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MATH 136 (145)
Lecture

# lect136_7_w14.pdf

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School
University of Waterloo
Department
Mathematics
Course
MATH 136
Professor
Robert Andre
Semester
Spring

Description
3 Monday, January 20 ▯ Lecture 7 : Projections onto lines and planes in ▯ Concepts: 1. projection of a vector b onto a vector a (or onto a line L) in ▯ and ▯ .2 3 2. representation of the vector w = perp b. a 3. computing proj b. a 4. shortest distance between a point b and a line L in ▯ . 3 5. point on a line L which is closest to a point . 6. projection of a vector onto a plane. 7. Point on a plane which is closest to a point. 3 8. finding the distance between a point b and a plane P(n, a) in ▯ . 7.1 Projections ▯ Occasionally, one might want to express a vector x as a sum of two orthogonal vectors (perpendicular vectors), called orthogonal component vectors of x. Suppose we are given a vector b in ▯ and another vector a. Suppose we want to express b as the sum of a scalar multiple of a, say ta, and another vector w, such that ta ▯ w. In this lecture we study how this can be done. 2 3 7.2 Definition ▯ Let b be a vector in ▯ or ▯ and let a be any other vector in the same space. We define the projection of b onto a as being the vector p = proj (b) = ta, ahere t is such the vector b ▯ p is orthogonal to p. The point, proj (b), as referred to as being the point on the line L = {ta : t ▯▯▯} which is the closest to b. The expression b ▯ proj (b) = a b – p is denoted by perp b. (This expression is pronounced “perp b onto a”). a a A few remarks: - So vector p = proj (b)ais a scalar multiple of the directed line segment a. - If in ▯ , we have defined p = proj (b) ao that the points (0, 0), p, and b form a right triangle. (The directed line segment of b forming the hypotenuse). - So it is always true that b ▯ proj ab) + perp (b)a Another remark. So proj (b) is a vector used to decompose the vector b into a sum of two orthogonal a vectors proj (a) and perp (b)a There is a theorem called the Projection theorem (also referred to as the Best approximation theorem) which guarantees that proj (b) is indead the point on the line ta which is closest to b. Optional reading: The proof of the Projection theorem is based on the Pythagorean theorem: If sa ▯▯proj b aelongs to the line L : ta , then b – sa = b – proj ba+ proj b a sa Since proj ba– sa belongs to the line L, and b ▯ proj (b) =aperp (b) thea proj b – sa aad b ▯ proj (b) are perpendicular. By the Pythagorean theorem a 2 2 2 || b – sa || = || b – proa b || + || proa b – sa || 2 2 This implies || b – proj ba|| < || b – sa || and so proj b as closer to b than sa. 7.2.1 Example ▯ Consider the vector b = (▯4, 3) and let a = (2, 0). Compute proj (b). a Solution: It is easy, in this case, to graphically visualize, and then identify what proj (b) as. Then p = proj b a ▯2(2, 0) = (▯4, 0) and perp b = b a p = (▯4, 3) ▯ (▯4, 0) = (0, 3). 7.3 How to compute proj b. Weawill show that the (orthogonal) projection of b onto a is given by the expression We will assume neither a nor b is the zero-vector. Suppose p = ta and the scalar t is such that perp a = b ▯ p is orthogonal to p = ta. Then b ▯ p is orthogonal to a. That is, a ▯ (b ▯ ta) = 0. Applying dot product properties Then Hence So the point on the line L = {ta : t ▯▯▯} which is the closest to b is proj b a 7.3.1 Example ▯ Find the projection, proj u, of u = (2, 3, 1) onto v = (1, 2, ▯6) . v Also find the norm o
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