Notes 1 – Geometric vectors in the plane and in the space L1.1 Applied vectors and free vectors. The typical treasure map gives information to you using applied vectors. From the oak three walk 30 steps est and then walk 40 steps north. So, in the first part of our treasure hunt, we start from a point O and then we move along the west-est direction, in the verse towards est, walking a distance of magnitude 30 steps. This task can be encoded by an arrow → (vector) starting at the point O (application point). This is an example of an applied vector. In Physics applied vectors are widely used. Think of a force applied to body, of the speed of a point, and so on. In Mathematics we want to separate the information encoded in applied vectors in two pieces: application point + vector. − Definition. A free vector → v is completely determined by − (i) the direction of the line containing → v, − (ii) the verse following which we move along the line according to → v, → − → − (iii) the magnitude (or length) of v denoted as || v ||. Definition. An applied vector is a free vector with an application point. Note that for each free vector there are infinitely applied vectors, one for each possible application point. This means that different applied vectors can correspond to the same free vector. Exercise. When is it the case that different applied vectors correspond to the same free vector? Example. A common way to produce applied vectors is to take two points, say A and B . Then −−−−−→ −−→ −−→ we write AB , or (B − A) , to represent the applied vector going from A to B . Notice that BA −−→ as the same direction and magnitude of AB but opposite verse. −→ Exercise. What kind of vector is AA? There are different ways to visualize the set of free vectors of the plane; one is the following. Fix a point O and apply all free vectors to this point. Then, the free vectors of magnitude R are in bijection with the points of the circumference of center O and radius R. Exercise. How can we visualize free vectors in three dimensional space? L1.2 Operation with vectors. Natural numbers arose naturally to count objects, and for a long while there was no place for the number zero (no objects). However, if we want not only to count, but also to perform operations, zero is of crucial importance. → − Definition. The zero vector 0 is the only vector of null magnitude. It does not have direction or verse. We first see how to multiply a vector by a real number c ∈ R which we call a scalar. → − − − − Definition. Let c ∈ R and → v be a free vector. If c = 0 then c→ v = 0→ v = 0 . If c 6= 0 then − c→ v is the vector: − (i) having the same direction of → v, → − (ii) having the same verse of v if c > 0 and opposite verse if c < 0, − − (iii) having magnitude ||c→ v || = |c| ||→ v ||. Exercise. Define the multiplication by a scalar for applied vectors. − − Note that c→ v is obtained by contracting or dilating the vector → v . Also, take note of the − absolute value of c appearing in the definition. What would be the problem with c||→ v ||? Let’s go back to our treasure hunt. To reach X , we have to start from O , walk 30 steps est and 40 steps north. Or, we can move 50 steps north-est from O and then reach X . In other −−→ words, we can move along the diagonal OX instead of following the two sides. −→ − − Definition. (Parallelogram Rule) Given free vectors → u and → v , their sum − u−+ v is obtained → − → − in the following way. Apply u and v at the same point O and consider the parallelogram −→ − − having → u and → v as two consecutive sides. Then, − u−+ v is the diagonal of the parallelogram starting from O . Exercise. Work out the sum of two parallel vectors. Exercise. What can we say about the sum of applied vectors? Is it always defined? The multiple by a scalar and the addition of vectors have many useful properties. These properties make computing expressions with vectors very similar to computing expressions with real numbers. − − − Proposition. (Basic Properties) Let c, d ∈ R and → u,→ v and → z be free vectors, then the following hold: − − (i) c(d→ u ) = (cd)→ u, → − − − (ii) (c + d) u = c→ u + d→ u, − − − → − − − → (iii) u + v = v + u, − − − − − − (iv) → u + (→ v +→ z ) = (→ u +→ v)+→ z, − − − − (v) c(→ u +→ v ) = c→ u + c→ v. Proof. Let’s see how to prove some of the properties above. (i) If cd = 0, then both vectors are the zero vector by definition. Assume c and d are not − − − − zero. We need to show that the vectors → w = c(d→ u ) and → z = (cd)→ u are equal. To do this we show that their direction, verse, and magnitude are the same. By definition of multiplication − − − − by a scalar, → w has the direction of → u and similarly for → z . If cd > 0, then the verse of → w is − − the one of → u and similarly for → z ; if cd < 0 the same argument applies. Finally, − − − − − ||→ w || = |c| ||d→ u || = |d| |c| ||→ u || = |dc| ||→ u || = ||→ z ||, and we are done. − − − − − − − − (v) Let → w = c(→ u +→ v ) and → z = c→ u + c→ v . Again, we want to show that → w = → z and → − → − − this is obvious if c = 0. Thus, we assume c 6= 0. To determine w we make u and → v to → − → − − share their tail, then we consider the parallelogram having u and v as sides. Then, → w has the direction of the diagonal of the parallelogram starting in the common tail, same or opposite verse depending on the sign of c and the magnitude is obtained multiplying by |c| − the magnitude of the diagonal. The vector → z is determined similarly, but considering the previous parallelogram scaled by |c|. Hence the desired equality follows. QED Exercise. Prove the remaining properties. − − − If we want to sum the vectors → u,→ v and → z , the result does not depend on the order in − which we compute. A handy way of doing this is concatenation. Apply → u in O , then → − → − → − → − − apply v in the head of u , and finally apply z in the head of u . If the head of → z is now − → − − − in the point A, then → u +→ v +→ z = OA. Exercise. Using twice the Parallelogram Rule, prove that the previous way of computing the sum of vectors produce the right answer. − − Using the previous Proposition, we can show that (−1)→ v can be reasonably called −→ v . In → − → − → − → − → − fact, (−1) v differs from v only for the verse which is the opposite. Thus, (−1) v + v = 0 Exercise. If the previous argument seems too obvious, read it again and be sure to understand each step. Exercise. Take a triangle and see its sides as vectors by assigning arrows of your choice. Now take the sum of these three vectors. What do you have to do to obtain the zero vector? Repeat for a square. L1.3 Components. Vectors of magnitude one play a special role and they deserve a name. Definition. A vector of magnitude one is called a versor or unit vector. → − u is a unit vector. − Exercise. Given a vector → u , show that → ||− u || To describe a point in the plane or in three space we can use coordinates. In order to do this in the plane, we fix two orthogonal, oriented axes and we choose a way to measure lengths on these axes. Then the position of each point is completely determined by taking orthogonal projections. The same can be done with vectors. Each coordinate, oriented axis provides us with a unit → − → − → − vector. In three space these are usually called i , j , and k . Using the concatenation description of the sum of vectors, we can represent each free vector as a linear combination of the same three unit vectors. → − − − Definition/Proposition. Any free vector → v in three space can be written as → v = vx i + → − → − vy j + vz k , for vx , vy , vz ∈ R. The numbers vx , vy , and vz are called the components of → − v with respect to the coordinate axes. In two space, i.e in the plane, simply there is no vz component. Exercise. Use orthogonal projection to find the expression in components of any vector. → − → − → − − Example. If we think the vector → v = i + 2 j + 3 k as applied in the intersection of the − coordinate axes, then the head of → v is in the point of coordinate (1, 2, 3). Operation on vectors can be easily performed taking advantage of components and of the Basic Properties. → − → − → − → − → − → − − − Example. Consider the vectors → v = i + 2 j + 3 k and → w = 3 i + 2 j + 1 k . Then, by applying the basic properties, we get → − → − → − − −→ − u−+ v = 4 i + 4 j + 4 k = 4→ z, − → − → − → − where → z = i + j + k. From this example it is possible to derive a general rule. Proposition. (Linear Combination of Vectors) Let a, b ∈ R. If we consider the vectors → − → − → − → − → − → − − → − w = wx i + wy j + wz k , then v = vx i + vy j + vz k , and → → − → − → − − − a→ v + b→ w = (avx + bwx ) i + (avy + bwy ) j + (avz + bwz ) k . Proof. We propose to arguments, one algebraic and the other geometric. Algebraic The result simply follows by applying the Basic Properties (i)-(v) to the expressions in components of → − − v and → w . Geometric First we notice that the proof can be split in two steps: the case a = 0 and then the case a = b = 1. To prove the proposition, it is enough to use orthogonal projections to determine the components. For example, in the case a = 0 and b > 0, we − − − have to determine the components of b→ w . As b→ w is a dilation, or contraction, of → w by the factor b the result on the components follows. QED Exercise. Why it is enough to prove the Proposition for a = 0 and then the case a = b = 1?
