Vectors Calaulus 1.3

Lines and Planes Lines Our goal is to come up with the equation of a line given a vector v parallel to the line and a point (a,b,c) on the line.  The figure (shown in 2D for simplicity) shows that if P is a point on the line then         = P + tv for some number t.           The picture is the same for 3D.  The formula is given below.          Parametric Equations of a Line The parametric equations for the line through the point (a,b,c) and parallel to the vector v are       = + tv  Example: Find the parametric equations of the line that passes through the point (1, 2, 3) and is parallel to the vector <4, -2, 1> Solution: We write:         = <1, 2, 3> + t <4, -2, 1> = <1 + 4t, 2 - 2t, 3 + t> or         x(t) = 1 + 4t,   y(t) = 2 - 2t,   z(t) = 3 + t Exercise Find the parametric equations of the line through the two points (2,1,7) and (1,3,5). Hint: a vector parallel to the line has tail at (2,1,7) and head at (1,3,5). Planes If S is a plane then a vector n is normal (perpendicular) to the plane if it is orthogonal to every vector that lies on the plane.  Suppose that n is a normal vector to a plane and (a,b,c) is a point on the plane.  Let (x,y,z) be a general point on the plane, then           is parallel to the plane, hence n . = 0 this defines the equation of the plane. Example: Find the equation of the plane that contains the point (2,1,0) and has normal vector <1,2,3> Solution: We have         <1,2,3> .   =  0 so that         1(x - 2) + 2(y - 1) + 3z  =  0 or         x + 2y + 3z  =  4 Example Find the equation of the plane through the points         P  =  (0,0,1)      Q  =  (2,1,0)     and      R  =  (1,1,1) Solution Let          v  =  Q - P  =  <2, 1, -1> and          w = R - P = <1, 1, 0> then to find a vector normal to the plane, we find the cross product of v and w:         or          <1, -1, 1> We can now use the formula:         <1, -1, 1> .   =  0 or         x - y + z - 1  =  0 or         x - y + z  =  1 Distance Between a Point and a Plane Let P be a point and Q be a point on a plane with normal vector n, then the distance between P and the plane is given by  Distance Between a Point P and a Plane With Normal Vector n Let Q be a point on the plane with normal vector n.  The the distance from the point P to this plane is given by                                    ||PQ . n||               ProjnPQ  =                                                         ||n|| Example   Find the distance between the point (1,2,3) and the plane         2x - y - 2z = 5 Solution The normal vector can be read off from the equation as         n  =  <2, -1, -2> Now find a convenient point on the plane such as Q  =  (0, -5, 0).  We have          PQ  =  <-1, -7, -3> and          n . PQ  =  -2 + 7 + 6  =  11 We find the magnitude of n by taking the square root of the sum of the squares.  The sum is         4 + 1 + 4  =  9 so          || n ||  =  3 Hence the distance from the point to the plane is 11/3.          The Angle Between 2 Planes The angle between two planes is given by the angle between the normal vectors. Example Find the angle between the two planes         3x - 2y + 5z  =  1      and      4x + 2y - z  =  4 We have the two normal vectors are         n = <3,-2,5>      and      m = <4,2,-1> We have         n . m  =  3,                     hence the angle is