Calculus and Vectors – How to get an A+ 8.5 Cartesian Equation of a Plane A Normal Equation of a Plane A plane may be determined by a point P0 ( x0 , y0 , z 0 ) and a vector perpendicular to the r plane n called the normal vector. B Cartesian Equation of a Plane Let write the normal vector of a plane in the form: r n = ( A, B, C ) Then, the normal equation (1) may be written as: ( x − x0 , y − y0 , z − z 0 ) ⋅ ( A, B, C ) = 0 Ax + By + Cz − Ax0 − By0 − Cz 0 = 0 or: Ax + By + Cz + D = 0 If P( x, y, z ) is a generic point on the plane, then: r P0 P ⊥ n and: r P0 P ⋅ n = 0 (1) This is the normal equation of a plane. Ex 1. Consider the plane π defined the Cartesian equation π : 2 x − 3 y + 6 z + 12 = 0 . a) Find a normal vector to this plane. r n = (2,−3,6) b) Find two points on this plane. If x = 0 and y = 0 then z = −2 . So (0,0,−2) ∈ π . If x = 0 and z = 0 then y = 4 . So (0,4,0) ∈ π . c) Find if the point P(1,2,3) is a point on this plane. 2(1) − 3(2) + 6(3) + 12 = 26 ≠ 0 ⇒∴ P ∉ π (2) equation which is called the Cartesian equation of a plane. Note. A normal vector to the plane is: r r r n = u ×v r r where u and v are the direction vectors of the plane. Ex 2. Find the Cartesian equation of a plane π that passes through the points A(1,−1,0) , B(0,0,1) , and C (0,−2,1) . r r u = AB = (−1,1,1); v = AC = (−1,−1,1) r r r r r i j k i j r r r − 1 1 = (2,0,2) = ( A, B, C ) n = u × v = −1 1 1 −1 −1 1 −1 −1 2x + 2z + D = 0 A(1,−1,0) ∈ π ⇒ 2(1) + 2(0) + D = 0 ⇒ D = −2 ∴ 2 x + 2 z − 2 = 0 or x + z − 1 = 0 Ex 3. Find parametric and vector equations for the plane: π : x − 2 y + 3z − 6 = 0 Ex 4. Find the intersections with the coordinate axes for the plane π : 3x + 2 y + z − 6 = 0 . Represent the plane graphically. Method #1 ⎧ x = 6 + 2s − 3t ⎪ ; s, t ∈ R ⎨y = s ⎪z = t ⎩ r r = (6,0,0) + s (2,1,0) + t (−3,0,1), s, t ∈ R Let A = π ∩ x − axis . y A = z A = 0 ⇒ x A = 2 . ∴ x − int = A(2,0,0) . Let B = π ∩ y − axis . x B = z B = 0 ⇒ y B = 3 . ∴ y − int = B(0,3,0) . Let C = π ∩ z − axis . xC = yC = 0 ⇒ zC = 6 . ∴ z − int = C (0,0,6) . Method #2 A(0,0,2) ∈ π ; B(0,−3,0) ∈ π ; C (6,0,0) ∈ π r r u = AB = (0,−3,−2); v = AC = (6,0,−2) r ∴ r = (0,0,2) + s (0,−3,−2) + t (6,0,−2); s, t ∈ R ⎧ x = 6t ⎪ ⎨ y = −3s ⎪ z = 2 − 2 s − 2t ⎩ 8.5 Cartesian Equation of a Plane ©2010 Iulia & Teodoru Gugoiu - Page 1 of 2 Calculus and Vectors – How to get an A+ Ex 5. Find the Cartesian equation of a plane with x − int = −1 , y − int = 2 , and z − int = −3 . Ax + By + Cz + D = 0 D = −1 ⇒ A = D A D D y − int = − = 2 ⇒ B = − B 2 D D z − int = − = −3 ⇒ C = C 3 D D Dx − y + z + D = 0 2 3 y z ∴ x − + + 1 = 0 or ∴ 6 x − 3 y + 2 z + 6 = 0 2 3 x − int = − F Angle between two Planes The angle between two plane s is defined as the angle between their normal vectors: r r n1 ⋅ n 2 cos θ = r r | n1 || n 2 | Note. Using this formula, you may get an acute or an obtuse angle depending on the normal vectors which are used. Ex 6. Find the angle between each pair of planes. a) π 1 : x + 2 y + 3z + 1 = 0 , π 2 : 3 x + 2 y + z + 2 = 0 r r n1 = (1,2,3); n2 = (3,2,1) r r n ⋅n (1)(3) + (2)(2) + (3)(1) 10 cosθ = r 1 r2 = = 2 2 2 2 2 2 | n1 || n2 | 14 1 + 2 + 3 3 + 2 +1 ∴θ = cos −1 (10 / 14) ≅ 44.42° b) π 1 : x + y + z + 1 = 0 , π 2 : x − y − 1 = 0 r r n1 = (1,1,1); n2 = (1,−1,0) r r n ⋅n (1)(1) + (1)(−1) + (1)(0) 0 = =0 cosθ = r 1 r2 = 2 2 2 2 2 2 | n1 || n2 | 3 2 1 + 1 + 1 1 + (−1) + 0 ∴θ = 90° Reading: Nelson Textbook, Pages 461-468 Homework: Nelson Textbook: Page 468 #1, 5, 7, 8, 9a, 11, 13, 17 8.5 Cartesian Equation of a Plane ©2010 Iulia & Teodoru Gugoiu - Page 2 of 2