1 Vector Algebra We will now define addition, subtraction and scalar multiplication of vectors. Let v; w 2 Rn then, v ± w = hv1 ± w1; v2 ± w2; ··· ; vn ± wni λv = λhv1; v2; :::; vni = hλv1; λv2; :::; λvni where v = (v1; :::; vn) and w = (w1; :::; wn) and λ 2 R. There is some underlying geometry to the addition of vectors. By our equivalence definition, we can proceed without loss of generality assuming v = ha; bi; w = hc; di and their base-points are at the origin. 2 Due to the geometry above, we say that the sum of two vectors obey the parallelogram law. We will now record some basic properties of vector algebra which will given as exercises. From now on I will choose not to bold vectors i.e v = v. Fact: (Basic Properties of Vector Algebra) For all vector v; u; w and for all scalars λ, • v + w = w + v • u + (v + w) = (u + v) + w • λ(v + w) = λv + λw Exercises (1) Sketch the vectors for v + tw where t = 0; ±1; ±2; ±3. Is there a geometric object in which the heads are describing? (2) Let v be a nonzero vector and λ be a nonzero scalar. Show that kλvk = jλjkvk. Why do we need the absolute value? (3) A unit vector v is a vector such that kvk = 1. Given any vector nonzero vector v in the plane, can you think of a scalar λ such that kλvk = 1? Hint: Use the question above! (4) Let i = h1; 0i and j = h0; 1i. Show that any vector v in the plane can be written as a linear combination of i; j i.e there exists s; t 2 R s.t ai + bj = v. (5) Verify the facts for the basic properties of vector algebra. 3 Vectors in Three Dimensions A point in R3 = R × R × R is of the form (a; b; c) where it is clear from the decomposition that a; b; c 2 R. A vector in R3 is of the form ha; b; ci. Fact: Given P = (a1; b1; c1) and Q = (a2; b2; c2) then d(P; Q) i.e kP − Qk is given by, p 2 2 2 kP − Qk = (a2 − a1) + (b2 − b1) + (c2 − c1) Parametric Equations of a Line Given P0 = (p0; p1; p2) and v = hv0; v1; v2i we have that a parametrization for the line through P0 in the direction of v is given by, r(t) = OP0 + tv = hp0; p1; p2i + thv0; v1; v2i 1 Exercises (1) Use vector geometry to show that a line can be parametrized in the way which is given in the beginning of this section. Verify this for both plane and space lines. (2) Show that parameterizations for space lines aren't unique. Hint: Think of an explicit example! (3) Let r1(t); r2(s) be two parametrizations for lines in 3-space. Come up with a definition for the intersection of two lines. With this definition verify whether r1(t) = h1; 0; 1i + th3; 3; 5i and r2(s) = h3; 6; 1i + sh4; −2; 7i intersect. (4) Let P; Q 2 R3. Show that g(t) = (1 − t)P + tQ where −∞ < t < 1 is a parametrization for the line through P; Q. With this definition show that the midpoint of the segment from P to Q is g(:5). 2 Dot Product and the Angle Between Two Vectors Let v = ha1; b1; c1i and w = ha2; b2; c2i then the dot product of v; w is defined by, v · w = a1a2 + b1b2 + c1c2 Theorem: Properties of the Dot Product (1) 0 · v = v · 0 = 0 (2) v · w = w · v (3) (λv) · w = v · (λw) = λ(v · w) (4) u · (v + w) = u · v + u· and (v + w) · u = v · u + w · u (5) v · v = kvk2 Theorem: (Dot Product and the Angle) Let 0 ≤ θ ≤ π be the angle between two nonzero vectors v; w then we have that, v · w = kvkkwk cos(θ) Theorem (Projection) Assume v is a nonzero vector. The projection of u along v is the vector, u · v u = v jj v · v Lemma (Decomposition) If v is a nonzero vector, then every vector u can be written as the sum of the projection ujj and a vector u? that is orthogonal to v. 1 Exercises (1) Prove the angle between vectors formula. Hint: If you have a triangle with sides being the vectors v; w, what is the third side? (2) If two nonzero vectors u; v are perpendicular then what is u · v ? (3) Use the above question to prove the projection formula? (4) In the case of projecting u onto a vector v, show that u may be decomposed as suggested in the lemma. 2 The Cross Product To being this section we will just recall some facts from basic linear algebra. We will start with the determinant of a 2x2 and 3x3 matrix. Let A be a 2x2 matrix with real entries then the determinant of A is given by, a b det(A) = = ad − bc c d The determinant of a 3x3 matrix is expressed in terms of 2x2 determinant called minors. a1 b1 c1 b2 c2 a2 c2 a2 b2 det(A) = a2 b2 c2 = a1 − b1 + c1 b3 c3 a3 c3 a3 b3 a3 b3 c3 Definition (The Cross Product) The cross product of vector v = ha1; b1; c1i and w = ha2; b2; c2i is the vector, i j k b1 c1 a1 c1 a1 b1 v × w = a1 b1 c1 = i − j + k b2 c2 a2 c2 a2 b2 a2 b2 c2 Theorem (Geometric Description of the Cross Product) The cross product v × w is the unique vector with the following three properties; (1) v × w is orthogonal to v; w (2) kv × wk = kvkkwk sin θ where θ is the angle between v; w and 0 ≤ θ ≤ π (3) fv; w; v × wg forms a right-handed system Theorem (Basic Properties of the Cross Product) Below are some properties of the cross product which you will verify in the exercises. (1) w × v = −v × w (2) v × v = 0 (3) (λv) × w = v × (λw) = λ(v × w) (4) (u + v) × w = u × w + v × w (5) u × (v + w) = u × v + u × w 1 Cross Products, Area and Volume We will now give some more geometric properties of the cross product and the determinant function. Below are images of a plane spanned by two vectors and a parallelpiped spanned by three vectors. Theorem (Area and Volume via Cross Products and Determinants) Let u; v; w be nonzero vectors in R3. Then we have the following relations (show in exercises). (1) The parallelogram P spanned by v; w has area A = kv × wk 0 1 u (2) The parallelepiped P spanned by u; v; w has volume V = ju · (v × w) = det @v A w Exercises (1) Verify all the basic properties of the cross product (2) Compute the area of a general parallelogram in the plane and volume of a parallelopiped in 3-space. (2) Prove the theorem above in reference to area of parallelograms and parallelpipeds spanned by nonzero vectors. (3) Prove the scalar triple product and show that v × w is orthogonal to both v; w where each are nonzero. 2 Planes in Three-Space Before exploring planes in 3-space, let us revise our knowledge of line equations in the plane using vectors. Let p0 be a point in the plane and ~p0 be the corresponding position vector. If we choose v to be the direction of the line passing through p0 we have the vector parametrization r(t) = ~p0 + tv. Here if we let v = hv1; v2i then for (x; y) on the line parametrized by r(t) we have h(x; y) − (x0; y0)i ·h−v2; v1i = 0. | {z } ~u This follows from the fact ~u is in the direction of v which is orthogonal to h−v2; v1i. Simplifying the dot product above we recover −v2x + v2x0 + v1y − v1y0 = 0 ) ax + by = c where a = −v2; b = v1 and c = −v2x0 − v1y. This standard representation of a line in the plane that we are used to. We will now show that this formulation carries on into 3-space i.e the equation of a plane is given by ax + by + cz = d. Consider a plane P passing through a point (x0; y0; z0). Let ~n = ha; b; ci be a nonzero vector such that for 3 any (x; y; z) 2 R we have h(x; y; z) − (x0; y0; z0)i · ~n = 0. Geometrically, the set of (x; y; z) such that the above equation is true gives a plane passing through the point (x0; y0; z0). If we simplify the notation above we have that ax + by + cz = ax0 + by0 + cz0. | {z } = d Given three points P; Q; R which are not collinear, there is a unique plane which passes through all of them. The uniqueness of this plane will come from the fact that it contained the lines which pass through P; R and P; Q i.e having common intersection P . Choosing P; Q; R as described above then the plane is given by h(x; y; z) − P i · (PQ~ × PR~ ). Notice that to compute this equation we needed a point on the plane, therefore we could have chosen P; Q or R. 1 Exercises (1) Try to come up with a definition which says when two planes are parallel i.e they don't intersect. Hint: Think of normal vectors! (2) Verify that the planes which use P; Q or R as points in the above image, describe the same plane.