THREE DIMENSIONAL GEOMETRY SEBASTIAN VATTAMATTAM 1. Introduction Notation: The triplet (x, y, z) may denote both a point and the vector xˆi+ yˆj+ zˆk. Each time of use, the sense will be specified. Vector Operations If ¯a = (a1, a2, a3),¯b= (b1, b2, b3), ¯c = (c1, c2, c3), 2 R, then ¯a +¯b= (a1 + b1, a2 + b2, a3 + b3) ¯a = (a1, a2, a3) ¯a ·¯b= a1b1 + a2b2 + a3b3 ¯a ׯb= (a2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1) Box Product [¯a ¯b¯c] = ¯a ·¯b× ¯c = (a1, a2, a3) · (b2c3 − b3c2, b3c1 − b1c3, b1c2 − b2c1) = a1(b2c3 − b3c2) + a2(b3c1 − b1c3) + a3(b1c2 − b2c1) This can also be written in the determinant form as [¯a ¯b¯c] =a1 a2 a3 b1 b2 b3 c1 c2 c3 Date: 12-March-2010. 12 SEBASTIAN VATTAMATTAM 2. Equations of a plane We say, a vector is normal to a plane if it is perpendicular to any vector in the plane. Equation of a plane passing through the point ¯a and having a normal vector ¯n See Figure ?? Figure 1. Plane Through ¯a and Normal to ¯n Let P(¯r) be any point in the plane. Then ¯r − ¯a is perpendicular to ¯n. Therefore, (¯r − ¯a) · ¯n = 0 If ¯a = (x1, y1, z1), ¯r = (x, y, z), ¯n = (a, b, c) then we get a(x − x1) + b(y − y1) + c(z − z1) = 03D GEOMETRY 3 Putting d = −ax1−by1−cz1 we get the general equation of a plane as ax + by + cz + d = 0 where the vector (a, b, c) is normal to the plane. Definition 2.1. Suppose a plane meets the coordinate axes at the points P(a, 0, 0),Q(0, b, 0),R(0, 0, c). a, b, c are called the intercepts of the plane on the axes. Plane in the Intercept Form See Figure ?? Figure 2. Plane Through the Points (a, 0, 0), (0, b, 0), (0, 0, c) Let a, b, c be the intercepts of a plane on the axes. Suppose the equation of the plane is lx + my + nz + d = 04 SEBASTIAN VATTAMATTAM Since it passes through the point P(a, 0, 0), we have al + d = 0 ) l = −d a Similarly, m = −d b , n = −d c So, the equation of the plane is is −dx a + −dy b + −dz c + d = 0 That is xa + yb + zc = 1 Plane Containing the Point ¯a and Parallel to the Vectors ¯c, ¯ d Let P(¯r) be any point in the plane. Then ¯r − ¯a, ¯c, ¯ d are coplanar. Therefore the box product [¯r − ¯a ¯c ¯ d] = 0 Cartesian Form If ¯r = (x, y, z), ¯a = (x1, y1, z1), ¯c = (c1, c2, c3), ¯ d = (d1, d2, d3) the equation isx − x1 y − y1 z − z1 c1 c2 c3 d1 d2 d3 = 0 Plane containing the points ¯a,¯band parallel to the vector ¯n Let P(¯r) be any point in the plane. Then ¯r − ¯a,¯b− ¯a, ¯n are coplanar. Therefore the box product [¯r − ¯a ¯b− ¯a ¯n] = 0 Cartesian Form3D GEOMETRY 5 If ¯r = (x, y, z), ¯a = (x1, y1, z1),¯b= (x2, y2, z2)¯n = (l,m, n) the equation isx − x1 y − y1 z − z1 x2 − x1 y2 − y1 z2 − z1 l m n = 0 Plane containing the points ¯a, ¯b, ¯c Let P(¯r) be any point in the plane. Then ¯r − ¯a,¯b− ¯a, ¯c − ¯a are coplanar. Therefore the box product [r − ¯a ¯b− ¯a ¯c − ¯a] = 0 Plane containing the point ¯a and perpendicular to the planes ¯r · ¯c = k, ¯r · ¯ d = l Let P(¯r) be any point in the plane. Then ¯r − ¯a, ¯c, ¯ d are coplanar. Therefore the box product [r − ¯a ¯c ¯ d] = 0 Plane containing the points ¯a,¯band perpendicular to the plane ¯r · ¯c = d Let P(¯r) be any point in the plane. Then ¯r − ¯a,¯b− ¯a, ¯c are coplanar. Therefore the box product [r − ¯a ¯b− ¯a ¯c] = 0 The Normal Form See Figure ?? Suppose a plane is normal to a vector with direction cosines l = cos ,m = cos , n = cos . That is the unit vector ˆn = (l,m, n) is normal to the plane. If P(¯r) is any point in the plane, then the shortest distance of the plane from the origin p = Proj(¯r) on ˆn = ¯r · ˆn6 SEBASTIAN VATTAMATTAM Figure 3. Plane in the Normal Form Thus the equation of the plane is x cos + y cos + z cos = p 3. Equations of a Line Definition 3.1. If a line makes angles , , , respectively with OX,OY,OZ then ˆn = (cos , cos , cos ) is a unit vector parallel to the line and cos , cos , cos are the direction cosines of the lime. If ¯n = (a, b, c) is parallel to the line, then, a, b, c are called direction ratios of the line. In that case the direction cosines will be ± a pa2 + b2 + c2 ,± b pa2 + b2 + c2 ,± c pa2 + b2 + c23D GEOMETRY 7 Line through the point ¯a and parallel to the vector ¯c. Let ¯r be any point on the line. Then ¯r − ¯a is parallel to ¯c. Therefore ¯r − ¯a = ¯c Hence the equation of the line is ¯r = ¯a + ¯c Now, let ¯r = (x, y, z), ¯a = (x1, y1, z1), ¯c = (c1, c2, c3) The equation becomes (x, y, z) = (x1, y1, z1) + (c1, c2, c3) Thus we get the Cartesian form, x = x1 + c1 y = y1 + c2 z = z1 + c3 Alternatively, x − x1 c1 = y − y1 c2 = z − z1 c3 This is called the Canonical form. Line through the points ¯a and ¯b. Let ¯r be any point on the line. Then ¯r − ¯a is parallel to ¯b− ¯a. Therefore ¯r − ¯a = (¯b− ¯a) Hence the equation of the line is ¯r = ¯a + (¯b− ¯a) Now, let ¯r = (x, y, z), ¯a = (x1, y1, z1),¯b= (x2, y2, z2) The equation becomes (x, y, z) = (x1, y1, z1) + (x2 − x1, y2 − y1, z2 − z1) Thus we get the Cartesian form, x = x1 + x2 − x1 y = y1 + y2 − y1 z = z1 + z2 − z18 SEBASTIAN VATTAMATTAM Alternatively,x − x1 x2 − x1 = y − y1 y2 − y1 = z − z1 z2 − z1 Line as the intersection of two non-parallel planes. The equations of a line may be given as a pair of equations of non-parallel planes, as a1x + b1y + c1z + d1 = 0 = a2x + b2y + c2z + d2 We can bring this to the standard form as follows: Put z = z1, a particular value, and then solve for x, y. If x = x1, y = y1 the line passes through the point ¯a = (x1, y1, z1) Now, the vectors (a1, b1, c1), (a2, b2, c2) are normal to the respective planes and hence ¯c = ¯a × ¯bis parallel to the required line. But, ¯c = (b1c2 − b2c1, c1a2 − c2a1, a1b2 − a2b1) So, the equations of the line are x − x1 b1c2 − b2c1 = y − y1 c1a2 − c2a1 = z − z1 a1b2 − a2b1 4. Problems Problem 4.1. Find the vector equation of a line through the point with position vectorˆi−2ˆj−3ˆk and parallel to the line joining the points with position vectors ˆi−ˆj+ 4ˆk and 2ˆi+ˆj+ 2ˆk. Also find the cartesian form of the equation. [Q 10 in [?]] Let ¯a = (1,−2,−3),¯b= (1,−1, 4), ¯c = (2, 1, 2). The required line passes through ¯a and is parallel to ¯c − ¯b= (1, 2,−2). So, the vector equation of the line is ¯r = ¯a + (¯c −¯b)3D GEOMETRY 9 That is ¯r =ˆi− 2ˆj− 3ˆk + (i + 2j − 2k) The Cartesian form is x − 1 1 = y + 2 2 = z + 3 −2 Problem 4.2. Find the vector equation of a plane which is at a distance of 7 units from the origin, and which is normal to the vector 3ˆi+ 5ˆj− 6ˆk. [Q 1 in [?]] Let ¯n = (3, 5,−6). Then the unit normal to the plane is ˆn = ¯n |¯n| = (3, 5,−6) p70 Distance of the plane from the origin is ¯r · ˆn = 7 So, the equation of the plane is ¯r · (3ˆi+ 5ˆj− 6ˆk) p70 Problem 4.3. Show that the plane whose vector equation is ¯r · (ˆi+2ˆj−ˆk) = 3 contains the line whose vector equation is ¯r =ˆi+ˆj+ (2ˆi+ˆj+ 4ˆk).[Q 10 in [?]] Equation of the plane is (4.1) x + 2y − z = 3 The line is (x, y, z) = (1, 1, 0) + (2, 1, 4) A general point on the line is (1 + 2, 1 + , 4) One can easily verify that these coordinates satisfy equation ??. Hence the conclusion.10 SEBASTIAN VATTAMATTAM Problem 4.4. Show that the line whose vector equation is ¯r = 2ˆi−2ˆj+3ˆk+(ˆi−ˆj+4ˆk) is parallel to the plane whose equation is ¯r · (ˆi+ 5ˆj+ ˆk) = 5.[Q 11 in [?]] Let ¯c = (1,−1, 4), ¯ d = (1, 5, 1). ¯c is the direction vector of the line and ¯ d is the normal vector to the plane. The line will be parallel to the plane iff ¯c, and ¯ d are perpendicular. ¯c · ¯ d = 0 Hence the conclusion. Problem 4.5. Find the point where the line x − 1 2 = y − 2 −3 = z + 3 4 meets the plane 2x + 4y − z = 1[Q 16 in [?]] A general point of the line is (1 + 2, 2 − 3,−3 + 4). If it is on the plane, we get 2(1 + 2) + 4(2 − 3) − (−3 + 4) = 1 ) = 1 So, the common point is (3,−1, 1) Problem 4.6. Find the equation of the line passing through the point with position vector (ˆi+ 2ˆj− 3ˆk) and having direction cosines (1/p3, 1/p3,−1/p3). The required line passes through ¯a = (1, 2,−3) and is parallel to the vector ¯c = (1/p3, 1/p3,−1/p3) So, its equation is x − 1 1/p3 = y − 2 1/p3 = z + 3 −1/p3 Multiplying throughout with p3 the equation becomes x − 1 1 = y − 2 1 = z + 3 −13D GEOMETRY 11 Problem 4.7. Find the equations of a line passing through the origin and equally inclined to the positive directions of the coordinate axes. Since the line is equally inclined to the axes, it passes through the point ¯b= (1, 1, 1). Thus, the line passes through the points ¯a = (0, 0, 0) and ¯b= (1, 1, 1). Equation of the line is¯r = ¯a + ¯b (x, y, z) = (0, 0, 0) + (1, 1, 1) ) x1 = y1 = z1 Problem 4.8. Find the equations of the line passing through the point with position vector (−ˆi− 2ˆj− 3ˆk) and perpendicular to the plane ¯r · (3ˆi− 4ˆj+ 5ˆk) = 11 Let ¯a = (−1,−2,−3). The vector ¯c = (3,−4, 5) is normal to the given plane. The required line is perpendicular to the plane and hence parallel to ¯c. The equation of the line is ¯r = ¯a + ¯b We get the cartesian form as x + 1 3 = y + 2 −4 = z + 3 5 Problem 4.9. Find the direction cosines of the line which is perpendicular to the lines whose direction ratios are (1,−1, 2), (2, 1,−1) The given lines are parallel to the vectors ¯a = (1,−1, 2) and ¯b= (2, 1,−1). The required line is parallel to ¯a ׯb= (−1, 5, 3) The direction cosines are ± −1 p35,± 5 p35,± 3 p3512 SEBASTIAN VATTAMATTAM Problem 4.10. Find the distance of the point (0, 2, 3) from the line x + 3 3 = y + 1 2 = z + 4 3 Hint: Let P(0, 2, 3) be the point and x+3 3 = y+1 2 = z+4 3 = k M(3k − 3, 2k − 1, 3k − 4) is a point on the line. If M is the foot of perpendicular, then PM is perpendicular to ¯c = (3, 2, 3). Therefore, ¯c · PM = 0 ) 3(3k − 3 + 2(2k − 3) + 3(3k − 7)) = 0 Find k and use distance formula to compute PM, the distance of P to the line. References [1] P.K.Jain,Hukum Singh,Mathematics, A Textbook for Class XII, Part I,NCERT, 2003 Mary Bhavan, Ettumanoor, Kerala, INDIA E-mail address: vattamattam@gmail.com