Distance between a point and a plane Let P be the given point and let the plane have the equation ax+ by+ cz= d. A normal vector to the plane is n= (a, b, c). Let Q be any point on the plane. The distance between the point P and the plane is the magnitude of the projection of QPon n. P Now nQPnproj QPnnn•=•JJJGJJJGand so the required distance = nQPprojJJJJG= QPQPPQQPn••••nnnJJJGJJJGJJJGJJJGnnn===•nnnnnsince QPPQ•=•nnJJJGJJJGExample Find the distance between the point P = (2, 1, 3) and the plane 2x+ y+ 2z= 3. D n Q 8

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Solution A point Q on the plane is Q = (1, 1, 0). Then QP(2,1, 3)(1,1, 0)(1, 0, 3)=−=JJJGand a normal vector to the plane is n= (2, 1, 2). The distance between the point P and the plane is QPn(1, 0, 3)(2,1, 2)206n(2,1, 2)41••+==++JG834+=JJ. Distance between a point and a line in R3Let lbe a line passing through a point Q and parallel to the vector v. Let P be a point not on the line l. The distance between P and lis the perpendicular distance of the point P from l. In the adjoining diagram the distance between P and l is the length of the linesegment PD. Let θbe the angle between vand QPJJJG. Then the required distance is QPsinθQPPQPDQPsinθ⋅⋅×=⋅===vvvvJJJGJJJGJ GJJJGJJJG×vvJJ. Example Find the distance between the point P = (5, 7, 10) and the line having parametric equations x= 2 + 3t, y= 3 – 3t, z= 4 + t. Solution The point Q = (2, 3, 4) is a point on the line so QP= (5, 7, 10) – (2, 3, 4) = (3, 4, 6). The distance is equal to QPv×JJJGvwhere v is the vector v= (3, −3, 1) and QPis the vector ×vJJJG463634QP,,(22,15,21)⎛⎞×=−=−⎜⎟vJJJG313133−−⎝⎠. The required distance is QPv×JJJGv(22, 15,21)(25)(46)4842254411150546(3,3, 1)991191919−++=====−++. lv PθQ D Vector Geometry and Linear Algebra MATH 1300 Unit 2 9

The distance between two lines in R3Let l1be a line through P that is parallel to the vector v1.Let l2be a line through Q that is parallel to the vector v2. The vector n= v1×v2is a normal vector to both lines l1and l2. The distance between the lines is the magnitude of the projection of QPon n. If θis the angle between QPand n, then the distance between the lines is QPcosθQPPQPcosθ⋅⋅•⋅==nnnnnJJJGJJJGJJJGQ•=nJJJG. Example Find the distance between the lines l1: x= 3 + 2t, y= 5 – 3t, z= 4 + 4tand l2: x= 1 + t, y= 2 + 2t, z= 3 −3t. SolutionThe point P = (3, 5, 4) is on line l1and the point Q = (1, 2, 3) is on line l2. The vector QP= (3, 5, 4) – (1, 2, 3) = (2, 3, 1). The vector v1= (2, −3, 4) is parallel to line l1and the vector v2= (1, 2, −3) is parallel to line l2. The normal vector to lines l1and l2is the vector JJJG123423=×=nvv2423,,(1,10, 7)1312⎛⎞−−−=⎜⎟−−⎝⎠. The distance is QPnn•JJJG= (2, 3, 1)(1, 10, 7)23073939.(1,10,7)11004915056•++===++2.4 Problems 1. Find the distance between the given point P and the given plane. a. P = (1, 3, 2) 2x+ 2y+ z= 1 b. P = ( 3, 5, 4) 3x+ 4y+ 5z= 12 c. P = (1, 0, 2) 2x+ y+ 2z= 1 d. P = (2, 2, 1) 3x+ y + z= 9 2. Find the distance between the following pairs of parallel planes by selecting a point on one plane and then finding the distance between that point and the other plane.