U of the LU decomposition that we saw earlier. The square root of the entry in the lower right corner is the distance you are looking for!!! EXAMPLE.6. Find the distance from v = 6 to the plane 5 generated by u = 6 5 and u = 6 5. Let A = 6 5. Then AT A = 8 5. Now 8 9 row reduction gives 8 5 5/ /5 5/ /5 8 9 / 5/ /5 So the distance from v to the span of u and u is p /5. But there is more information given by this reduced matrix. The square root of the first diagonal entry is the distance of u from the origin. The square root of the second diagonal entry is the distance from u to the line generated by u. The square root of the third diagonal entry is the distance of v from the plane generated by u and u. Distance, distance, distance. Also the product of the three diagonal entries is the square of the volume of the parallelepiped formed by the columns of A. The product of the first two diagonal entries is the square of the area of the parallelogram formed by u and u. The first diagonal entry is is the square of the length of u. Length, area, volume!!! EXAMPLE.. Given the points A(,,,), B(,,, ), C(,,, ) in four dimensions find the distance from C to the line through A and B. The first step is to restate the problem in terms of vectors in R. It should be clear that the problem is equivalent to finding the distance from AC to the line generated by AB. Now AB = 6 5 and AC = 6 5. So let A = 6 5 and then AT A =». One elementary row operation is all that s needed to 6» put A T A in row echelon form giving so the distance 8/5 we are looking for is p 8/5. We can also see that (8/5) = 56 so the area of the parallelogram formed by AB and AC is 56. EXAMPLE.8. This example is trickier and it is left up to you to figure out why it works. Find the distance from the line x = t 6 5 to the plane x = r s 6 5. Let A = 6 5. So AT A = 6 5 which reduces to 6 5/ / / 8/5 /5 5. So the distance we want is / p /. There is actually a simple explanantion why this trick works but it a bit beyond the level of an introductory course. On the other hand if we look at the specific problem of finding the distance from a point to a line we can see why it works. Suppose we want to find the distance from v to the line generated by u. Then we would have A = ˆu v. The Gram matrix would be»» A T u T u u u v A = ˆu v = v T u v v v One row operation reduces this to» u u u v v v u v u u u v The entry in the bottom right can be written as v (u v) u = v v cos θ If you ve followed the reasoning so far you should be able to finish it yourself. Notice also the fact that the determinant of the Gram matrix gives the square of the volume implies the Cauchy-Scwarz theorem that was mentioned earlier in this chapter.

21 Problems Problems. Find Proj v u and Perp v u for»» a. u =, v = 5 b. u = 5 v = 5. Find the distance from x to the given line»» a. x =, x = t»»» b. x =, x = + t 5 c. x = 5, x = 5 + t 5 d. x = 6 5, x = t 6 5 c. u = 6 5, v = 6 5 d. u = i + j k, v = j 5k e. u = i + j, v = j + k. Find the distance from x to the given hyperplane» a. x =, x y = a. x = 5, x + y c. x = 6 5, x + x + z = b. x = x x = 5 5, x+y z = 6. Let v = 6 5, v = 6 5, v = 6 5, v = 6 5. Find the distance a. from v to the span of v. b. from v to the span of v and v. c. from v to the span of v, v and v. b. x = 5, x y z = c. x = 5, x + y + z = 9 d. x = 6 5, x + x + x + x =. Find the point on the given line that is closest to x»»» a. x =, x = + t b. x = 5, x = 5 + t 5 c. x = 5, x = 5 + t 5 5. Find the point on the given hyperplane that is closest to x

22 Chapter. Vector Geometry The Cross Product The cross product is defined only for vectors in R. Given vectors u = u u 5 and v = v v 5 then the cross product of these vectors, u v written u v, is defined by i j k u v = u u u v v v Expanding this determinant by cofactors along the first row we get v θ v sin (theta) u u v = (u v u v )i (u v u v )j + (u v u v )k 5 First of all notice that the cross product of two vectors in R is another vector in R. There are a few properties of the cross product that follow directly from this definition. Try to identify the property of determinants that result in the following. u u = for any u R. u v = v u. u (u v) = v (u v) =. u (kv) = ku v u (v + w) = u v + u w The first property implies that if two vectors in R are parallel then their cross product is. The second property says that changing the order of the factors reverses the sense of the cross product. The third property says that the cross product of two vectors is always orthogonal to both of the original factors. The right hand rule says that u v points in the direction of your right hand thumb if you point the fingers of your right hand along u so that the fingers curl in the direction of v. EXAMPLE.9. Let then u = 5 v = 5 5 i j k u v = = 8i + 5j + 9k 5 Notice that this cross product is orthogonal to each of the original factors as the following dot products show = = = = 9 Another important property of the cross product is u v = u v sin θ Figure.. where θ is the angle between u and v. The proof of this is left as an exercise. This formula means that the magnitude of the cross product of two vectors gives the area of the parallelogram determined by the two vectors. This is illustrated in Figure. EXAMPLE.. Given points A(,, ), B(5, 5, ) and C(, 5, ) find the area of triangle ABC. As usual, look at the sides of the triangle as vectors. The vectors AB = 5 and AC = 5 determine a parallelogram. The area of this parallelogram is AB AC = = 9 Triangle ABC is half of this parallelogram so the area of triangle ABC is 9. EXAMPLE.. We know that a plane in R can be represented by an equation of the form x = x + su + tv where u and v are linearly independent. Suppose we take the dot product of both sides of this equation with u v. We get x (u v) = (x + su + tv) (u v) = x (u v) + su (u v) + tv (u v) = x (u v) + + Let n = u v, this will be a normal vector to the plane. The above computations can then be written x n = x n or n T x = n T x, another familiar equation of the plane. There is another way of looking at the cross product. Let u = a b5 A = c b c a5 x = x x 5 c b a x

23 Problems then Ax = c b c a5 x x 5 = cx + bx cx ax 5 = u x b a x bx + ax So a cross product can be seen as a linear transformation. The scalar triple product Three linearly independent vectors u, v and w in R determine a parallelepiped as shown in Figure.. height v x w u This means that the absolute value of a determinant gives the volume of the parallepiped determined by the rows of the determinant (or the columns of the determinant since a matrix and its transpose have the same determinant). EXAMPLE.. Find the volume of the parallelepiped determined by u = 5, v = 5, w = 5. From the previous comments the volume can be easily computed using a determinant 5 with the given vectors as rows (or columns). = = 5 The volume is the absolute value of the determinant so the volume is. w Figure.. The parallelepiped determined by u, v and w. We want to find a formula for the volume of this parallelepiped. Basic geometry tells us that the volume is the area of the base times the height. Suppose the base is determined by vectors v and w. Then the area of the base is v w. To find the height of the parallelepiped we can project the third vector u onto a normal to the base. But v w is normal to the base so the height is the length of u (v w) Proj v w u = v w v w This means that the height is u (v w) v w The volume of the parallelpiped is therefore given by Volume = (area of base) u (v w) = v w v w = u (v w) v times (height) The value u (v w) is called the scalar triple product of u, v, and w. It s a bit surprising but the scalar triple product can be computed as a determinant. u (v w) = u i j k u 5 v v v u w w w = u u u v v v w w w EXAMPLE.. Find the distance from u to the plane generated by v and w where u = 5 v = 5 w = 5 There are several ways of solving this problem. The method we will use is based on the geometric insight that the distance we want is just the height of the parallelepiped with a base determined by v and w with the third side being u. The volume of this parallelepiped is u (v w) = = + 8 = 6 The area of the base is v w = = 8 The height is just the volume divided by the area of the base so the distance we are looking for is 6 8 = Another method would be to project u onto v w as illustrated in Fig.. This gives u (v w) (v w) (v w) v w = The distance we want is the length of this projection which is 6 9 = = 9 This is equivalent to the answer found above.