MATH1002 Lecture Notes - Lecture 3: Unit Vector

In proving the Pythagorean theorem in class, we saw that |u vector + v vector |^2 = ||u vector ||^2 + 2 u vector middot v vector + ||v vector ||^2. Moreover, from the "physics'" definition of the dot-product, i.e. u vector middot v vector = ||u vector || ||v vector || cos (theta) and the fact that cos (theta) lessthanorequalto 1, it is obvious that u vector middot v vector lessthanorequalto ||u vector|| ||v vector|| (called the Schwarz inequality). Use both of the above properties of dot-products and norms to prove the triangle inequality. All stops must follow loyally from one another, and must be justified.

Let V be a nonemp an addition operation and a scalar multiplication operation with scalars V a vector space over F, provided the following ten pty set (whose elements are called vectors) on which are defined in F. We call conditions are satisfied: AL. Closure under addition: For each pair of vectors u and v in V, the sum u under addition: For each pair of vectors u and v in V, the sumu+v is also in V. We say that V is closed under addition. 2. Closure under scalar multiplication: For each vector v in V and each scalar k n F, the scalar multiple kv is also in V. We say that V is closed under scalar multiplication A3. Commutativity of addition: For all u, v V, we have u+v=v+u. A4. Associativity of addition: For all u, v, w e V, we have (u + v) + w = u + (v + w). A5. Existence of a zero vector in V: In V there is a vector, denoted 0, satisfying v+0=v, for all v E V. A6. Existence of additive inverses in V: For each vector v In V, there is a vector denoted -V, in V such that v+1-v) = 0. A7. Unit property: For all veV Iv=v. A8. Associativity of scalar multiplication: For all v e V and all scalars r, sEF. (rs)vr(sv). A9. Distributive property of scalar multiplication over vector addition: For all u V E V and all scalars r EF r(u + v) = ru + rv. A10. Distributive property of scalar multiplication over scalar addition: For all v e V and all scalars r,SEF (r+s)v = rv + sv.

One way of describing a plane is by giving a point P in the plane and two directions v and w in the plane. From this data we can describe the plane as the points you can get by adding to P every possible weighted sum of v and w, i.e. av + bw for various scalars a and b. This description is much like our parametric description of a line, except we use two directions instead of one. [4 points] We have seen that you get the same line if you rescale the direction vector by a nonzero scalar. What is the analogue for planes? That is, if you have two vectors v and w describing a plane, which other pairs of direction vectors will give the same plane? [2 points] Explain why the osculating plane is the plane determined by the velocity and acceleration vectors. [4 points] We call a set of vectors linearly dependent when we can write one vector in the set as a weighted sum of the others. For example, {(1,0,0), (0,1,0), (1,2,0)} is linearly dependent since (1,2,0) = (1,0,0) + 2(0,1,0). So while it's easy to show that a set is linearly dependent, it's trickier to show that it isn't. A set which is not linearly dependent is called linearly independent. To show that a set {u, v, w} is linearly independent, we must show that the only way to have an equation au + bv + cw = 0 is if all the scalars a,b,c are 0. This is because if a is nonzero, then we'd have u = b/a v + c/a w, so the set would be linearly dependent. So here's the problem: If two 3-dimensional vectors v and w are orthogonal, show they must also be linearly independent. (Hint: We have to show that if av + bw = 0), then a and b are both 0. We can do this using the dot product.)