Problems in 3D Geometry: Lines and Planes

THREE DIMENSIONAL GEOMETRY SEBASTIAN VATTAMATTAM 1. Introduction Notation: The triplet (x; y; z) may denote both a point and the vector x^i+ y^j+ z^k: Each time of use, the sense will be specied. Figure 1. Plane Through a and Normal to n Date: 12-March-2010. 12 SEBASTIAN VATTAMATTAM 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 b= (a2b3 􀀀 a3b2; a3b1 􀀀 a1b3; a1b2 􀀀 a2b1) Box Product [a bc] = 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 bc] =a1 a2 a3 b1 b2 b3 c1 c2 c32. 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 2 Let P(r) be any point in the plane. Then r 􀀀 a is per- pendicular 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 Figure 2. Plane Through a and Normal to n 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. Denition 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 3 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 Figure 3. Plane Through the Points (a; 0; 0); (0; b; 0); (0; 0; c) 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 = 13D GEOMETRY 5 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 Form 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 = l6 SEBASTIAN VATTAMATTAM 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 4 Figure 4. Plane in the Normal Form3D GEOMETRY 7 Suppose a plane is normal to a vector with direction cosines l = cos ;m = cos ; n = cos : That is the unit vector ^n = (l; m; n) is normal to the plane. If P(r) is any point in the plane, then the shortest dis- tance of the plane from the origin p = Proj(r) on ^n = r ^n Thus the equation of the plane is x cos + y cos + z cos = p 3. Equations of a Line Denition 3.1. If a line makes angles ; ; ; respectively with OX;OY;OZ then ^n = (cos ; cos ; cos ) is a unit vector parallel to the line and cos ; cos ; cos are the di- rection 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 + c2 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)8 SEBASTIAN VATTAMATTAM 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 􀀀 z1 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:3D GEOMETRY 9 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 vector^i 􀀀2^j 􀀀3^k and parallel to the line joining the points with position vectors ^i 􀀀^j+ 4^k and 2^i+^j+ 2^k. Also nd the cartesian form of the equation. [Q 10 in [1]] 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) That is r =^i 􀀀 2^j 􀀀 3^k + (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 3^i+ 5^j 􀀀 6^k: [Q 1 in [1]]10 SEBASTIAN VATTAMATTAM Let n = (3; 5;􀀀6): Then the unit normal to the plane is ^n = n jnj = (3; 5;􀀀6) p70 Distance of the plane from the origin is r ^n = 7 So, the equation of the plane is r (3^i+ 5^j 􀀀 6^k) p70 Problem 4.3. Show that the plane whose vector equation is r (^i+2^j 􀀀^k) = 3 contains the line whose vector equation is r =^i+^j+ (2^i+^j+ 4^k):[Q 10 in [1]] 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 equa- tion 4.1. Hence the conclusion. Problem 4.4. Show that the line whose vector equation is r = 2^i 􀀀2^j+3^k+(^i 􀀀^j+4^k) is parallel to the plane whose equation is r (^i+ 5^j+ ^k) = 5.[Q 11 in [1]] 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 ic; and d are per- pendicular. c d = 03D GEOMETRY 11 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 [1]] 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 (^i+ 2^j 􀀀 3^k) and having di- rection 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 􀀀1 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 isr = a + b (x; y; z) = (0; 0; 0) + (1; 1; 1) ) x1 = y1 = z112 SEBASTIAN VATTAMATTAM Problem 4.8. Find the equations of the line passing through the point with position vector (􀀀^i 􀀀 2^j 􀀀 3^k) and perpen- dicular to the plane r (3^i 􀀀 4^j+ 5^k) = 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 b= (􀀀1; 5; 3) The direction cosines are 􀀀1 p35;5 p35;3 p35 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 perpendic- ular to c = (3; 2; 3). Therefore, c PM = 0 ) 3(3k 􀀀 3 + 2(2k 􀀀 3) + 3(3k 􀀀 7)) = 03D GEOMETRY 13 Find k and use distance formula to compute PM, the distance of P to the line. For any clarication please contact vattamattam@gmail.com 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