advertisement

Math 114 – Rimmer 12.5 Equations of Lines and Planes 12.5 Equations of Lines and Planes In order to find the equation of a line, we need : A ) a point on the line P0 ( x0 , y0 , z0 ) B ) a direction vector for the line v = a, b, c r = r0 + tv vector equation of line L z x, y, z = x0 , y0 , z0 + t a, b, c P0 P = tv P ( x, y, z ) P0 ( x0 , y0 , z0 ) equating components: r L parametric equations of the line L r0 v r0 = x0 , y0 , z0 r = x, y , z x x = x0 + at , y = y0 + bt , z = z0 + ct P0 P = tv = t a, b, c y eliminating the parameter t (solve for t in each, then equate the results) x − x0 y − y0 z − z0 = = a b c symmetric equations of the line L Math 114 – Rimmer 12.5 Equations of Lines and Planes Find parametric equations of the line containing ( 5,1,3) and ( 3, −2, 4 ) . In order to find the equation of a line, we need : A ) a point on the line P0 ( x0 , y0 , z0 ) Pick either point B ) a direction vector for the line v = a, b, c Use the vector from one point to the other L: x = y = z = point direc. x0 + at y0 z0 + bt + ct v = 3 − 5, −2 − 1, 4 − 3 v = −2, −3,1 L: x = 5 − 2t y = 1 − 3t z = 3 + t 1 Two lines in 3 space can interact in 3 ways: Math 114 – Rimmer 12.5 Equations of Lines and Planes A) Parallel Lines - their direction vectors are scalar multiples of each other B) Intersecting Lines - there is a specific t and s, so that the lines share the same point. C) Skew Lines their direction vectors are not parallel and there is no values of t and s that make the lines share the same point. Math 114 – Rimmer 12.5 Equations of Lines and Planes Determine whether the lines L1 and L2 are parallel, skew or intersecting. If they intersect, find the point of intersection. L1 L2 x = 3 − t y = 5 + 3t z = −1 − 4t x = 8 + 2s y = −6 − 4 s z = 5 + s Check to make sure that the z values are equal for this t and s. Set the x ' s = 3 − t = 8 + 2s −t = 5 + 2 s t = −5 − 2 s Set the y ' s = −1 − 4t = 5 + s −1 − 4 ( −1) = 5 + ( −2 ) 3 = 3 check This should happen at the same time t , so plug in t = −5 − 2 s 5 + 3t = −6 − 4s 5 + 3 ( −5 − 2 s ) = −6 − 4s 5 − 15 − 6s = −6 − 4s −10 − 6s = −6 − 4s t = −5 − 2 ( −2 ) −2 s = 4 t = −1 s = −2 Now find the pt. of intersection. using L1 x = 3 − ( −1) y = 5 + 3 ( −1) z = −1 − 4 ( −1) ( 4, 2,3) 2 Math 114 – Rimmer 12.5 Equations of Lines and Planes Determine whether the lines L1 and L2 are parallel, skew or intersecting. If they intersect, find the point of intersection. L1 L2 x = 4 + t y = −8 − 2t z = 12t x = 3 + 2s y = −1 + s z = −3 − 3s Check to make sure that the z values are equal for this t and s. Set the x ' s = 4 + t = 3 + 2s t = −1 + 2 s ?? 12 ( −3 ) =− 3 − 3 ( −1) −36 ≠ 0 Set the y ' s = −8 − 2t = −1 + s −8 − 2 ( −1 + 2s ) = −1 + s This should happen at This means that the x and y values are the same for t = −3 and s = −1, but the z values are different. the same time t , so plug in t = −1 + 2 s −8 + 2 − 4 s = −1 + s t = −1 + 2 s −6 − 4s = −1 + s −5s = 5 s = −1 t = −3 The lines are skew Planes Math 114 – Rimmer 12.5 Equations of Lines and Planes In order to find the equation of a plane, we need : A ) a point on the plane P0 ( x0 , y0 , z0 ) B ) a vector that is orthogonal to the plane n = a, b, c n z P0 P = r − r0 n ⋅ ( r − r0 ) = 0 n ⋅ r = n ⋅ r0 P ( x, y, z ) vector equation of the plane r P0 ( x0 , y0 , z0 ) n ⋅ ( r − r0 ) = 0 ⇒ r0 x this vector is called the normal vector to the plane y r0 = x0 , y0 , z0 r = x, y , z P0 P = x − x0 , y − y0 , z − z0 n = a , b, c a ( x − x0 ) + b ( y − y0 ) + c ( z − z0 ) = 0 scalar equation of the plane ax − ax0 + by − by0 + cz − cz0 = 0 ax + by + cz − ( ax0 + by0 + cz0 ) = 0 ax + by + cz + d = 0 linear equation of the plane 3 Determine the equation of the plane that contains the lines L1 and L2 . L1 L2 x = 3 − t y = 5 + 3t z = −1 − 4t x = 8 + 2s y = −6 − 4 s z = 5 + s We have two points in the plane: Math 114 – Rimmer 12.5 Equations of Lines and Planes In order to find the equation of a plane, we need : A ) a point on the plane B ) a vector that is orthogonal to the plane n = a, b, c from L1 : ( 3,5, −1) and from L2 : ( 8, −6,5 ) We have two vectors in the plane: from L1 : −1,3, −4 and from L2 : 2, −4,1 Let u = −1,3, −4 and v = 2, −4,1 Find u × v. i u × v = −1 2 j 3 k −4 = −4 1 3 −4 −4 1 i + ( −1) −1 −4 2 1 j+ −1 3 2 −4 k = ( 3 − 16 ) i + ( −1)( −1 + 8 ) j + ( 4 − 6 ) k u × v = −13i − 7 j − 2k = n −13 x − 7 y − 2 z + 72 = 0 −13 x − 7 y − 2 z + d = 0 −13 ( 3) − 7 ( 5 ) − 2 ( −1) + d = 0 13 x + 7 y + 2 z − 72 = 0 −39 − 35 + 2 + d = 0 ⇒ d = 72 Math 114 – Rimmer 12.5 Equations of Lines and Planes Determine the equation of the plane that passes through (1, 2,3) , ( 3, 2,1) , and ( −1, −2, 2 ) . R Q P PQ = 3 − 1, 2 − 2,1 − 3 PQ = 2, 0, −2 In order to find the equation of a plane, we need : A ) a point on the plane we have 3 to choose from PR = −1 − 1, −2 − 2, 2 − 3 B ) a vector that is orthogonal to the plane PR = −2, −4, −1 n = a, b, c we need 2 vectors in the plane Let u = 2, 0, −2 and v = −2, −4, −1 Find u × v. i u×v = 2 −2 j 0 −4 k 0 −2 = −4 −1 −2 −1 = ( 0 − 8) i i + ( −1) 2 −2 −2 −1 j + 2 0 −2 −4 k + ( −1)( −2 − 4 ) j + ( −8 − 0 ) k u × v = −8i + 6 j − 8k = n or n = 4i − 3 j + 4k ( in lowest terms ) 4x − 3 y + 4z + d = 0 4 (1) − 3 ( 2 ) + 4 ( 3) + d = 0 4 − 6 + 12 + d = 0 4 x − 3 y + 4 z − 10 = 0 ⇒ d = −10 4 Math 114 – Rimmer 12.5 Equations of Lines and Planes Two distinct planes in 3-space either are parallel or intersect in a line. z y x Find the line of intersection of the two planes Use the elimination method x − 2y + z = 0 2 ( x − 2 y + z = 0) 2x + 3y − 2z = 0 2x + 3y − 2z = 0 Math 114 – Rimmer 12.5 Equations of Lines and Planes 2x − 4 y + 2z = 0 2x + 3y − 2z = 0 4x − y = 0 y = 4x Take the first equation: x − 2 y + z = 0 and plug in y = 4 x x − 2 ( 4x ) + z = 0 ⇒ z = 7 x x can be anything and we have y and z as functions of x. Let x = t then y = 4t and z = 7t L: x = t y = 4t z = 7t 5 Math 114 – Rimmer 12.5 Equations of Lines and Planes If two planes intersect, then you can determine the angle between them. between planes = between their normal vectors n1 n2 cos θ = θ n1 ⋅ n 2 n1 n 2 θ Find the angle between the planes x − 2y + z = 0 2x + 3 y − 2z = 0 cos θ = n1 ⋅ n 2 n1 n 2 6 102 = θ ≈ 54 ⇒ n1 = 1, −2,1 and n 2 = 2,3, −2 n1 = 6 and n 2 = 17 ⇒ n1 ⋅ n 2 = 2 − 6 − 2 = −6 Distance between a point and a plane: Math 114 – Rimmer 12.5 Equations of Lines and Planes 1. Take any point in the plane call it P0 ( x0 , y0 , z0 ) . 2. Let b = P0 P1 = x1 − x0 , y1 − y0 , z1 − z0 ( x1 , y1 , z1 ) projn b 3. Take projn b. The length of this vector = D Remember projnb = compn b = n⋅b n 4. Since D must be positive, take the absolute value. D= n ⋅b ax + by + cz + d = 0 n n ⋅ b = a, b, c ⋅ x1 − x0 , y1 − y0 , z1 − z0 = a ( x1 − x0 ) + b ( y1 − y0 ) + c ( z1 − z0 ) = ax1 + by1 + cz1 + ( − ax0 − by0 − cz0 ) ⇒ n ⋅ b = ax1 + by1 + cz1 + d D= d ax1 + by1 + cz1 + d a 2 + b2 + c2 6 Math 114 – Rimmer 12.5 Equations of Lines and Planes Distance between a point and a line: Math 114 – Rimmer 12.5 Equations of Lines and Planes 7 Math 114 – Rimmer 12.5 Equations of Lines and Planes 8