136 ANALYTIC GEOMETRY i.e., if their direction vectors R 1 and R 2 are collinear. If the straight lines are given by canonical equations, then the condition that they are parallel can be written as l 1 l 2 = m 1 m 2 = n 1 n 2 .(4.6.3.2) Remark. If parallel lines have a common point (i.e., r 1 = r 2 in parametric equations), then they coincide. Example 2. Let us show that the lines x – 1 2 = y – 3 1 = z 2 and x – 3 4 = y + 1 2 = z 4 are parallel to each other. Indeed, condition (4.6.3.2) is satisfied, 2 4 = 1 2 = 2 4 , and hence the lines are parallel. 4.6.3-3. Conditions for two lines to be perpendicular. Two straight lines given by vector parametric equation r = r 1 + tR 1 and r = r 2 + tR 2 are perpendicular if R 1 ⋅ R 2 = 0.(4.6.3.3) If the lines are given by canonical equations, then the condition that they are perpendicular can be written as l 1 l 2 + m 1 m 2 + n 1 n 2 = 0,(4.6.3.3a) which coincides with formula (4.6.3.3) written in coordinate form. Example 3. Let us show that the lines x – 1 2 = y – 3 1 = z 2 and x – 2 1 = y + 1 2 = z –2 are perpendicular. Indeed, condition (4.6.3.3a) is satisfied, 2 ⋅ 1 + 1 ⋅ 2 + 2 ⋅ (–2)=0, and hence the lines are perpendicular. 4.6.3-4. Theorem on the arrangement of two lines in space. THEOREM ON THE ARRANGEMENT OF TWO LINES IN SPACE. Two lines in space can: a) be skew; b) lie in the same plane and not meet each other, i.e., be parallel; c) meet at a point; d) coincide. A general characteristic of all four cases is the determinant of the matrix x 2 – x 1 y 2 – y 1 z 2 – z 1 l 1 m 1 n 1 l 2 m 2 n 2 ,(4.6.3.4) 4.6. LINE AND PLANE IN SPACE 137 whose entries are taken from 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 of the lines. In cases a–d of the theorem, for the matrix (4.6.3.4) we have, respectively: a) the determinant is nonzero; b) the last two rows are proportional to each other but are not proportional to the first row; c) the last two rows are not proportional, and the first row is their linear combination; d) all rows are proportional. 4.6.3-5. Angles between planes. Consider two planes given by the general 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, (4.6.3.5) 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). N N 1 2 φ Figure 4.49. Angles between planes. The angle between two planes (see Fig. 4.49) is defined as any of the two adjacent dihedral angles formed by the planes (if the planes are parallel, then the angle between them is by definition equal to 0 or π). One of these dihedral angles is equal to the angle ϕ between the normal vectors N 1 =(A 1 , B 1 , C 1 )andN 2 =(A 2 , B 2 , C 2 ) to the planes, which can be determined by the formula cos ϕ = A 1 A 2 + B 1 B 2 + C 1 C 2 A 2 1 + B 2 1 + C 2 1 A 2 2 + B 2 2 + C 2 2 = N 1 ⋅ N 2 |N 1 ||N 2 | .(4.6.3.6) If the planes are given by vector parametric equations r = r 1 + R 1 t + R 2 s or r = r 1 + R 1 t + R 2 s,(4.6.3.7) then the angle between the planes is given by the formula cos ϕ = (R 1 × R 2 ) ⋅ (R 1 × R 2 ) |R 1 × R 2 ||R 1 × R 2 | .(4.6.3.8) 138 ANALYTIC GEOMETRY 4.6.3-6. Conditions for two planes to be parallel. Two planes given by the general equations (4.6.3.5) in coordinate form are parallel if and only if the following condition for the planes to be parallel is satisfied: A 1 A 2 = B 1 B 2 = C 1 C 2 ≠ D 1 D 2 ;(4.6.3.9) in this case, the planes do not coincide. For planes given by the general equations (4.6.3.5) in vector form, the condition becomes N 2 = λN 1 or N 2 × N 1 = 0;(4.6.3.10) i.e., the planes are parallel if their normals are parallel. Example 4. Let us show that the planes x – y + z = 0 and 2x – 2y + 2z + 5 = 0 are parallel. Since condition (4.6.3.9) is satisfied, 1 2 = –1 –2 = 1 2 , we see that the planes are parallel to each other. 4.6.3-7. Conditions for planes to coincide. Two planes coincide if they are parallel and have a common point. Two planes given by the general equations (4.6.3.5) coincide if and only if the following condition for the planes to coincide is satisfied: A 1 A 2 = B 1 B 2 = C 1 C 2 = D 1 D 2 .(4.6.3.11) Remark. Sometimes the case in which the planes coincide is treated as a special case of parallel straight lines and is not distinguished as an exceptional case. 4.6.3-8. Conditions for two planes to be perpendicular. Planes are perpendicular if their normals are perpendicular. Two planes determined by the general equations (4.6.3.5) are perpendicular if and only if the following condition for the planes to be perpendicular is satisfied: A 1 A 2 + B 1 B 2 + C 1 C 2 = 0 or N 1 ⋅ N 2 = 0,(4.6.3.12) where N 1 =(A 1 , B 1 , C 1 )andN 2 =(A 2 , B 2 , C 2 ) are the normals to the planes. Example 5. Let us show that the planes x – y + z = 0 and x – y – 2z + 5 = 0 are perpendicular. Since condition (4.6.3.12) is satisfied, 1 ⋅ 1 +(–1) ⋅ (–1)+1 ⋅ (–2)=0, we see that the planes are perpendicular to each other. 4.6. LINE AND PLANE IN SPACE 139 4.6.3-9. Intersection of three planes. The point of intersection of three planes given by the 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, A 3 x + B 3 y + C 3 z + D 3 = 0, has the following coordinates: x 0 =– 1 Δ D 1 B 1 C 1 D 2 B 2 C 2 D 3 B 3 C 3 , y 0 =– 1 Δ A 1 D 1 C 1 A 2 D 2 C 2 A 3 D 3 C 3 , z 0 =– 1 Δ A 1 B 1 D 1 A 2 B 2 D 2 A 3 B 3 D 3 ,(4.6.3.13) where Δ is given by the formula Δ = A 1 B 1 C 1 A 2 B 2 C 2 A 3 B 3 C 3 .(4.6.3.14) Remark. Three planes are concurrent at a single point if Δ ≠ 0.IfΔ = 0 and at least one of the second- order minors is nonzero, then all planes are parallel to a single line. If all minors are zero, then the planes are concurrent in a single line. 4.6.3-10. Intersection of four planes. If four planes given by the 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, A 3 x + B 3 y + C 3 z + D 3 = 0, A 4 x + B 4 y + C 4 z + D 4 = 0, are concurrent at a single point, then xyz1 x 1 y 1 z 1 1 x 2 y 2 z 2 1 x 3 y 3 z 3 1 = 0.(4.6.3.15) To find the points of intersection, it suffices to find the point of intersection of any three of them (see Paragraph 4.6.3-9). The remaining equation follows from the three other equations. 4.6.3-11. Angle between straight line and plane. Consider a plane given by the general equation Ax + By + Cz + D = 0,orr ⋅ N + D = 0 (4.6.3.16) and a line given by the canonical equation x – x 1 l = y – y 1 m = z – z 1 n ,or(r – r 1 ) × R = 0,(4.6.3.17) where N =(A, B, C) is the normal to the plane, r and r 1 are the respective position vectors of the points (x, y, z)and(x 1 , y 1 , z 1 ), and R =(l, m, n) is the direction vector of the line. The angle between the line and the plane (see Fig. 4.50) is defined as the complementary angle θ of the angle ϕ between the direction vector R of the line and the normal N to the plane. For this angle, one has the formula sin θ =cosϕ = Al + Bm + Cn √ A 2 + B 2 + C 2 √ l 2 + m 2 + n 2 = N ⋅ R |N||R| .(4.6.3.18) 140 ANALYTIC GEOMETRY N R φ θ M 0 Figure 4.50. Angle between straight line and plane. 4.6.3-12. Conditions for line and plane to be parallel. A plane given by the general equation (4.6.3.16) and a line given by the canonical equa- tion (4.6.3.17) are parallel if Al + Bm + Cn = 0, Ax 1 + By 1 + Cz 1 + D ≠ 0, or N ⋅ R = 0, N ⋅ r 1 + D ≠ 0; (4.6.3.19) i.e., a line is parallel to a plane if the direction vector of the line is perpendicular to the normal to the plane. Conditions (4.6.3.19) include the condition under which the line is not contained in the plane. 4.6.3-13. Condition for line to be entirely contained in plane. A straight line given by the canonical equation (4.6.3.17) is entirely contained in a plane given by the general equation (4.6.3.16) if Al + Bm + Cn = 0, Ax 1 + By 1 + Cz 1 + D = 0, or N ⋅ R = 0, N ⋅ r 1 + D = 0. (4.6.3.20) Remark. Sometimes the case in which a line is entirely contained in a plane is treated as a special case of parallel straight lines and is not distinguished as an exception. 4.6.3-14. Condition for line and plane to be perpendicular. A line given by the canonical equation (4.6.3.17) and a plane given by the general equa- tion (4.6.3.16) are perpendicular if the line is collinear to the normal to the plane (is a normal itself), i.e., if A l = B m = C n ,orN = λR,orN × R = 0.(4.6.3.21) 4.6.3-15. Intersection of line and plane. Consider a plane given by the general equation (4.6.3.16) and a straight line given by the parametric equation x = x 1 + lt, y = y 1 + mt, z = z 1 + nt,orr = r 1 + tR. 4.6. LINE AND PLANE IN SPACE 141 The coordinates of the point M 0 (x 0 , y 0 , z 0 ) of intersection of the line with the plane (see Fig. 4.50), if the point exists at all, are determined by the formulas x 0 = x 1 + lt 0 , y 0 = y 1 + mt 0 , z 0 = z 1 + nt 0 ,orr = r 1 + t 0 R,(4.6.3.22) where the parameter t 0 is determined from the relation t 0 =– Ax 1 + By 1 + Cz 1 + D Al + Bm + Cn =– N ⋅ r 1 + D N ⋅ R .(4.6.3.23) Remark. To obtain formulas (4.6.3.22) and (4.6.3.23), one should rewrite the equation of the straight line in parametric form and replace x, y,andz in equation (4.6.3.16) of the plane by their expressions via t.From the resulting expression, one finds the parameter t 0 and then the coordinates x 0 , y 0 ,andz 0 themselves. Example 6. Let us find the point of intersection of the line x/2 =(y – 1)/1 =(z + 1)/2 with the plane x + 2y + 3z – 29 = 0. We use formula (4.6.3.23) to fi nd the value of the parameter t 0 : t 0 =– 1 ⋅ 0 + 2 ⋅ 1 + 3 ⋅ (–1)–29 1 ⋅ 2 + 2 ⋅ 1 + 3 ⋅ 2 =– –30 10 = 3. Then, according to (4.6.3.22), we finally obtain the coordinates of the point of intersection in the form x 0 = 0 – 2 ⋅ 3 = 6, y 0 = 1 – 1 ⋅ 3 = 4, z 0 =–1 – 2 ⋅ 3 = 5. 4.6.3-16. Distance from point to plane. The deviation of a point from a plane is defined as the number δ equal to the length of the perpendicular drawn from this point to the plane and taken with sign + if the point and the origin lie on opposite sides of the plane and with sign – if they lie on the same side of the plane. Obviously, the deviation is zero for the points lying on the plane. To obtain the deviation of a point M 1 (x 1 , y 1 , z 1 ) from a given plane, one should re- place the current Cartesian coordinates (x, y, z) on the left-hand side in the normal equa- tion (4.6.1.7) of this plane by the coordinates of the point M 1 : δ = x 1 cos α + y 1 cos β + z 1 cos γ – p = r 1 ⋅ N 0 – p,(4.6.3.24) where N 0 =(cosα,cosβ,cosγ) is a unit vector and r 1 is the position vector of the point M 1 (x 1 , y 1 , z 1 ). If the plane is given by the parametric equation (4.6.1.5), then the deviation of the point M 1 from the plane is equal to δ = [(r 1 – r 0 )R 1 R 2 ] |R 1 × R 2 | .(4.6.3.25) The distance from a point to a plane is defined as the nonnegative number d equal to the absolute value of the deviation; i.e., d = |δ| = |x 0 cos α + y 0 cos β + z 0 cos γ – p |.(4.6.3.26) Let us write out some more representations of the distance for the cases in which the plane is given by the general equation (4.6.1.3) and the parametric equation (4.6.1.5): d = |Ax 0 + By 0 + Cz 0 + D| √ A 2 + B 2 + C 2 = [(r 1 – r 0 )R 1 R 2 ] |R 1 × R 2 | .(4.6.3.27) Example 7. Let us find the distance from the point M(5, 1,–1) to the plane x – 2y – 2z + 1 = 0. Using formula (4.6.3.27), we obtain the desired distance d = |1 ⋅ 5 +(–2) ⋅ 1 +(–2) ⋅ (–1)+1| 1 2 +(–2) 2 +(–2) 2 = 6 3 = 2. 142 ANALYTIC GEOMETRY 4.6.3-17. Distance between two parallel planes. We consider two parallel planes given by the general equations Ax +By +Cz + D 1 = 0 and Ax + By + Cz + D 2 = 0. The distance between them is d = |D 1 – D 2 | √ A 2 + B 2 + C 2 .(4.6.3.28) 4.6.3-18. Distance from point to line. The distance from a point M 0 (x 0 , y 0 , z 0 ) to a line given by the canonical equation (4.6.2.3) is determined by the formula d = R × (r 0 – r 1 )| |R = mn y 1 – y 0 z 1 – z 0 2 + nl z 1 – z 0 x 1 – x 0 2 + lm x 1 – x 0 y 1 – y 0 2 √ l 2 + m 2 + n 2 .(4.6.3.29) Note that the last formulas are significantly simplified if R is the unit vector (l 2 +m 2 +n 2 = 1). Remark. The numerator of the fraction (4.6.3.29) is the area of the triangle spanned by the vectors r 0 – r 1 and R, while the denominator of this fraction is the length of the base of the triangle. Hence the fraction itself is the altitude d of this triangle. Example 8. Let us find the distance from the point M 0 (3, 0, 4) to the line x/1 =(y – 1)/2 = z/2. We use formula (4.6.3.29) to obtain the desired distance d = 22 1 – 00– 4 2 + 21 0 – 40– 3 2 + 12 0 – 31– 0 2 √ 1 2 + 2 2 + 2 2 = √ 153 3 . 4.6.3-19. Distance between lines. Consider two nonparallel lines (see Fig. 4.51) given in the canonical form x – x 1 l 1 = y – y 1 m 1 = z – z 1 n 1 ,or(r – r 1 ) × R 1 = 0 and x – x 2 l 2 = y – y 2 m 2 = z – z 2 n 2 ,or(r – r 2 ) × R 2 = 0. R R 1 2 M M 1 2 Figure 4.51. Distance between lines. . R .(4.6.3.23) Remark. To obtain formulas (4.6.3.22) and (4.6.3.23), one should rewrite the equation of the straight line in parametric form and replace x, y,andz in equation (4.6.3.16) of the plane by their. the plane, r and r 1 are the respective position vectors of the points (x, y, z )and( x 1 , y 1 , z 1 ), and R =(l, m, n) is the direction vector of the line. The angle between the line and the plane. length of the perpendicular drawn from this point to the plane and taken with sign + if the point and the origin lie on opposite sides of the plane and with sign – if they lie on the same side of