Unit vectors. A unit vector is a vector whose length is 1. We can interpret unit vectors as being directions, and we can use them in place of angles Length, dot products, and since they carry the same information as an angle. Unit vectors can be identified with points on the 3 cross products in R unit sphere Math 131 Multivariate Calculus D Joyce, Spring 2014 S2 = {(x, y, z) | x2 + y2 + z2 = 1}. Length, dot products, and cross products to- Now that we can talk about the length of a vec- gether allow us to do geometry in three dimensions. tor, we can construct a unit vector in the same This page is a quick review of them, not a complete direction as a given vector simply by dividing by study. its length. If a is a vector in R3, the unit vector in the same direction is Length and distance. The length k(a , a , a )k 1 2 3 a of a vector a = (a1, a2, a3) (also called its norm) is u = . defined as kak q 2 2 2 We can rewrite that equation as k(a1, a2, a3)k = a1 + a2 + a3. The length of a vector a is just the distance from a a = kak u to the origin 0 = (0, 0, 0). The distance between two vectors a = (a1, a2, a3) which says that a vector a is the product of its and b = (b1, b2, b3) is the length of the displacement length and its direction. vector b − a between them

k(b1, b2, b3) − (a1, a2, a3)k = Dot products. The dot products a · b of vec- p 2 2 2 (b1 − a1) + (b2 − a2) + (b3 − a3) . tors (sometimes called inner products and denoted Scaling a vector, that is, multiplying it by a con- ha|bi) is the sum of the products of corresponding stant c changes its length by a factor of the absolute coordinates, that is, value of c a · b = (a , a , a ) · (b , b , b ) kcak = |c| kak. 1 2 3 1 2 3 = a b + a b + a b The triangle inequality holds for vectors 1 1 2 2 3 3 ka − bk ≤ kak + kbk. Notice right away that we can interpret the square of the length of the vector as an inner prod- b uct. Since H ¡ H 2 2 2 2 H kak = a1 + a2 + a3, ¡ HH ¡ H ¡ HHj therefore 1 a 2 ¡  kak = a · a. ¡   ¡  Because of this connection between norm and inner ¡  0 product, we can often reduce computations involv- ing length to simpler computations involving inner It expresses the fact that one side of a triangle products. a − b is no longer than the sum of the other two Inner products are commutative: kak + kbk. See Euclid’s Elements Proposition I.20 for Euclid’s proof of this inequality. a · b = b · a.

1 Also, inner products distribute over addition,

H a · (b + c) = (a · b) + (a · c), ¡ H ¡ HHb − a H ¡ HH ¡ Hj and over subtraction, 1 a¡  ¡   b ¡θ  a · (b − c) = (a · b) − (a · c), ¡  and the inner product of any vector and the 0 vec- 2 2 2 tor is 0 kb − ak = kak + kvk − 2kak kbk cos θ. a · 0 = 0. We can convert the distances to dot products to simplify this equation. Furthermore, dot products and scalar products kb − ak2 = (b − a) · (b − a) have a kind of associativity, namely, if c is a scalar, then = b · b − 2b · a + a · a 2 2 (cu) · v = c(u · v) = u · (cv). = kbk − 2b · a + kak Now, if we subtract kak2 + kbk2 from both sides of These last few statements can be summarized by our equation, and then divide by −2, we get saying that dot products are linear in each coordi- nate, or that dot products are bilinear operations. a · b = kak kbk cos θ. That gives us a way of geometrically interpreting The dot product of two vectors and the co- the dot product. We can also solve the last equation sine of the angle between them. The law of for cos θ, a · b cosines for oblique triangles says that given a tri- cos θ = , angle with sides a, b, and c, and angle θ between kak kbk sides a and b, which will allow us to do trigonometry by means of linear algebra. Note that  a · b  ¡HH θ = arccos . ¡ HHc kak kbk H ¡ HH ¡ H  Orthogonal vectors. The word “orthogonal” is a¡  ¡  synonymous with the word “perpendicular,” but for  b ¡θ  ¡  some reason is preferred in many branches of math- ematics. We’ll write w ⊥ v if the vectors w and v are orthogonal, or perpendicular. c2 = a2 + b2 − 2ab cos θ.

Now, start with two vectors a and b, and place them in the plane with their tails at the same point. BBM B Let θ be the angle between these two vectors. The 1 v B  vector that joins the head of a to the head of b is B  w B  b − a. Now we can use the law of cosines to see B that

2 Two vectors are orthogonal if the angle between area is the length of u times the length of v times them is 90◦. Since the cosine of 90◦ is 0, that means the sine of the angle θ between them. Thus

w ⊥ v if and only if hw|vi = 0 ku × vk = kuk kvk cos θ

Two vectors are orthogonal, a ⊥ b, if and only if The direction of u × v will be orthogonal to the their dot product is 0. plane of u and v in a direction determined by a right-hand rule (when the coordinate system is right-handed). Projections of one vector onto another. The The easiest way to define cross products is to use projection of one vector b onto another vector a is the standard unit vectors i, j, and k for R3. If defined to be the vector u = (u , u , u ) = u i + u j + u k, b · a 1 2 3 1 2 3 projab = a. kak2 and v = (v , v , v ) = v i + v j + v k, It’s the component of b in the direction of a. This 1 2 3 1 2 3 projection of one vector on another is useful in an- then u × v is defined as swering some questions about geometry using linear algebra. u×v = (u2v3−u3v2)i+(u3v1−u1v3)j+(u1v2−u2v1)k which is much easier to remember when you write The definition of cross products. The cross it as a determinant product × : R3 × R3 → R3 is an operation that u u u u u u takes two vectors u and v in space and determines u × v = 2 3 i − 1 3 j + 1 2 k v v v v v v another vector u × v in space. (Cross products are 2 3 1 3 1 2 sometimes called outer products, sometimes called i j k vector products.) Although we’ll define u × v al- = u1 u2 u3 gebraically, its geometric meaning is more under- v1 v2 v3 standable. Properties of cross products. There are a whole lot of properties that follow from this defi- COC C nition. First of all, it’s anticommutative C  C  ¡ v × u = −(u × v), C  ¡  C  ¡ C  ¡ so any vector cross itself is 0 C ¡ ¡ u × v C ¡ ¡ C ¡ ku × vk ¡ v × v = 0. C v ¡ ¡ 1 C ¡  It’s bilinear, that is, linear in each argument, so it C ¡ θ  ¡C  u distributes over addition and subtraction, 0 acts as C C¡C  C¡ C zero should, and you can pass scalars in and out of arguments

The cross product u×v is determined by its length u × (v ± w) = (u × v) ± (u × w) and its direction. It’s length is equal to the area of the parallelgram whose sides are u and v, and that (u ± v) × w = (u × w) ± (v × w)

3 0 × v = 0 = v × 0 angle between u and v. Therefore, the area of this triangle is c(u × v) = (cu) × v = u × (cv) 1 1 A couple more properties you can check from the Area = 2 kuk kvk sin θ = 2 ku × vk. definition, or from the properties already found are that hu×v|ui = 0 and hu×v|vi = 0. Those imply (In general, the area of a any triangle is half the that the vector u × v is orthogonal to both vectors product of two adjacent sides and the sine of the u and v, and so it is orthogonal to the plane of u angle between them.) and v. Area of a parallelogram in R3. Now consider Standard unit vectors and cross products. a parallelogram in 3-space where two of the sides Interesting things happen when we look specifically are u and v. at the cross products of standard unit vectors. Of  course  ¡  ¡  i × i = j × j = k × k = 0,  ¡  ¡ H since any vector cross itself is 0. But ¡ HH ¡ ¡ H ¡ ¡ HH ¡ i × j = k, j × k = i, k × i = j, ¡ HH¡ 1 u¡  ¡  and  v ¡θ  ¡ j × i = −k, k × j = −i, i × k = −j, all of which follows directly from the definition. Of course, if the triangle is doubled to a parallelo- gram, then the area of the parallelogram is ku×vk. Thus, the norm of a cross product is the area of Length of the cross product, areas of trian- the parallelgram bounded by the vectors. gles and parallelograms. A direct computation We now have a geometric characterization of the (which we’ll omit) shows that cross product. The cross product u×v is the vector ku × vk = kuk kvk sin θ orthogonal to the plane of u and v pointing away from it in a the direction determined by a right- where θ is the angle between the vectors u and v. hand rule, and its length equals the area of the Consider a triangle in 3-space where two of the parallelgram whose sides are u and v. sides are u and v. Note that u × v is 0 if and only if u and v lie in a line, that is, they point in the same direction or ¡HH the directly opposite directions. H ¡ HH ¡ H Math 131 Home Page at ¡ HH 1 http://math.clarku.edu/ djoyce/ma131/ u¡  ~ ¡   v ¡θ  ¡ 

Taking u to be the base of the triangle, then the height of the triangle is kvk sin θ, where θ is the

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