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planes and distances

E
Elias Dinsa

The document discusses planes and distances in R3. It begins by explaining that a plane can be represented by a normal vector n and a reference point P0 on the plane. The equation of a plane is derived as the dot product of any point P on the plane and the normal vector n being equal to 0. Examples are given of finding the equation of a plane given information like the normal vector or three points on the plane. The document also discusses finding the distance between planes, points and lines by using properties of orthogonality and projections.

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4. Planes and distances How do we represent a plane in R3 ? In fact the best way to specify a plane is to give a normal vector 鐃n to the plane and a point P0 on the plane. Then if we are given any point P on the plane, the vector P0P is a vector in the plane, so that it must be orthogonal to the normal vector 鐃n. Algebraically, we have P0P 鐃n = 0.揃 Lets write this out as an explicit equation. Suppose that the point P0 = (x0, y0, z0), P = (x, y, z) and 鐃n = (A, B, C). Then we have (x x0, y y0, z z0) (A, B, C) = 0.揃 Expanding, we get A(x x0) + B(y y0) + C(z z0) = 0, which is one common way to write down a plane. We can always rewrite this as Ax + By + Cz = D. Here D = Ax0 + By0 + Cz0 = (A, B, C) (x0, y0, z0) = 鐃 n OP0.揃 揃 This is perhaps the most common way to write down the equation of a plane. Example 4.1. 3x 4y + 2z = 6, is the equation of a plane. A vector normal to the plane is (3, 4, 2). Example 4.2. What is the equation of a plane passing through (1, 1, 2), with normal vector 鐃n = (2, 1, 1)? We have (x 1, y + 1, z 2) (2, 1, 1) = 0.揃 So 2(x 1) + y + 1 (z 2) = 0, so that in other words, 2x + y z = 1. A line is determined by two points; a plane is determined by three points, provided those points are not collinear (that is, provided they dont lie on the same line). So given three points P0, P1 and P2, what is the equation of the plane containing P0, P1 and P2? Well, we would like to 鍖nd a vector 鐃n orthogonal to any vector in the plane. Note that and are two vectors in the plane, which by assumption areP0P1 P0P2 1
鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 not parallel. The cross product is a vector which is orthogonal to both vectors, 鐃n = P0P2.P0P1 So the equation we want is ( P0P2) = 0.P0P P0P1 揃 We can rewrite this a little. = OP Expanding and rear足P0P OP0. ranging gives OP ( P0P2) = ( P0P1 OP0 揃 P0P1 P0P2).揃 Note that both sides involve the triple scalar product. Example 4.3. What is the equation of the plane through the three points, P0 = (1, 1, 1), P1 = (2, 1, 0) and P2 = (0, 1, 1)? and P0P2 = (1, 2, 2).P0P1 = (1, 2, 1) Now a vector orthogonal to both of these vectors is given by the cross product: 鐃n = P0P1 P0P2 脹 j k 1 2 1 1 2 2 = + k 2 1 2 2 1 1 2 1 2 2 = 脹 j 1 1 = 2脹 + 3 k.j 4 Note that 鐃n P0P1 = 2 6 + 4 = 0, 揃 as expected. It follows that the equation of is 2(x 1) + 3(y 1) 4(z 1) = 0, so that 2x + 3y 4z = 1. For example, if we plug in P2 = (0, 1, 1), then 2 0 + 3 揃 1 + 4 = 1,揃 as expected. 2
Example 4.4. What is the parametric equation for the line l given as the intersection of the two planes 2x y + z = 1 and x + y z = 2? Well we need two points on the intersection of these two planes. If we set z = 0, then we get the intersection of two lines in the xy-plane, 2x y = 1 x + y = 2. Adding these two equations we get 3x = 3, so that x = 1. It follows that y = 1, so that P0 = (1, 1, 0) is a point on the line. Now suppose that y = 0. Then we get 2x + z = 1 x z = 2. As before this says x = 1 and so z = 1. So P1 = (1, 0, 1) is a point on l. P0P = t P0P1, for some parameter t. Expanding (x 1, y 1, z) = t(0, 1, 1), so that (x, y, z) = (1, 1 t, t). We can also calculate distances between planes and points, lines and points, and lines and lines. Example 4.5. What is the distance between the plane x 2y + 3z = 4 and the point P = (1, 2, 3)? Call the closest point R. Then PR is orthogonal to every vector in the plane, that is, n = (1, 2, 3)PR is normal to the plane. Note that 鐃 is normal to the plane, so that PR is parallel to the plane. Pick any point Q belonging to the plane. Then the triangle PQR has a right angle at R, so that PR = 賊 proj鐃 PQ.n When x = z = 0, then y = 2, so that Q = (0, 2, 0) is a point on the plane. PQ = (1, 4, 3). Now 鐃緒申n鐃2 = 鐃n. 揃 鐃n. = 12 + 22 + 32 = 14 and 鐃n 揃 PQ = 4. So proj鐃n PQ = 2 7 (1, 2, 3). 3
鐃 鐃 鐃 鐃 鐃 鐃 So the distance is 2 14. 7 Here is another way to proceed. The line through P, pointing in the direction 鐃n, will intersect the plane at the point R. Now this line is given parametrically as (x 1, y 2, z 3) = t(1, 2, 3), so that (x, y, z) = (t + 1, 2 2t, 3 + 3t). The point R corresponds to (t + 1) 2(2 2t) + 3(3 + 3t) = 4, so that 14t = 2 that is t = 2 7 . So the point R is 1 the same answer as before (phew!). 鐃 鐃 鐃 鐃 鐃 鐃 (9, 10, 27). 7 It follows that PR = 1 (2, 4, 6) = 2 (1, 2, 3), 7 7 Example 4.6. What is the distance between the two lines (x, y, z) = (t2, 3t+1, 2t) and (x, y, z) = (2t1, 23t, t+1)? If the two closest points are R and R鐃 then RR鐃 is orhogonal to the direction of both lines. Now the direction of the 鍖rst line is (1, 3, 1) and the direction of the second line is (2, 3, 1). A vector orthogonal to both is given by the cross product: 脹 j k 1 3 1 2 3 1 = 3 k.j 9 To simplify some of the algebra, lets take 鐃n = k,j + 3 which is parallel to the vector above, so that it is still orthogonal to both lines. It follows that RR鐃 is parallel to 鐃n. Pick any two points P and P鐃 on the two lines. Note that the length of the vector proj鐃n P鐃 P, 4
is the distance between the two lines. Now if we plug in t = 0 to both lines we get P鐃 = (2, 1, 2) and P = (1, 2, 1). So P鐃 P = (1, 1, 1). Then 鐃 鐃緒申n鐃2 = 12 + 32 = 10 and n P鐃 P = 2.揃 It follows that proj鐃n P鐃 P = 2 (0, 1, 3) = 1 (0, 1, 3). 10 5 and so the distance between the two lines is 1 10. 5 5
MIT OpenCourseWare http://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

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planes and distances

  • 1. 4. Planes and distances How do we represent a plane in R3 ? In fact the best way to specify a plane is to give a normal vector 鐃n to the plane and a point P0 on the plane. Then if we are given any point P on the plane, the vector P0P is a vector in the plane, so that it must be orthogonal to the normal vector 鐃n. Algebraically, we have P0P 鐃n = 0.揃 Lets write this out as an explicit equation. Suppose that the point P0 = (x0, y0, z0), P = (x, y, z) and 鐃n = (A, B, C). Then we have (x x0, y y0, z z0) (A, B, C) = 0.揃 Expanding, we get A(x x0) + B(y y0) + C(z z0) = 0, which is one common way to write down a plane. We can always rewrite this as Ax + By + Cz = D. Here D = Ax0 + By0 + Cz0 = (A, B, C) (x0, y0, z0) = 鐃 n OP0.揃 揃 This is perhaps the most common way to write down the equation of a plane. Example 4.1. 3x 4y + 2z = 6, is the equation of a plane. A vector normal to the plane is (3, 4, 2). Example 4.2. What is the equation of a plane passing through (1, 1, 2), with normal vector 鐃n = (2, 1, 1)? We have (x 1, y + 1, z 2) (2, 1, 1) = 0.揃 So 2(x 1) + y + 1 (z 2) = 0, so that in other words, 2x + y z = 1. A line is determined by two points; a plane is determined by three points, provided those points are not collinear (that is, provided they dont lie on the same line). So given three points P0, P1 and P2, what is the equation of the plane containing P0, P1 and P2? Well, we would like to 鍖nd a vector 鐃n orthogonal to any vector in the plane. Note that and are two vectors in the plane, which by assumption areP0P1 P0P2 1
  • 2. 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 鐃 not parallel. The cross product is a vector which is orthogonal to both vectors, 鐃n = P0P2.P0P1 So the equation we want is ( P0P2) = 0.P0P P0P1 揃 We can rewrite this a little. = OP Expanding and rear足P0P OP0. ranging gives OP ( P0P2) = ( P0P1 OP0 揃 P0P1 P0P2).揃 Note that both sides involve the triple scalar product. Example 4.3. What is the equation of the plane through the three points, P0 = (1, 1, 1), P1 = (2, 1, 0) and P2 = (0, 1, 1)? and P0P2 = (1, 2, 2).P0P1 = (1, 2, 1) Now a vector orthogonal to both of these vectors is given by the cross product: 鐃n = P0P1 P0P2 脹 j k 1 2 1 1 2 2 = + k 2 1 2 2 1 1 2 1 2 2 = 脹 j 1 1 = 2脹 + 3 k.j 4 Note that 鐃n P0P1 = 2 6 + 4 = 0, 揃 as expected. It follows that the equation of is 2(x 1) + 3(y 1) 4(z 1) = 0, so that 2x + 3y 4z = 1. For example, if we plug in P2 = (0, 1, 1), then 2 0 + 3 揃 1 + 4 = 1,揃 as expected. 2
  • 3. Example 4.4. What is the parametric equation for the line l given as the intersection of the two planes 2x y + z = 1 and x + y z = 2? Well we need two points on the intersection of these two planes. If we set z = 0, then we get the intersection of two lines in the xy-plane, 2x y = 1 x + y = 2. Adding these two equations we get 3x = 3, so that x = 1. It follows that y = 1, so that P0 = (1, 1, 0) is a point on the line. Now suppose that y = 0. Then we get 2x + z = 1 x z = 2. As before this says x = 1 and so z = 1. So P1 = (1, 0, 1) is a point on l. P0P = t P0P1, for some parameter t. Expanding (x 1, y 1, z) = t(0, 1, 1), so that (x, y, z) = (1, 1 t, t). We can also calculate distances between planes and points, lines and points, and lines and lines. Example 4.5. What is the distance between the plane x 2y + 3z = 4 and the point P = (1, 2, 3)? Call the closest point R. Then PR is orthogonal to every vector in the plane, that is, n = (1, 2, 3)PR is normal to the plane. Note that 鐃 is normal to the plane, so that PR is parallel to the plane. Pick any point Q belonging to the plane. Then the triangle PQR has a right angle at R, so that PR = 賊 proj鐃 PQ.n When x = z = 0, then y = 2, so that Q = (0, 2, 0) is a point on the plane. PQ = (1, 4, 3). Now 鐃緒申n鐃2 = 鐃n. 揃 鐃n. = 12 + 22 + 32 = 14 and 鐃n 揃 PQ = 4. So proj鐃n PQ = 2 7 (1, 2, 3). 3
  • 4. 鐃 鐃 鐃 鐃 鐃 鐃 So the distance is 2 14. 7 Here is another way to proceed. The line through P, pointing in the direction 鐃n, will intersect the plane at the point R. Now this line is given parametrically as (x 1, y 2, z 3) = t(1, 2, 3), so that (x, y, z) = (t + 1, 2 2t, 3 + 3t). The point R corresponds to (t + 1) 2(2 2t) + 3(3 + 3t) = 4, so that 14t = 2 that is t = 2 7 . So the point R is 1 the same answer as before (phew!). 鐃 鐃 鐃 鐃 鐃 鐃 (9, 10, 27). 7 It follows that PR = 1 (2, 4, 6) = 2 (1, 2, 3), 7 7 Example 4.6. What is the distance between the two lines (x, y, z) = (t2, 3t+1, 2t) and (x, y, z) = (2t1, 23t, t+1)? If the two closest points are R and R鐃 then RR鐃 is orhogonal to the direction of both lines. Now the direction of the 鍖rst line is (1, 3, 1) and the direction of the second line is (2, 3, 1). A vector orthogonal to both is given by the cross product: 脹 j k 1 3 1 2 3 1 = 3 k.j 9 To simplify some of the algebra, lets take 鐃n = k,j + 3 which is parallel to the vector above, so that it is still orthogonal to both lines. It follows that RR鐃 is parallel to 鐃n. Pick any two points P and P鐃 on the two lines. Note that the length of the vector proj鐃n P鐃 P, 4
  • 5. is the distance between the two lines. Now if we plug in t = 0 to both lines we get P鐃 = (2, 1, 2) and P = (1, 2, 1). So P鐃 P = (1, 1, 1). Then 鐃 鐃緒申n鐃2 = 12 + 32 = 10 and n P鐃 P = 2.揃 It follows that proj鐃n P鐃 P = 2 (0, 1, 3) = 1 (0, 1, 3). 10 5 and so the distance between the two lines is 1 10. 5 5
  • 6. MIT OpenCourseWare http://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.