Section 13.5: Equations of Lines and Planes 1 Objectives 1. Find vector, symmetric, or parametric equations for a line in space given two points on the line, given a point on the line and a vector parallel to the line, or given the equations of intersecting planes in space. (3,4,5,7,9,13) 2. Determine if two lines are parallel, skew, or intersecting; find the point of intersection if possible. (10,11,15,17) 3. Find the equation of a plane given three points on the plane, given a point on the plane and a vector normal to the plane, given two points on the plane and the equation of a plane perpendicular to it. (19,25,29,33) 4. Find the point of intersection of a line and a plane. (35,37) 5. Determine if two planes are parallel, perpendicular, or neither and compute the angle between planes. (41,43,45,47,49) 6. Compute the distance between a line in space and a point off the line. (59) 7. Compute the distance between a plane in space and a point off the plane. (61) 2 Assignments 1. Read Section 13.5 2. Problems: 1,3,9,11,17,21,23,27,31,35,45,50,53,63,68,70 3. Read Section 13.6 3 Lecture Notes In this section, we ll learn how to find equations for lines and planes in 3 dimensions. Equations for lines in two dimensions are easy; we learned how to do that in algebra. Equations for lines in three dimensions are not that different; in each case, all we want to do is find a convenient way to describe the set of points that constitute the line. In two dimensions, we only need a point and a slope to define a line. If we have two points, then we can recover the slope, so two points are all that is needed to define a line. It works much the same way in three dimensions, where now the slope is replaced by the idea of the direction vector of the line. Planes are the really new concept here. Note that a point and a vector parallel to the direction of the plane are not enough to describe a plane in three dimensions. However, if we have a point and the direction of a vector that is perpendicular to the plane, we can find the equation for the plane easily. 1
3.1 Equations of Lines We can find the equation of a line in three dimensions by using either two points on the line or by using one point on the line and a vector that is parallel to the line. 3.1.1 Vector Equations In three dimensions, we can equate a point P 0 (x 0, y 0, z 0 ) with a direction vector starting at the origin OP =< x 0, y 0, z 0 >. This notation will be useful when finding the equation of the line. Assume that we have two points, A(x 0, y 0, z 0 ) and B(x 1, y 1, z 1 ) on a line L. Then OA =< x 0, y 0, z 0 > and OB =< x 1, y 1, z 1 >. The direction of the line L is then the same as the direction AB =< x 1 x 0, y 1 y 0, z 1 z 0 >, that is, AB is parallel to L. We can move from A to B by noting the OB = OA + AB. We can get to a general point P (x, y, z) on the line, where OP =< x, y, z >, by moving from A a scalar multiple in the direction AB, so that OP = OA + tab. This is the vector equation for the line L, where t is called the parameter. We can generate all of the points on the line L by choosing different values for t and using the vector equation. Your text explains the vector equation in a different (and probably better) way. Let P 0 (x 0, y 0, z 0 ) be a point on L, and let v be a vector that is parallel to L. Tnen any point on the line P (x, y, z) is some distance away from P 0 in the direction of L. We can represent the points P 0 and P by their position vectors, i.e., the vectors with starting points at the origin and endpoints at the coordinates of the points. Label these vectors r 0 and r for the position vectors for P 0 and P, respectively. If P 0 P is a, then a = tv for some t IR. We can use the triangle law for the addition of two vectors to see that r = r 0 + tv, which gives the vector equation for the line L. (There s a good illustration in your text that I won t try to repeat here.) 3.1.2 Parametric Equations We can get the parametric equations for L by writing out and equating the components of the vector equation for L. Thus we get r = r 0 + tv < x, y, z > = < x 0, y 0, z 0 > +t < a, b, c > x = x 0 + ta y = y 0 + tb z = z 0 + tc. The components of the direction vector v are the direction numbers of our line L. 2
3.1.3 Symmetric Equations We can solve each of the parametric equations for t to obtain the symmetric equations x x 0 a = y y 0 b = z z 0, c as long as none of the direction numbers are 0. If a direction number is zero, it simply means that the line is constant along that component. If, for instance, we have a = 0, then the line is constant along the x component and the symmetric equations become x = x 0, y y 0 b = z z 0. c 3.2 Example 1 In Section 13.1, we had a homework question which asked you to determine if three points were on a line. There were two sets of points in the problem: 1. P 1 (5, 1, 3), P 2 (7, 9, 1), P 3 (1, 15, 11) 2. P 1 (0, 3, 4), P 2 (1, 2, 2), P 3 (3, 0, 1) First, find the line L between P 1 and P 2. Then, try to find a t that would give P 3. The direction of the line between P 1 and P 2 is < 7 5, 9 1, 1 3 >=< 2, 8, 4 >. The vector equation for L is r =< 5, 1, 3 > +t < 2, 8, 4 >, which gives the parametric equations The symmetric equations are x = 5 + 2t y = 1 + 8t z = 3 4t. x 5 2 = y 1 8 = z 3 4. Thus, if x = 1 as in P 3, the other coordinates for the point on the line must be y = 15 and z = 11, which is what we have for P 3. Thus, all three points lie on the same line. Class Questions 1. Repeat the above exercise for the second set of three points. 2. When does the line from the previous example intersect the xy plane? The yz plane? The xz plane? Insert example 16 here about skew/parallel lines 3
3.3 Planes A plane in space is determined by a point P 0 (x 0, y 0, z 0 ) in the plane and a vector n perpendicular to the plane. The vector n is called the normal vector. Let P (x, y, z) be an arbitrary point in the plane. Since n is orthogonal to the plane, it is orthogonal to every vector in the plane. In particular, n is orthogonal to the vector defined by P 0 P, which is clearly in the plane. If we again use position vectors n and r 0 to represent our points P and P 0, respectively, then we can see that P 0 P =< x x 0, y y 0, z z 0 >= n r 0, and since n is orthogonal to this vector, we must have n (n r 0 ) = 0. This last equation defines the vector equation for the plane. The scalar equation for the plane is obtained by writing out the vector equation of the plane, using n =< n x, n y, n z >. This gives n (n r 0 ) = 0 < n x, n y, n z > < x x 0, y y0, z z 0 > = 0 n x (x x 0 ) + n y (y y 0 ) + n z (z z 0 ) = 0. The above equation is the scalar equation of the plane through the point P 0 (x 0, y 0, z 0 ) with normal vector n =< n x, n y, n z >. We can simplify the scalar equation of the plane to get the linear equation of the plane. The linear equation of the plane is where d = (n x x 0 + n y y 0 + n z z 0 ). Class Questions: n x x + n y y + n z z + d = 0, 1. What are the vector equations for the xy, yz and xz planes? 2. What is the normal vector to each of these planes? 3.3.1 Example 3: Problems 13.5.20, 13.5.22 3.3.2 Example 4: Problem 13.5.28 If we know two vectors in a plane, then we can also find the equation of the plane. Problem 13.5.28 is a good problem to work because you are given three point in the plane and asked to find the equation for the plane. In this case, you need to find two vectors that define the plane and a vector that is orthogonal to the plane. The plane through the origin P 0 and P 1 (2, 4, 6) and P 2 (5, 1, 3) contains the vectors P 0 P 1 and P 0 P 2. These vectors are v 1 =< 2, 4, 6 > and v 2 < 5, 1, 3 >, respectively. (Why?) A vector orthogonal to the plane containing these vectors is n = v 1 v 2. Thus, n = 18i + 24j + 22k. Thus, the equation of the plane is given by 18(x 0) + 24(y 0) + 22(z 0) = 0. 4
3.3.3 Other facts about planes In three-dimensional space, either two planes are parallel or they intersect. Two planes are parallel if their corresponding normal vectors are parallel, that is, if n A = αn B. If two planes intersect, then the angle between the planes is the acute angle between their normal vectors. 3.3.4 Example 5: Problem 13.5.40 To find the cosine of the angle between the two planes x+y+z = 0 and x+2y+3z = 1, we first note that the vectors normal to the respective planes are n 1 =< 1, 1, 1 > and n 2 =< 1, 2, 3 >. (Why?) Then, we know that cos θ = n 1 n 2 n 1 n 2 = 1 + 2 + 3 3 14 = 6 7. 5