MAT 1341: REVIEW II SANGHOON BAEK


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1 MAT 1341: REVIEW II SANGHOON BAEK 1. Projections and Cross Product 1.1. Projections. Definition 1.1. Given a vector u, the rectangular (or perpendicular or orthogonal) components are two vectors u 1 and u 2 such that u = u 1 + u 2 and u 1 u 2 ( u 1 and u 2 are perpendicular to each other). Definition 1.2. Given two vectors u and v, the projection of u on v, written by proj v u, is the rectangular component of u in the direction of v if 0 θ < π/2 or of v if π/2 < θ π. Remark 1.3. By definition of projection vector, we have the following rectangular decomposition: for any two nonzero vectors u and v. u = proj v u + ( u proj v u). Theorem 1.4. proj v u ( u v ) v = v. v v 2 Proof. Case 1: 0 θ < π/2 proj v u = u cos θ v = u ( u v ) u v v v v. v v Case 2: π/2 < θ π 1
2 Example 1.5. proj v u = ( u cos θ) ( v ) = u ( u v ) v u v v v. v v (1) Find the orthogonal projection of [1, 2, 3] along [ 4, 0, 5]. (2) Find a rectangular decomposition of u = [4, 0, 1], one of which is collinear with v = [1, 1, 0] Cross Product. Definition 1.6. Given two vectors u and v in threedimensional space, the cross product, written by u v, is a vector defined by u v := ( u sin θ) n, where 0 θ π is the angle between u and v, and n is the unit vector orthogonal to both u and u satisfying the righthand rule (i.e., as the fingers on your right hand curl from u to v, the direction of the thumb is the direction of n, thus of u v). As n = 1, we have Theorem 1.7. u v = u sin θ. The magnitude of the cross product u v can be interpreted as the positive area of the parallelogram having u and v as sides. In other words, Corollary 1.8. (1) The area of the parallelogram defined by u and v = u v. (2) The area of the triangle defined by u and v = 1 u v. 2 Example 1.9. Let i = [1, 0, 0], j = [0, 1, 0], and k = [0, 0, 1] be unit vectors. Verify the following relations: (1) i j = k, j k = i, and k i = j. (2) j i = k, k j = i, and i k = j. (3) i i = j j = k k = 0. Theorem Let u, v and w be vectors. Then the following properties hold. (1) (AntiCommutativity) u v = ( v u). 2
3 (2) (Distributivity) and u ( v + w) = u v + u w ( u + v) w = u w + v w. (3) (Scalar Multiplication) If t is a scalar, then t( u v) = (t u) v = u (t v). (4) (Orthogonality) If u and u are nonzero vectors, then u v = 0 if and only if u and v are collinear. Remark In general, the associativity does not hold in cross product (i.e., ( u v) w u ( v w)). For example, if u = i, v = i, and w = j, then we have ( u v) w = 0 and u ( v w) = j. Theorem Let u = [a 1, a 2, a 3 ] and v = [b 1, b 2, b 3 ]. Then Example u u = [a 2 b 3 a 3 b 2, a 3 b 1 a 1 b 3, a 1 b 2 a 2 b 1 ]. (1) Let u = [3, 4, 2] and v = [1, 3, 7]. Find all vectors in R 3 which are orthogonal to both u and v. (2) Let u = [0, 1, 2], v = [1, 2, 2], and w = [1, 0, 1]. Compute (2 u + v) w. Theorem The volume of the parallelepiped defined by three vectors u, v, and w is ( u v) w. Proof. By definition of the dot product, ( u v) w = u v w cos θ where θ is the angle between u v and w. Note that u v is the area of the base parallelogram. As w cos θ is the parallelepiped s height, u v w cos θ represents the volume of the parallelepiped. By treating the planes of v and w and of w and u as the base planes of the parallelepiped defined by u, v, and w, we get Example ( u v) w = ( v w) u = ( w u) v. (1) Find the area of the parallelogram defined by u = [1, 2, 2] and v = [2, 3, 1]. (2) Find the area of the triangle defined by P = (1, 1, 1), Q = (2, 3, 4), and R = (3, 2, 5). (3) Find the volume of the parallelepiped determined by u = [1, 2, 1], v = [ 2, 0, 3], and w = [0, 1, 2]. 3
4 2. Lines Suppose that L is a line in 3dimensional space passing through a point P 0 = (x 0, y 0, z 0 ) parallel to a nonzero vector v = [m 1, m 2, m 3 ]. Then, for any point P = (x, y, z) on the line L, the vector OP = [x, y, z] is equal to OP 0 + t v for some t R. In other words, the line L is represented by (1) [x, y, z] = [x 0, y 0, z 0 ] + t[m 1, m 2, m 3 ] for all t R. This is called a vector equation of the line L. From (1), we get (2) x = x 0 + tm 1 y = y 0 + tm 2. z = z 0 + tm 3. This is called a parametric equation of the line L. Example 2.1. Let L be a line passing through the points (1, 4, 3) and (2, 6, 5). (1) Write a vector equation for L. (2) Write a parametric equation for L. Example 2.2. Consider a line which is parallel to v = [1, 3, 3] and passes through a point (3, 2, 6). (1) Write a vector equation for L. (2) Write a parametric equation for L Intersection of Two Lines in 3dimensional space. Geometrically there are 4 possibilities for the intersection of two lines in R 3 : (1) Intersect at a point (2) Parallel (3) Coincident (4) Skew (=two different lines that are neither parallel nor intersecting). On the other hand, the intersection of two lines in R 3 can be represented by the solutions of the system of line equations: { [x, y, z] = [x 0, y 0, z 0 ] + t[m 1, m 2, m 3 ] (3) [x, y, z] = [x 0, y 0, z 0] + s[m 1, m 2, m 3], and the solutions of (3) correspond to above 4 possibilities as follows:  Unique solution Intersect at a point  No solution Parallel or Skew  Infinitely many solutions Coincident. 4
5 Example 2.3. Find the intersection of given two lines: [x, y, z] = [1, 2, 3] + t[4, 5, 6] and [x, y, z] = [0, 2, 1] + s[3, 4, 4] 2.2. The Distance from a Point to a Line. When we talk about distances, we usually mean the shortest distance. To find the distance from a point S to a line passing through a point P 0 parallel to a vector v, we find the length of the component of P 0 S normal to the line. As the length is P 0 S sin θ, where θ is the angle between v and P 0 S, we get Theorem 2.4. In R 3, the distance d from a point S to a line through P 0 parallel to v is d = P 0 S v. Proof. d = P 0 S sin θ = P 0 S v. Alternatively, one can find the distance d as follows: let S 0 be the point on the line such that S 0 S is normal to the vector v. Then we can find the point S 0 by solving the following system of equations: (1) S 0 is on the line. (2) S 0 S v = 0 Finally, we get the distance d = S 0 S. Example 2.5. Find the distance from a point (1, 1, 5) to a line [x, y, z] = [1, 3, 0]+t[1, 1, 2]. 5
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