Lines and Planes in R 3

.3 Lines and Planes in R 3 P. Daniger Lines in R 3 We wish to represent lines in R 3. Note that a line may be described in two different ways: By specifying two points on the line. By specifying one point on the line and a vector parallel to it. If we are given two points, P and Q on a line, then a vector parallel to it is P Q. There are three ways of representing a line algebraically. Vector Representation of a Line Given a point P = ( 0, y 0, 0 ) on the line and a vector v = (a, b, c) parallel to it. An arbitrary point X = (, y, ) on the line will be given by the vector equation: y OX = OP + tv. 0 = y 0 + t 0 If we are given two points, P and Q on the line, then take v = P Q. Parametric Representation of a Line Given a point P = ( 0, y 0, 0 ) on the line and a vector v = (a, b, c) parallel to it. An arbitrary point X = (, y, ) on the line will be given by the system of equations: = 0 + ta y = y 0 + tb = 0 + tc If we are given two points, P and Q = (, y, ) on the line, then take v = becomes. = 0 + t( 0 ) y = y 0 + t(y y 0 ) = 0 + t( 0 ) Symmetric Representation of a Line Solving for t gives the Symmetric Representation of a Line. t = 0 a = y y 0 b Note that this is only definied if a, b and c are non-ero. a b c = 0 c P Q and this

.3 Lines and Planes in R 3 P. Daniger Eample. Find the equation of the line l joining P = (,, ) to Q = (, 0, ) in vector form. P Q = (0,, 0), so the equation os given by y = + t 2. Find the equation in parametric form of the line l above. = y = t = 3. Find the equation in symmetric form of the line l above. This does not eist. 4. Does R = (, 2, 2) lie on l? Substituting R = (, 2, 2) for X = (, y, ) we get = 2 = t 2 = Which has no solution, so R does not lie on the line. 5. Does S = (, 2, ) lie on l? Substituting S = (, 2, ) for X = (, y, ) we get = 2 = t = Which is true when t =, so this lies on the line. 6. With the parameteriation above at what point will we be when t = 2 When t = 2 we will be at (, 3, ). 0 0 2

.3 Lines and Planes in R 3 P. Daniger 2 Planes in R 3 We wish to represent planes in R 3. Note that a plane may be described in three different ways: By specifying three points on the plane. By specifying one point in the plane and two vectors parallel to it. By specifying one point in the plane and a vector perpendicular to it. The third form is preferable since it needs the least information. Let π be a plane described by a vector n = (a, b, c) orthogonal to it and a point P = ( 0, y 0, 0 ) which lies in it. Consider a point Q = (, y, ) on the plane π. Since n is orthogonal to the plane n v = 0 for any vector v parallel to the plane. Now P Q = ( 0, y y 0, 0 ) is in the plane. So n P Q = 0, or a( 0 ) + b(y y 0 ) + c( 0 ) = 0 This is called the point normal form of the equation of a plane. Setting d = a 0 + by 0 + c 0 = n OP, we get a + by + c = d This is called the standard form of the equation of a plane. Eample 2. Find the equation of the plane π which is orthogonal to the vector n = (,, 2) and through the point P = (, 0, ) d = n OP = (,, 2) (, 0, ) = 3 Equation is n OX = d + y + 2 = 3 2. Is Q = (, 2, 3) π? Put X = Q in the equation to get + 2 + 2 3 = 9 3 So equation is inconsistent and Q π. 3. Is Q = (3, 2, ) π? Put X = Q in the equation to get 3 + 2 + 2 ( ) = 3 So equation is consistent and Q π. 3

.3 Lines and Planes in R 3 P. Daniger If to vectors u and v parallel to the plane are given we can take solve the equations n u = 0 n v = 0. Generally this will result in an infinite solution set (a line through the origin). Any vector parallel to this line will work for n as all are perpendicular to the plane. The magnitude of n will affect the value of d. Alternately we can use the cross product n = u v. (see Section 4.3). If a plane is defined three points P, Q and R in the plane then P Q, P R and QR are all vectors parallel to the plane and the method outlined above may be used. Eample 3 Find the equation of the plane π parallel to u = (, 0, ) and v = (2,, 2), through P = (,, ). Let n = (n, n 2, n 3 ) be the normal vector to π. u n = 0 (, 0, ) (n, n 2, n 3 ) = n + n 3 = 0. v n = 0 (2,, 2) (n, n 2, n 3 ) = 2n n 2 + 2n 3 = 0. The solution to the set of simultaneous equations n + n 3 = 0. 2n n 2 + 2n 3 = 0. are vectors of the form (t, 0, t), for any t R. We arbitrarily pick t =, so n = (, 0, ). Alternately, using cross product i j k n = u v = 0 2 2 = i k Now, d = n OP = (, 0, ) (,, ) = = 0. So the equation of the plane is = 0 Eample 4 Find the equation of the plane through the points P = (,, ), Q = (0,, 2) and R = (,, 2) P Q = (, 0, ) and QR = (, 0, 0) are two vectors in the plane. Let n = (n, n 2, n 3 ) be the normal vector to the plane. P Q n = 0 (, 0, ) (n, n 2, n 3 ) = n + n 3 = 0. QR n = 0 (, 0, 0) (n, n 2, n 3 ) = n = 0. 4

.3 Lines and Planes in R 3 P. Daniger The solution to the set of simultaneous equations n + n 3 = 0, and n = 0 are vectors of the form (0, t, 0), for any t R. We arbitrarily pick t =, so n = (0,, 0). Alternately, using cross product: n = P Q QR i j k = 0 = (0,, 0) 0 0 Now, d = n OP = (0,, 0) (,, ) =. So the equation of the plane is y = Eample 5. Find the point of intersection of the plane = 0 with the line l: 2 2 y = + t The line can be epressed as = 2 + 2t, y = t, = + t. Intersection (, y, ) for, y and which satisfy both the equation of the line and that of the plane. Substitute equation for the line into the equation for the plane and solve for t. So the solution is t =. (2 + 2t) ( + t) = 0 } {{ } } {{ } t + = 0 Substituting back in the equation for the line l we get the point of intersection: (2 2( ), ( ), ) = (0, 2, 0) 2. Find the point of intersection of π with the line l: y = + t 2 Intersect when these are the same, y and. = + t, y = + 2t, = + t, substitute in equation for plane and solve for t Many solutions, so the line is in the plane. ( + t) ( + t) = 0 } {{ } } {{ } 0 = 0 5

.3 Lines and Planes in R 3 P. Daniger 3. Find the point of intersection of π with the line l: y = 2 + t Intersect when these are the same, y and. = 2 + t, y = + 2t, = + t, substitute in equation for plane and solve for t (2 + t) ( + t) = 0 = 0 No solution, so the line is parallel to the plane, but not in it. 2 6