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Lecture

# lect136_6_w14.pdf

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Department
Mathematics
Course Code
MATH 136
Professor
Robert Andre

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3 Friday, January 17 ▯ Lecture 6 : Planes in ▯ Concepts: 1. plane n ▯(x ▯ a) = 0 in ▯ where n is a norm to the plane. n 2. hyperplane in ▯ . 3. scalar equation of a plane n x + 1 1 + n 2 2 d =3 3a + n a +1 1a . 2 2 3 3 4. vector equation x = sv + tw + a, of a plane 5. finding the vector and standard (scalar) equation of a plane. 6. orthogonal line to a plane In this lecture we discuss two equations containing vectors which represent a plane in ▯ . 3 3 6.1 Intuitive description of a plane P in ▯ . We will assume an intuitive understanding of a “plane” in 3-space. We want to formulate a mathematical representation of a plane P in 3-space in such a way that mathematical representation corresponds to our intuitive perception of what the set of points (vectors) in P are. Given a plane P we can imagine a vector n = (n , n , n 1 w2ose3directed line segment is perpendicular (orthogonal) to all line segments which lie on P. Say a = (a ,1a ,2a )3is any point (vector) on P. - Note that a does not represent a directed line segment which “lies”on P and so it need not be orthogonal to n. (We use the word point (a , a , a1) o2 P 3o avoid confusion so as we don't think it is a vector whose directed line segment “lies” on P.) - We want to find a way to “recognize” all points (vectors) x which are on P. - Observe that for any x on P, since a is also on P, x▯▯ a is a vector whose directed line segment is parallel to the line segment joining a to x and so is orthogonal to n. - This holds true for all vectors x on the plane P. - Hence the vectors on P are precisely those vectors x that satisfy “ x ▯ a is orthogonal to n.” - Thus for any point x on P, (x▯▯ a) ▯ n. 6.1.1 Definition ▯ Let a = (a , a 1 a2) b3 a vector in ▯ and n = (n , n , n 1 be2a n3n- 3 zero vector. We define a plane, P(n, a) in ▯ , as being all vectors x such that x ▯▯a▯▯▯n. Equivalently, P(n, a) = {x ▯ ▯ : n3 ▯ (x ▯ a) = 0 } The vector n is referred to as the normal to the plane. The equation n ▯(x ▯ a) = 0 will be referred to as a “normal vector equation of the plane” where n and a are given and x is a “vector variable”. Recall that, for vectors v and w which are non-zero and linearly independent we previously referred to the set of vectors represented by vector equation x = sv + tw + 3 a of ▯ as a plane. We will soon show that the set both P(n, a) and the equation x = sv + tw + a can be used to describe the same plane. 6.1.2 Remarks According to the definition n ▯(x ▯ a) = 0, any two vectors define a particular plane. One vector, n, can be used as the normal to the plane while the other vector, a, represents a point on the plane. - There are many possibilities for the normal n of a plane for if n is orthogonal to all lines on a plane then so is any scalar multiple of n. That is, for any non-zero scalar t, P(n, a) = P(tn, a) since, n ▯(x ▯ a) = 0 ▯▯▯tn ▯(x ▯ a) = 0 (Invoking DP3) - Observe that, according to this definition, the vector a does indeed belong to P, since n ▯ (a ▯ a) = n ▯0 = 0. - Also note that if b is a point on P(n, a) then P(n, a) and P(n, b) are the same plane since - The definition of “plane” in ▯ generalizes to a definition of a plane in higher n n dimensions ▯ . For example, if n and a are two vectors in ▯ , where n > 3, a set of vectors in n P(n, a) = {x▯▯ ▯ : n ▯(x ▯ a) = 0 } is called a hyperplane. We will see that this such sets P(n, a) and be represented in the form given in definition 2.10. - Note that the vectors n and a need not be linearly independent. It is possible to have as normal vector a vector n which is collinear with a. For example, the set {x : (0,0,1)▯▯(x ▯ (0,0,1)) = 0 } is the horizontal plane containing a = n = (0, 0, 1). 3 6.2 Definition ▯ The plane in ▯ , whose normal vector is n = (n , n , n )1and2con3ains the point a = (a ,1a 2 a 3 can be described by a scalar equation. To see this: The plane P(n, a) is the set of all x = (x , x1, x2) 3uch that n ▯(x ▯ a) = 0, that is the set of all x = (x , x , x ) such that 1 2 3 n 1x 1 a )1+ n (x2▯▯2 ) +2n (x ▯3a 3 = 03 Equivalently, it is the set of all vectors x = (x , x1, 2 ) 3uch that n1 1+ n x2 2n x =3 3= n a + n1 1+ n a2 2 3 3 Expressions of the form n1 1+ n x2 2n x 3 3 are “scalar (standard) equation of the plane”. These planes have norm (n , n , n ). S1e 2 3 the following example. 6.2.1 Observation ▯ Let d be a real number and let T = {(x, y, z) : ax + by + cz = d }. If a ▯▯0, we see that d = (d , d , d ) = (d/a, 0, 0) ▯▯T , that is, ad + bd + cd = d. Let 1 2 3 1 2 3 n = (a, b, c). This example show that, that given any scalar equation of the form ax + by + cz = d, we can easily find one point d = (d , d , d ) which satisfies this equation, and then 1 2 3 setting n = (a, b, c), the set of all x = (x, y, z) that satisfies this equation is P(n , d) = {x : n ▯(x ▯ d) = 0}. So scalar equations for the plane can easily be expressed in the form of a “normal vector equation” n ▯(x ▯ d) = 0. 3 The following theorem shows that sets in ▯ represented by the equation x = sv + tw + a (remember, we also called these “planes” in 2.9) can also be represented by n ▯ (x ▯ a) = 0, and so are planes as we have now defined them. 6.3 Theorem ▯ Suppose v and w are linearly independent vectors in. If x = sv + tw + a for some s, t in ▯ then x belongs to the plane represented by (v ▯ w) ▯(x ▯ a) = 0. Proof : First note (from properties in 5.2) that, since n = v ▯ w, n ▯▯sv = 0 and n ▯tw = 0 for any s and t. Suppose x = sv + tw + a. Then So sv + tw + a ▯{x ▯ ▯ : n 3 ▯ (x ▯ a) = 0} as required. 3 6.4 Theorem ▯ Let a ▯ ▯ and let P be the plane represented by n ▯(x ▯ a) = 0 (where n is not 0). Then there exists linearly independent vectors v and w such that v ▯ w = kn and 3 a vector d in P such that P ▯ T = {x ▯ ▯ : x = sv + tw + d for some s, t in ▯ }. Proof : Let a = (a, b, c
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