4.3 Equations in 3-space

Transcription

1 4.3 Equations in 3-space istance can be used to define functions from a 3-space R 3 to the line R. Let P be a fixed point in the 3-space R 3 (say, with coordinates P (2, 5, 7)). Consider a function f : R 3 R that assigns to any point Q R 3 the distance between P and Q: f(q) = P Q. This function can be written using the formula for the distance: if the point Q has coordinates (x, y, z), then f(q) = f(x, y, z) = (x 2) 2 + (y + 5) 2 + (z 7) 2 = P Q

2 4.3. EQUATIONS IN 3-SPACE 87 What are the level sets of this function? Since distance is never negative, the level sets are not empty only if we consider a positive level (say, 3). The 3-level set is the set of all points Q(x, y, z) R 3 such that the distance P Q is equal to 3. We know that geometrically this set is a sphere centered at the point P of radius 3. The algebraic description of this set is provided by the equation (x 2)2 + (y + 5) 2 + (z 7) 2 = 3 After we square both sides (x 2) 2 + (y + 5) 2 + (z 7) 2 = 3 2 we see a typical equation of a sphere in 3-space. As opposed to curves in a plane, typically no curve in 3-space can be a level set of a function from the 3-space to R. A typical level set of such a function is a surface in the 3-space. A good example is a sphere as we just described Linear equations in 3-space We will now look closely at a linear equation with three variables: Ax + By + Cz = where A, B, C, and are some numbers. This equation defines a set the set of all points P (x, y, z) in 3-space with the coordinates satisfying the equation. Of course, if we multiply both sides of this equation by a constant, all three numbers will get multiplied but the new equation will represent the same object. So, we cannot extract any useful geometric information about the object from one of the numbers A, B, C,. Consider any two points Q (q 1, q 2, q 3 ) and R (r 1, r 2, r 3 ) on the set defined by the equation Ax + By + Cz =. This means the following equalities are satisfied: Aq 1 + Bq 2 + Cq 3 = and Ar 1 + Br 2 + Cr 3 =. Subtract the second equality from the first: A (q 1 r 1 ) + B (q 2 r 2 ) + C (q 3 r 3 ) = 0 This equation has a nice interpretation in terms of vectors: it says that the ot Product of the vectors A, B, C and RQ q1 r 1, q 2 r 2, q 3 r 3 is zero. It means that the vectors A, B, C and RQ are perpendicular. We conclude that for any points Q and R in the set, the vector RQ is perpendicular to the (fixed!) vector A, B, C. What is the meaning of the number? The equation says that is equal to the dot product of the vectors A, B, C and x, y, z : A, B, C x, y, z = If we divide both sides by the magnitude of A, B, C, then the left hand side A, B, C x, y, z A, B, C

3 88 CHAPTER 4. EUCLIEAN 3-SPACE: NEW CONCEPTS has a perfect geometric meaning: it is the scalar component of the vector x, y, z along the vector A, B, C. So, the equation basically says that for all points (x, y, z) of the set the scalar component of the vector x, y, z along the direction A,B,C of A, B, C is equal to. So, the equation describes a plane in 3-space as the plane perpendicular to the vector A, B, C and intersecting the A, B, C - line through the origin at the point A,B,C units away from the origin in the direction of the vector A, B, C. Problem. Plane through 3 points Find an equation of the plane containing the points P (1, 1, 12), Q(2, 6, 9), and R( 8, 4, 7). Generalization. Find an equation of the plane containing the points P (p 1, p 2, p 3 ), Q(q 1, q 2, q 3 ), and R(r 1, r 2, r 3 ). Strategy. If we denote the vector P Q by v = v1, v 2, v 3 and the vector P R by w = w1, w 2, w 3, the parametric presentation is x = p 1 + v 1 t + w 1 s y = p 2 + v 2 t + w 2 s z = p 3 + v 3 t + w 3 s Computation. P Q = 1, 5, 3, P R = 9, 3, 5 x = 1 + t 9s Answer: y = 1 + 5t + 3s z = 12 3t 5s Problem. 4 points in plane etermine whether the points lie on a plane: P (1, 2, 0), Q( 1, 0, 5), R(0, 2, 1), S(1, 0, 1). Generalization. etermine whether the given 4 points lie on a plane. Strategy. 1. Find an equation of the plane through the points P, Q, and R. 2. Check if the point S belongs to that plane. Computation. 2. x = 1 2t s y = 2 2t z = 5t + 1s 1 = 1 2t s 0 = 2 2t 1 = 5t + 1s 1. P Q = 2, 2, 5, P R = 1, 0, 1 s = 2 t = 1 1 = 5t + 1s So, the points are not in one plane. False.

4 4.3. EQUATIONS IN 3-SPACE 89 Problem. Intersection of two planes x = 1 s Find the intersection of the planes y = 2t z = 1 t + 4s and x = 1 2t s y = 2 2t z = 5t + s Generalization. Find the intersection of two planes given in parametric presentations f 1, f 2 : R 2 R 3. Strategy. Restate the questions: Find all pairs (t 1, s 1 ) and (t 2, s 2 ) such that f 1 (t 1, s 1 ) = f 2 (t 2, s 2 ). 1. Solve the system f 1 (t 1, s 1 ) = f 2 (t 2, s 2 ). 2. If the system has no solution, then the planes do not intersect. If there is one independent variable remaining in the system, then the intersection is a line and that variable can play the role of parameter in a parametric presentation of the line. If there are two independent variables remaining in the system, then the planes coincide. Computation. 1. s 1 = 1 2 s 2 t 1 = s 2 t 2 = 3 4 s 2 1 s 1 = 1 2t 2 s 2 2t 1 = 2 2t 2 1 t 1 + 4s 1 = 5t 2 + s 2 s 1 = 2t 2 + s 2 t 1 = 1 t 2 1 (1 t 2 ) + 4(2t 2 + s 2 ) = 5t 2 + s 2 2. There is one independent variable (s 2 ) remaining in the system. So the intersection is a line with parametric presentation y = 2 2( 3 x = 1 2( 3 4 s 2) s 2 4 s 2) z = 5( 3 4 s 2) + s 2 x = s 2 y = s 2 z = 11 4 s 2 Problem. Vector orthogonal to a plane Find a nonzero vector orthogonal to the plane through the points P (2, 3, 0), Q ( 1, 0, 2), and R (0, 7, 0). Generalization. Find a nonzero vector orthogonal to the plane through the points P (p 1, p 2, p 3 ), Q (q 1, q 2, q 3 ), and R (r 1, r 2, r 3 ). Strategy. 1. A vector v is orthogonal to the plane if and only if it is orthogonal to two non-parallel vectors in that plane. Consider vectors P Q and P R in the plane. Then the vector v is orthogonal to the plane if and only if P Q v = 0 and P R v = 0

5 90 CHAPTER 4. EUCLIEAN 3-SPACE: NEW CONCEPTS 2. Since the system { P Q v = 0 P R v = 0 has two equations and three unknowns (x, y, and z), we expect to find infinitely many solutions. Any solution of the system provides a vector orthogonal to the plane. Computation P Q = 3, 3, 2, P R = 2, 10, 0. P Q v = 3x 3y + 2z = 0 { 3x 3y + 2z = 0 2x 10y = 0 { 3 ( 5y) 3y + 2z = 0 x = 5y If we choose y = 1, then x = 5 and z = 6. Answer: 5, 1, 6. P R v = 2x 10y = 0 { z = 6y x = 5y If we have to find a unit vector (i.e. a vector with magnitude 1) orthogonal to the plane, we can divide the vector we got 5, 1, 6 by its magnitude ( 5) ( 6) 2 = 62: 5 62, 1 62, 6 62 If we need the first coordinate of the vector to be positive, multiply the whole vector by 1: 5 1 6,, Problem. istance to a plane Find the distance from the point P (1, 2, 0) to the plane 2x + y + z = 3. Generalization. Find the distance from the point P (p 1, p 2, p 3 ) to the plane Ax+ By + Cz =. Strategy. 1. Take any point Q on the plane. Consider the vector QP. 2. We find the scalar component of the vector QP in the direction of the vector v = A, B, C : comp v QP = QP v v 3. The distance from the point P to the plane is the absolute value of the scalar component comp v QP Computation. 2. comp v QP = v 1. The point Q (0, 0, 3) belongs to the plane. QP = 1, 2, 3 QP v = = 1 6

6 4.3. EQUATIONS IN 3-SPACE The distance from the point P to the plane is 1 6 = 1 6. When a plane is given by an equation Ax+By+Cz =, we can immediately see a vector perpendicular to the plane: A, B, C. If another plane is given A x+b y +C z =, we can define the angle between these planes as the angle between two vectors perpendicular to the planes: A, B, C and A, B, C. Problem. Angle between planes Find the angle between the given planes x+2y 3z = 35 and 8y+5z = 68. Generalization. Find the angle between the given planes Ax + By + Cz = and ax + by + cz = d. Strategy. The angle between the planes is equal to the angle between normal vectors to the planes. We find the angle between normal vectors u = A, B, C and w = a, b, c using ot Product: cos θ = u w u w Computation. u = 1, 2, 3, w = 0, 8, 5 cos θ = u w u w = = θ = cos rad Problem. Parallel/perpendicular planes etermine whether the planes x + 2y 3z = 35 and x + 8y + 5z = 68 are parallel, perpendicular, or neither. Generalization. etermine whether the planes Ax + By + Cz = and ax + by + cz = d are parallel, perpendicular, or neither. Strategy. 1. The planes are parallel if and only if their normal vectors are parallel (if and only if the normal vectors are proportional). 2. The planes are perpendicular if and only if their normal vectors are perpendicular (if and only if the ot Product of normal vectors is equal to 0). Computation. 1. The vectors 1, 2, 3 and 1, 8, 5 are normal to the corresponding planes. These vectors are proportional if and only if 1, 2, 3 = λ 1, 8, 5 for some λ. 1 = λ 2 = 8λ 3 = 5λ

7 92 CHAPTER 4. EUCLIEAN 3-SPACE: NEW CONCEPTS This system has no solutions. So, the vectors are not proportional and the planes are not parallel. 2. 1, 2, 3 1, 8, 5 = = 0 Answer: the planes are perpendicular. Two planes given by equations Ax+By +Cz = and A x+b y +C z = are parallel if and only if the vectors A, B, C and A, B, C are parallel (which means the vectors are proportional: A, B, C = λ A, B, C for some constant λ). In this case the second equation can be rewritten in the form Ax + By + Cz = for some number. Thus all the points of the first plane have the scalar component A,B,C along the vector A, B, C while all the points of the second plane have the scalar component A,B,C along the same vector A, B, C. Now we can find the distance between the planes as A, B, C A, B, C Problem. istance between parallel planes Find the distance between the given parallel planes x + 2y 3z = 35 and x 2y + 3z = 21. Generalization. Find the distance between the given parallel planes Ax + By + Cz = and A x + B y + C z =. Strategy. Multiply (or divide) the second equation by a constant to have it in the form Ax + By + Cz =. We find the scalar components of each plane in the direction of the vector A, B, C normal to both planes. These components are A,B,C and A,B,C. The distance between the planes is A, B, C A, B, C Computation. The second plane has equation x + 2y 3z = 21. A, B, C = 35 = A, B, C = 21 = The distance between the planes is = 56 =

8 4.3. EQUATIONS IN 3-SPACE 93 Problem. istance to a plane Find the distance from the point P (1, 2, 0) to the plane 2x + y + z = 3. Generalization. Find the distance from the point P (p 1, p 2, p 3 ) to the plane Ax+ By + Cz =. Strategy. 1. Find the scalar component of the plane in the direction of the vector v = A, B, C normal to the plane. This scalar component is equal to A,B,C. 2. Find the scalar component of the vector OP = p1, p 2, p 3 in the direction of the vector A, B, C : comp v OP = OP v. v 3. The distance from the point P to the plane is A, B, C comp v OP Computation. 1. A,B,C = = 3 6 OP v 2. comp v OP = v = The distance from the point P to the plane is 3 4 =