Math1300:MainPage/three-space

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1 Contents 1 Planes in 3-space 2 The Equation of a plane through Three Points 2.1 Synopsis (Computation of the Equation of a Plane through Three Points) 3 Lines in 3-space 4 The Line as the Intersection of Two Planes 5 Distance from a Point to a Plane 5.1 The Distance between Parallel Planes Planes in 3-space The equation of a line in 2-space allows us to identify the points that are on the line: a point (x 0,y 0 ) if on the line if and only if it satisfies the corresponding equation. There are several forms that such an equation can take: Point-slope: Slope-intercept: y = mx + b A similar situation exists for planes in 3-space. We can find an equation which determines the points that are in the plane. The point (x,y,z ) is a point in the plane, and the vector in coordinate form is (a,b,c). The points in the plane are all 0 (x,y,z) 0 0 so that the vector from (x 0,y 0,z 0 ) to (x,y,z) is orthogonal to In other words, Point-normal form of a plane (not all of a, b and c are zero): Contents 1

2 a(x x0) + b(y y0) + c(z z0) = 0 We can expand the point-normal form to get a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = ax + by + cz (ax 0 + by 0 + cz 0 ) = ax + by + cz + d = 0 General form of the equation of a plane (not all of a, b and c are zero): ax + by + cz + d = 0 Examples: Math1300:MainPage/three-space 1. Suppose The equation of the plane through (1, 1,2) normal to is 1(x 1) + 2(y + 1) + 3(z 2) = x + 2y + 3z 5 = 0 2. The following two equations are equations of the same plane since any triple (x,y,z) satisfying the first equation also satisfies the second equation, and vice-versa: x + 2y + 3z 5 = 0 2x + 4y + 6z 10 = 0 Clearly, if is orthogonal to a plane, then so is for any 3. Suppose that (a,b,c ) and (a,b,c ) are proportional, that is, (a,b,c ) = r(a,b,c ) for some Consider the planes 1 1 that 1 satisfy 2 the 2 2 equations a x + b 1 y + c 1 z 1 + d 1 = 0 and a 2 x + b 2 y + c 2 z + d 2 = 0. Then these planes are either equal or parallel, for if (x 0,y 0,z 0 ) is in the second plane, then a 1 x 0 + b 1 y 0 + c 1 z 0 + d 1 = ra 2 x 0 + rb 2 y 0 + rc 2 z 0 + d 1 = r(a 2 x 0 + b 2 y 0 + c 2 z 0 ) + d 1 = rd 2 + d 1. Hence (x 0,y 0,z 0 ) is also in the first plane if and only if d 1 = rd 2. This implies (a 1,b 1,c 1,d 1 ) = r(a 2,b 2,c 2,d 2 ) The Equation of a plane through Three Points Suppose we want to find an equation of the plane through the points (1,1, 2), (1,2,1) and (2,1, 4). There are several approaches we could take: The general equation of a line is ax + by + cz + d = 0. We need to determine appropriate values of a,b, c and d. Using the three points, we get a homogeneous system of linear equations So we take the augmented matrix and put it in reduced row echelon form: Planes in 3-space 2

3 The general solution is then d = t, b = t, and or is 2x 3y + z + 3 = 0 which in turn implies that the equation of the plane A second approach is to use the point-normal form of the equation of a plane. Using the first of the three points, we have a(x 1) + b(y 1) + c(z + 2) = 0, and using the other two points we have with augemented matrix and reduced row echelon form This has a general solution of c = t, b = 3t and a = 2t, so a(x 1) + b(y 1) + c(z + 2) = t(2(x 1) 3(y 1) + (z + 2)) = 0, and so 2(x 1) 3(y 1) + (z + 2) = 0. Expanding this equation gives the general form: 2x 3y + z + 3 = 0. A third approach is to use the cross product. Let P = (1,1, 2), R = (1,2,1) and Q = (2,1, 4) be the three points. The cross product of and will be orthogonal to the both vectors, and hence to the plane containing the three points. It's an easy computation: and the equation of the plane is 2x 3y + z + d = 0 Using any one of the given points in the plane, we get d = 3. Synopsis (Computation of the Equation of a Plane through Three Points) The following three methods may be used to find the plane through P = (x 1,y 1,z 1 ), Q = (x 2,y 2,z 2 ) and R = (x 3,y 3,z 3 ): 1. Use the form ax + by + cz + d = 0 and the three points to get three equations in four unknowns. 2. Use the normal form with P as a point in the plane: a(x x 1 ) + b(y y 1 ) + c(z z 1 ) = 0 and the remaining two points to get two equations in three unknowns. 3. Use to get a normal to the plane, and any one of the points to find d in the general form ax + by + cz + d = 0. Lines in 3-space As we have seen previously, if we consider a vector in coordinate form, then then the set of all where t is a real number, will be all the points on the line through and If consider where t is any real number. The Equation of a plane through Three Points 3

4 The geometry of the parallelogram rule reveals the situation: the line through and has been translated to a parallel one through Indeed, the line consists of all vectors of the form Parametric Equation of a Line in 3-space Let P = (x0,y0,z0) be a point and let be a vector in coordinate form. Then the line through P in the direction of consists of all points Example: Which of the points (3, 2,1,) and (2, 3,1) is on the line through (1, 1,2) in the direction of ( 1,2,1): For (x,y,z) to be on the line, there must be a t so that (x,y,z) = (1, 1,2) + t( 1,2,1) = (1 t, 1 + 2t,2 + t). If (x,y,z) = (3, 2,1), then the three coordinates respectively imply that t = 2, have only one value, the point (3, 2,1), is not on the line. and t = 1. Since t can If (x,y,z) = (2, 3,1),. then the three coordinates respectively imply t = 1, t = 1 and t = 1. Hence (2, 3,1) = (1, 1,2) + ( 1)( 1,2,1), and so is (2, 3,1) on the line. Example: Give the equation of the line in 3-space containing the two points (x 0,y 0,z 0 ) and (x 1,y 1,z 1 ): Let P = (x,y,z ), Q = (x,y,z ) and Then in coordinate notation is (x points 0 0 on 0 the line 1 are 1 all 1 those of the form 0 x,y y,z z ), and so the And so the line joining P and Q consists of all tp + (1 t)q, the points Q and P respectively. Note that t = 0 and t = 1 gives The Line as the Intersection of Two Planes Suppose we have two planes with equations Finding the line of intersection is the same as finding all solutions to the system of linear equations with augmented matrix Lines in 3-space 4

5 The reduced row echelon form for this system of equations is The solutions to this system of equations satisfy z = t, y = t 5 and x = t 3 or (x,y,z) = (t 3,t 5,t) = ( 3, 5,0) + t(1,1,1) which is the parametric equation of a line. It's easy to verify that every such point satisfies both of the equations defining the planes. Here is a plot of the two planes and their line of intersection: The more general situation is now pretty clear. Given two planes with equations a 1 x + b 1 y + c 1 z + d 1 = 0 and a 2 x + b 2 y + c 2 z + d 2 = 0 then the line of intersection, when existing, is found by solving the system of linear equations with augmented matrix The Line as the Intersection of Two Planes 5

6 If both rows have leading ones, then there is one free variable, and this leads to the parametric equation of the line of intersection. If at least one row has no leading one, then the last row is of the form If the * is nonzero, there are no solutions (this corresponds to parallel planes). If the * is zero, then the two equations do not correspond to different planes, and so there is no line of intersection. Example: The line of intersection of the two planes given above consist of all points of the form (x,y,z) = (t 3,t 5,t) = ( 3, 5,0) + t(1,1,1). Where does this line intersect the y-z plane? A point is in that plane if the x-coordinate is zero. This implies t = 3, and so the desired point is (0, 2,3). Distance from a Point to a Plane We have already computed the distance from a point to a line in 2-space. The problem in 3-space is almost identical. We want to compute the distance from a point (x 0,y 0,z 0 ) to a plane with equation ax + by + cz + d = 0. We will use the properties of the dot product in 3-space. Let (x 1,y 1,z 1 ) be any point in the plane, and let with initial vertex (x 1,y 1,z 1 ) and terminal vertex (x 1 + a,y 1 + b,z 1 + c). This vector is orthogonal to the plane. Let have the same initial vertex as and (x 0,y 0,z 0 ) as terminal vertex. Then d, the distance from the point (x 0,y 0,z 0 ) to the plane, is the same as the length of the projection of onto Compare this result with the one for 2-space. Distance from a Point to a Plane 6

7 The Distance between Parallel Planes Math1300:MainPage/three-space The basic idea for computing the distance between parallel planes is simple. Take a point on one plane and find the distance from that point to the second plane. Consider the two planes with equations 2x y + 3z + 1 = 0 and 2x y + 3z 1 = 0. It is clear that these planes are parallel since both are orthogonal to the vector (2, 1,3). How do we find a point on the first plane? Make it as easy as possible: set all but one of the variables equal to zero, say, x = z = 0. then y = 1 and (0,1,0) is in the first plane. Now we compute the distance from (0,1,0) to the second plane: The Distance between Parallel Planes 7