Handbook of mathematics for engineers and scienteists part 24 pot

4.6. LINE AND PLANE IN SPACE 129 R R R R Mxyz(,,) M 1 1 1 2 2 Figure 4.40. Plane passing through a point and parallel to two straight lines. 4.6.1-9. Equation of plane passing through point and parallel to two straight lines. The equation of the plane passing through a point M 1 (x 1 , y 1 , z 1 ) and parallel to two straight lines with direction vectors R 1 =(l 1 , m 1 , n 1 )andR 2 =(l 2 , m 2 , n 2 ) (see Fig. 4.40) is      x – x 1 y – y 1 z – z 1 l 1 m 1 n 1 l 2 m 2 n 2      = 0,or  (r – r 1 )R 1 R 2  = 0,(4.6.1.13) where r and r 1 are the position vectors of the points M(x, y, z)andM 1 (x 1 , y 1 , z 1 ), respec- tively. Example 7. Let us find the equation of the plane passing through the point M 1 (0, 1, 0) and parallel to the straight lines with direction vectors R 1 =(1, 0, 1)andR 2 =(0, 1, 2). According to (4.6.1.13), the desired equation is      x – 0 y – 1 z – 0 101 012      = 0, whence –x – 2y + z + 2 = 0. 4.6.1-10. Plane passing through two points and perpendicular to given plane. The plane (see Fig. 4.41) passing through two points M 1 (x 1 , y 1 , z 1 )andM 2 (x 2 , y 2 , z 2 )and perpendicular to the plane given by the equation Ax + By + Cz + D = 0 is determined by the equation      x – x 1 y – y 1 z – z 1 x 2 – x 1 y 2 – y 1 z 2 – z 1 ABC      = 0,or  (r – r 1 )(r 2 – r 1 )N  = 0,(4.6.1.14) where r, r 1 ,andr 2 are the position vectors of the points M(x, y, z), M 1 (x 1 , y 1 , z 1 ), and M 2 (x 2 , y 2 , z 2 ), respectively. Remark. If the straight line passing through points M 1 (x 1 , y 1 , z 1 )andM 2 (x 2 , y 2 , z 2 ) is perpendicular to the original plane, then the desired plane is undetermined and equations (4.6.1.14) become identities. 130 ANALYTIC GEOMETRY N N M M Mxyz(,,) 2 1 Figure 4.41. Plane passing through two points and perpendicular to given plane. Example 8. Let us find an equation of the plane passing through the points M 1 (0, 1, 2)andM 2 (2, 2, 3) and perpendicular to the plane x – y + z + 5 = 0. According to (4.6.1.14), the desired equation is      x – 0 y – 1 z – 2 2 – 02– 13– 2 1 –11      = 0, whence 2x – y – 3z + 7 = 0. 4.6.1-11. Plane passing through point and perpendicular to two planes. The plane (see Fig. 4.42) passing through a point M 1 (x 1 , y 1 , z 1 ) and perpendicular to two (nonparallel) planes A 1 x + B 1 y + C 1 z + D 1 = 0 and A 2 x + B 2 y + C 2 z + D 2 = 0 is given by the equation      x – x 1 y – y 1 z – z 1 A 1 B 1 C 1 A 2 B 2 C 2      = 0,or  (r – r 1 )N 1 N 2  = 0,(4.6.1.15) where N 1 =(A 1 , B 1 , C 1 )andN 2 =(A 2 , B 2 , C 2 ) are the normals to the given planes and r and r 1 are the position vectors of the points M(x, y, z)andM 1 (x 1 , y 1 , z 1 ), respectively. Figure 4.42. Plane passing through a point and perpendicular to two planes. Remark 1. Equations (4.6.1.15) mean that the vectors −−−→ M 1 M, N 1 ,andN 2 are coplanar. 4.6. LINE AND PLANE IN SPACE 131 Remark 2. If the original planes are parallel, then the desired plane is undetermined. In this case, equations (4.6.1.15) become identities. Example 9. Let us find an equation of the plane passing through the point M 1 (0, 1, 2) and perpendicular to the planes x – y + z + 5 = 0 and –x + y + z – 1 = 0. According to (4.6.1.15), the desired equation is      x – 0 y – 1 z – 2 1 –11 –11 1      = 0, whence x + y – 1 = 0. 4.6.1-12. Equation of plane passing through line of intersection of planes. The planes passing through the line of intersection of the planes A 1 x +B 1 y + C 1 z + D 1 = 0 and A 2 x + B 2 y + C 2 z + D 2 = 0 are given by the equation α(A 1 x + B 1 y + C 1 z + D 1 )+β(A 2 x + B 2 y + C 2 z + D 2 )=0,(4.6.1.16) which is called the equation of a pencil of planes.Hereα and β are arbitrary parameters. Let α ≠ 0.Setβ/α = λ; then equation (4.6.1.16) becomes A 1 x + B 1 y + C 1 z + D 1 + λ(A 2 x + B 2 y + C 2 z + D 2 )=0.(4.6.1.17) By varying the parameter λ from –∞ to +∞, we obtain all the planes in the pencil. For λ = 1, we obtain equations of the planes that bisect the angles between the given planes provided that the equations of the latter are given in normalized form. Remark. The passage from equation (4.6.1.16) to equation (4.6.1.17) excludes the case α = 0. Equa- tion (4.6.1.17) does not define the plane A 2 x + B 2 y + C 2 z + D 2 = 0; i.e., equation (4.6.1.17) for various λ defines all the planes in the pencil but one (the second of the two given planes). 4.6.2. Line in Space 4.6.2-1. Parametric equation of straight line. The parametric equation of the line that passes through a point M 1 (x 1 , y 1 , z 1 ) and is parallel to a direction vector R =(l, m, n) (see Fig. 4.43) is x = x 1 + lt, y = y 1 + mt, z = z 1 + nt,orr = r 1 + tR,(4.6.2.1) where r = −−→ OM and r 1 = −−−→ M 1 M. As the parameter t varies from –∞to +∞, the point M with position vector r =(x, y, z) determined by formula (4.6.2.1) runs over the entire straight line in question. It is convenient to use the parametric equation (4.6.2.1) if one needs to find the point of intersection of a straight line with a plane. The numbers l, m,andn characterize the direction of the straight line in space; they are called the direction coefficients of the straight line. For a unit vector R = R 0 , the coefficients l, m, n are the cosines of the angles α, β,andγ formed by this straight line (the direction of the vector R 0 ) with the coordinate axes OX, OY ,andOZ. These cosines can be expressed via the coordinates of the direction vector R as cos α = l √ l 2 + m 2 + n 2 ,cosβ = m √ l 2 + m 2 + n 2 ,cosγ = n √ l 2 + m 2 + n 2 .(4.6.2.2) Example 1. Let us find the equation of the straight line that passes through the point M 1 (2,–3, 1)andis parallel to the direction vector R =(1, 2,–3). According to (4.6.2.1), the desired equation is x = 2 + t, y =–3 + 2t, z = 1 – 3t. 132 ANALYTIC GEOMETRY r tR R r 1 1 M Mxyz(,,) X O Y Z Figure 4.43. Straight line passing through a point and parallel to direction vector. 4.6.2-2. Canonical equation of straight line. The equation x – x 1 l = y – y 1 m = z – z 1 n ,or(r – r 1 ) × R = 0,(4.6.2.3) is called the canonical equation of the straight line passing through the point M 1 (x 1 , y 1 , z 1 ) with position vector r 1 =(x 1 , y 1 , z 1 ) and parallel to the direction vector R =(l, m, n). Remark 1. One can obtain the canonical equation (4.6.2.3) from the parametric equations (4.6.2.1) by eliminating the parameter t. Remark 2. In the canonical equation, all coefficients l, m,andn cannot be zero simultaneously, since |R| ≠ 0. But some of them may be zero. If one of the denominators in equations (4.6.2.3) is zero, this means that the corresponding numerator is also zero. Example 2. The equations (x – 1)/1 =(y – 3)/4 =(z – 3)/0 determine the straight line passing through the point M 1 (1, 3, 3) and perpendicular to the axis OZ. This means that the line lies in the plane z = 3,and hence z – 3 = 0 for all points of the line. Example 3. Let us find the equation of the straight line passing through the point M 1 (2,–3, 1) and parallel to the direction vector R =(1, 2,–3). According to (4.6.2.3), the desired equation is x – 2 1 = y + 3 2 = z – 1 –3 . 4.6.2-3. General equation of straight line. The general equation of a straight line in space defines it as the line of intersection of two planes (see Fig. 4.44) and is given analytically by a system of two linear equations A 1 x + B 1 y + C 1 z + D 1 = 0, A 2 x + B 2 y + C 2 z + D 2 = 0, or r ⋅ N 1 + D 1 = 0, r ⋅ N 2 + D 2 = 0, (4.6.2.4) where N 1 =(A 1 , B 1 , C 1 )andN 2 =(A 2 , B 2 , C 2 ) are the normals to the planes and r is the position vector of the point (x, y, z). The direction vector R is equal to the cross product of the normals N 1 and N 2 ; i.e., R = N 1 × N 2 ,(4.6.2.5) 4.6. LINE AND PLANE IN SPACE 133 Figure 4.44. Straight line as intersection of two planes. and its coordinates l, m,andn can be obtained by the formulas l =    B 1 C 1 B 2 C 2    , m =    C 1 A 1 C 2 A 2    , n =    A 1 B 1 A 2 B 2    .(4.6.2.6) Remark 1. Simultaneous equations of the form (4.6.2.4) define a straight line if and only if the coefficients A 1 , B 1 ,andC 1 in one of them are not proportional to the respective coefficients A 2 , B 2 ,andC 2 in the other. Remark 2. For D 1 = D 2 = 0 (and only in this case), the line passes through the origin. Example 4. Let us reduce the equation of the straight line x + 2y – z + 1 = 0, x – y + z + 3 = 0 to canonical form. We choose one of the coordinates arbitrarily; say, x = 0.Then 2y – z + 1 = 0,–y + z + 3 = 0, and hence y =–4, z =–7. Thus the desired line contains the point M(0,–4,–7). We find the cross product of the vectors N 1 =(1, 2,–1)andN 2 =(1,–1, 1) and, according to (4.6.2.5), obtain the direction vector R =(1,–2,–3) of the desired line. Therefore, with (4.6.2.3) taken into account, the equation of the line becomes x 1 = y + 4 –2 = z + 7 –3 . 4.6.2-4. Equation of line in projections. The equation of a line in projections can be obtained by eliminating first z and then y from the general equations (4.6.2.4): y = kx + a, z = hx + b.(4.6.2.7) Each of two equations (4.6.2.7) defines a plane projecting the straight line onto the planes OXY and OXZ (see Fig. 4.45). Remark 1. For straight lines parallel to the plane OYZ, this form of the equations cannot be used; one should take the projections onto some other pair of coordinate planes. Remark 2. Equations (4.6.2.7) can be represented in the canonical form x – 0 1 = y – a k = z – b h .(4.6.2.8) 134 ANALYTIC GEOMETRY x O y z Figure 4.45. Straight line with equation in projections. 4.6.2-5. Equation of straight line passing through two points. The canonical equation of the straight line (see Fig. 4.46) passing through two points M 1 (x 1 , y 1 , z 1 )andM 2 (x 2 , y 2 , z 2 )is x – x 1 x 2 – x 1 = y – y 1 y 2 – y 1 = z – z 1 z 2 – z 1 ,or(r – r 1 ) × (r 2 – r 1 )=0,(4.6.2.9) where r, r 1 ,andr 2 are the position vectors of the points M(x, y, z), M 1 (x 1 , y 1 , z 1 ), and M 2 (x 2 , y 2 , z 2 ), respectively. The parametric equations of the straight line passing through two points M 1 (x 1 , y 1 , z 1 ) and M 2 (x 2 , y 2 , z 2 ) in the rectangular Cartesian coordinate system OXY Z can be written as x = x 1 (1 – t)+x 2 t, y = y 1 (1 – t)+y 2 t, z = z 1 (1 – t)+z 2 t, or r =(1 – t)r 1 + tr 2 .(4.6.2.10) Remark. Eliminating the parameter t from equations (4.6.2.10), we obtain equations (4.6.2.9). r r r 2 1 2 1 M M Mxyz(,,) X O Y Z Figure 4.46. Straight line passing through two points. N M 0 Figure 4.47. Straight line passing through point and perpendicular to plane. 4.6.2-6. Equation of straight line passing through point and perpendicular to plane. The equation of the straight line passing through a point M 0 (x 0 , y 0 , z 0 ) and perpendicular to the plane given by the equation Ax + By+ Cz+ D = 0,orr ⋅ N + D = 0 (see Fig. 4.47), is x – x 0 A = y – y 0 B = z – z 0 C ,(4.6.2.11) where N =(A, B, C) is the normal to the plane. 4.6. LINE AND PLANE IN SPACE 135 4.6.3. Mutual Arrangement of Points, Lines, and Planes 4.6.3-1. Angles between lines in space. Consider two straight lines determined by vector parametric equations r = r 1 + tR 1 and r = r 2 + tR 2 . The angle ϕ between these lines (see Fig. 4.48) can be obtained from the formulas cos ϕ = R 1 ⋅ R 2 |R 1 ||R 2 | ,sinϕ = |R 1 × R 2 | |R 1 ||R 2 | . If the lines are given by the canonical equations x – x 1 l 1 = y – y 1 m 1 = z – z 1 n 1 and x – x 2 l 2 = y – y 2 m 2 = z – z 2 n 2 , then the angle ϕ between the lines can be found from the formulas cos ϕ = l 1 l 2 + m 1 m 2 + n 1 n 2  l 2 1 + m 2 1 + n 2 1  l 2 2 + m 2 2 + n 2 2 , sin ϕ =     m 1 n 1 m 2 n 2    2 +    n 1 l 1 n 2 l 2    2 +    l 1 m 1 l 2 m 2    2  l 2 1 + m 2 1 + n 2 1  l 2 2 + m 2 2 + n 2 2 , (4.6.3.1) which coincide with formulas (4.6.3.1) written in coordinate form. R R 2 1 φ Figure 4.48. Angles between lines in space. Example 1. Let us find the angle between the lines x 1 = y – 2 2 = z + 1 2 and x 0 = y – 2 3 = z + 1 4 . Using the first formula in (4.6.3.1), we obtain cos ϕ = 1 ⋅ 0 + 3 ⋅ 2 + 4 ⋅ 2 √ 0 2 + 3 2 + 4 2 √ 1 2 + 2 2 + 2 2 = 14 15 , and hence ϕ ≈ 0.3672 rad. 4.6.3-2. Conditions for two lines to be parallel. Two straight lines given by vector parametric equations r = r 1 + tR 1 and r = r 2 + tR 2 are parallel if R 2 = λR 1 or R 2 × R 1 = 0, . 0. 4.6.1-12. Equation of plane passing through line of intersection of planes. The planes passing through the line of intersection of the planes A 1 x +B 1 y + C 1 z + D 1 = 0 and A 2 x + B 2 y +. cosines of the angles α, β ,and formed by this straight line (the direction of the vector R 0 ) with the coordinate axes OX, OY ,andOZ. These cosines can be expressed via the coordinates of the. equations of the form (4.6.2.4) define a straight line if and only if the coefficients A 1 , B 1 ,andC 1 in one of them are not proportional to the respective coefficients A 2 , B 2 ,andC 2 in the