# Section 2.4: Equations of Lines and Planes

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1 Section.4: Equations of Lines and Planes An equation of three variable F (x, y, z) 0 is called an equation of a surface S if For instance, (x 1, y 1, z 1 ) S if and only if F (x 1, y 1, z 1 ) 0. x + y + z 1 is the equation of the unit sphere centered at the origin. The graph of a system of two equations F (x, y, z) 0, G (x, y, z) 0 represents the intersection of two surfaces represented by F (x, y, z) 0 and by G (x, y, z) 0, respectively, and is usually a curve. A) Lines in R 3. A line l is determined by two elements: one point P 0 on the line l and a direction v of l,i.e., any vector that is parallel to l. The goal here is to describe the line using algebra so that one is able to digitize it. Suppose that the coordinate of the point P 0 on the line and a direction v are given as: P 0 (x 0, y 0, z 0 ) is a given point on l v ha, b, ci is parallel to l. Consider any point P (x, y, z) in the space. Let! P 0 P be the vector connecting P 0 and P. If P is located exactly on the line, then! P 0 P is parallel to the line l,and thus it is parallel to v. On the other hand, if P is o the line, then, since P 0 is on the line,! P 0 P cannot possibly be parallel to the line. Therefore,! P 0 P cannot possibly be parallel to v. We just concluded that P is on l if and only if! P 0 P is parallel to v. Now! P 0 P hx, y, zi hx 0, y 0, z 0 i hx x 0, y y 0, z z 0 i v ha, b, ci. 1

2 z P(x,y,z) Po V l O y x So which is equivalent to! P 0 P // v ()! P 0 P t v (for a constant t) x x 0 a y y 0 b z z 0. c We called these three equation symmetric form of the system of equations for line l. If we set which is equivalent to x x 0 a y y 0 b z z 0 c t, x x 0 a y y 0 b z z 0 c t t t,

3 Or x x 0 + at y y 0 + bt z z 0 + ct, We call it the parametric form of the system of equations for line l. This system can be written in the form of vector equation: r! r 0 + t v, r hx, y, zi,! r 0 hx 0, y 0, z 0 i. Example 4.1. (a) Find the equation of the line passing through (5, 1, 3), having direction v h1, 0, i. Express answer in (i) symmetric form, (ii) vector form, and (iii) parametric form. (b) Find two other points on the line. Solution: (a) (i) x 5 y + 1 z (ii) r h5, 1, 3i + t h1, 0, i (iii) x 5 + t y 1 z 3 t (b) Take t 1, (x, y, z) (6, 1, 1). Take t 1, (x, y, z) (4, 1, 5). Example 4.. (a) Find the equation, in symmetric form, of the line l passing through A (, 4, 3) and B (3, 1, 1). (b) Determine where the line l intersects xy plane. Solution. (a) The line is parallel to vector! AB. So we choose this vector as the direction of l, v! AB h3, 1, 1i h, 4, 3i h1, 5, 4i, and A (, 4, 3) as the point on l. Thus, the equation is x 1 y 4 5 z

4 (b) If this line crosses xy plane somewhere at (x, y, z), then z 0. So this point (x, y, 0) satis es the line equation, i.e., x 1 y We solve this system to obtain x µ 4 3 y z 0. Example 4.3. Given two lines l 1 : x 1 + t, y + 3t, z 4 t l : x t, y 3 + t, z 3 + 4t. Determine whether they intersect each other, or they are parallel, or neither (skew lines). Solution: First of all, in each line equation, "t" is a parameter (or free variable) that can be chosen arbitrarily. Therefore, the parameter "t" in the equations for line l 1 is DIFFERENT from the parameter "t" in the equations for line l. To clarify this issue, we rewrite as l 1 : x 1 + t, y + 3t, z 4 t l : x s, y 3 + s, z 3 + 4s, and intersection of these two lines consists of solutions of the following system of six equations, x 1 + t, y + 3t, z 4 t x s, y 3 + s, z 3 + 4s, for ve variables: x, y, z, t, s. Two lines intersect each other if and only If this system has a solution. If, for instance, (x 0, y 0, z 0, t 0, s 0 ) is a solution, then the rst three components, (x 0, y 0, z 0 ) is a point of intersection. 4

5 We now proceed to solution the system by eliminating x,y,z: 1 + t s (1) + 3t 3 + s () 4 t 3 + 4s. (3) There are three equations with two unknowns. We start with two equations, for instance, the rst and the second equation: 1 + t s + 3t 3 + s. This can be easily solved as, by subtracting times the second equation from the rst equation, i.e., 7 5t 6 ) t 11 5, s 1 + t 8 5. We need to verify that the solution, t 11 5, s 8, from the rst two 5 equations (1) & (), satis es the third equation (3). So Apparently, LHS of (3) 4 t µ 8 RHS of (3) 3 + 4s t 11 5, s 8 5 is not a solution of the entire system (1)-(3). We thus conclude that these two line cannot possibly intersect. Answer: skew lines 5

6 N z Po P(x,y,z) Plane π l O y x B) Equations of Plane. De nition. Any vector that is perpendicular to a plane is called a normal vector to the plane. There are two normal directions (opposite to each other) to a given plane. For any given vector n, there are in nite many parallel planes that are all having n as their normal vector. If we also know a point on the plane, then, this plane is uniquely determined. In other words, a plane π can be determined by a point P 0 (x 0, y 0, z 0 ) on the plane and a vector as its normal vector n ha, B, Ci. For any point P (x, y, z), if this point P is on the plane π, then the line segment P 0 P entirely lies on the plane. Consequently, vector! P 0 P r! r 0 hx x 0, y y 0, z z 0 i, where r hx, y, zi,! r 0 hx 0, y 0, z 0 i, is perpendicular to the normal vector n. On the other hand, if P is o the 6

7 plane π, then r! P 0 P is not perpendicular to n. We conclude that or P π (P belongs to π) ()! P 0 P n 0, ( r! r 0 ) n 0. We call it vector equation of the plane π.in terms of components, hx x 0, y y 0, z z 0 i ha, B, Ci 0. We obtain scalar form of equation of plane π : (Vector Equation) A (x x 0 ) + B (y y 0 ) + C (z z 0 ) 0, (Scalar Equation) or Ax + By + Cz + D 0. (Linear Equation) In 3D spaces, any linear equation as above represents a plane with a normal vector n ha, B, Ci. (In D, any linear equation is a straight line.) Example 4.4. Find the equation of the plane passing through P 0 (, 4, 1) having a normal vector n h, 3, 4i. Solution: A, B 3, C 4.The equation is or (x ) + 3 (y 4) + 4 (z + 1) 0, x + 3y + 4z 1 0. Example 4.5. Find the equation of the plane passing through P (1, 3, ), Q (3, 1, 6), R (5,, 0). Solution: Let u! P R h5,, 0i h1, 3, i h4, 1, i v! P Q h3, 1, 6i h1, 3, i h, 4, 4i 7

8 Q V P U R The vector u v i j k i i 0 j 14 k ³ 6 i + 10 j + 7 k j k is perpendicular to both u and v.thus, ³ n 6 i + 10 j + 7 k is perpendicular to π. Now, we take this normal vector and one point P (1, 3, ) (you may choose Q or R,instead), and the equation is or 6 (x 1) + 10 (y 3) + 7 (z ) 0 6x + 10y + 7z Note that if we chose Q (3, 1, 6) as the known point, then the equation would be 6 (x 3) + 10 (y + 1) + 7 (z 6) 0 or 6x + 10y + 7z

9 Example 4.6. Find the intersection, if any, of the line x + 3t, y 4t, z 5 + t and the plane 4x + 5y z 18. Solution: We need to solve the system of all four equations x + 3t y 4t z 5 + t 4x + 5y z 18 for x, y, z, t. To this end, we substitute the rst three equations into the last one. This leads to which can be simpli ed to 4 ( + 3t) + 5 ( 4t) (5 + t) 18, 10t 18. So t, and x + 3t 4 y 4t 8 z 5 + t 3. Answer: the intersection is ( 4, 8, 3). Example 4.7. Given two planes π 1 : x + y + z 1 π : x y + 3z 1. Find (a) the line of intersection, and (b) the angle between two planes. 9

10 Plane π1 n1 n Line l Plane π as Solution: Plane π 1 and plane π have normal vectors n 1 and n,respectively, n 1 h1, 1, 1i n h1,, 3i. The line is on both planes and thus is perpendicular to both normal vectors. The direction of the line is i j k v n 1 n i j k 5 i j 3 k. To nd the equation of the line, we also need a point on the line, i.e., on both planes. So we look for one solution to the system x + y + z 1 x y + 3z 1. 10

11 This system has in nite many solutions (why). solution, we set z 0 to reduce the system to Since we only need one x + y 1 x y 1. Subtracting the second equation from the rst, we nd 3y 0 ) y 0 x 1. So P (1, 0, 0) l. The equation of the line, in parametric form, is x 1 + 5t y t z 3t. Solution #: Another way to nd the equation of this line is to solve the system x + y + z 1 x y + 3z 1 directly in terms of z. In other words, we choose z as parameter. To this end, we subtract the second equation from the rst one to get 3y z 0 ) y 3 z. Substituting this into plane π 1 : µ x + 3 z + z 1 ) x z, we obtain the equation of the line x z y 3 z z z. 11

12 Note that z is basically a free variable. If we set z 3t, this becomes x 1 + 5t y t z 3t which is identical to what we got earlier. (b) The angle between two planes is the same as the angle between their normal vectors. So cos θ n 1 n h1, 1, 1i h1,, 3i j n 1 j j n j jh1, 1, 1ij jh1,, 3ij p p p p µ θ arccos p p (rad) π (deg) 70. Practical advice in nding equations of lines or planes: Regardless what information a problem provides, we always look for a point and a direction. That would be su cient to solve the problem. More precisely (1) If we want equations of a line, then we look for one point on the line and a vector PARALLEL to the line. () If we want an equation of a plane, then we look for one point on the plane and a vector PERPENDICULAR to the plane. Cross product may be used to create a vector perpendicular given two vectors. 1

13 P 1 dist Line l V P 0 C) Distance between points, lines and planes: Let S and T be two sets of points. Then dist (S, T ) min fdist (P, Q) j P S, Q T g. In other words, the distance between two sets is de ned as the smallest distance between two points from di erent sets. 1) Distance between a point P 1 (x 1, y 1, z 1 ) and the line l: x x 0 + at y y 0 + bt z z 0 + ct. Pick a point on the line, say P 0 (x 0, y 0, z 0 ), and a direction (unit vector) of the line 1 v p ha, b, ci. a + b + c Then the cross product ³! P 0 P 1 v 13

14 by de nition, has the length ³! dist (P 1, l) P 0 P 1 v! P 0 P 1 sin θ. Example 4.8. Find the distance from P 1 (1,, 1) to the line l : x 1 + t y 3t z 4t. Solution. P 0 (1,, 0) is a point on l, and A unit direction of the line is! P 0 P 1 h1,, 1i h1,, 0i h0, 4, 1i. v 1 p 9 h, 3, 4i So ³! P 0 P 1 v p 1 i j k p µ 3 4 i p 1 ³ 13 i + j + 8 k, 9 j k and ³! dist (P 1, l) P 0 P 1 v p r p () Distance from a point P 1 (x 1, y 1, z 1 ) to a plane π : Ax+By+Cz+D Pick any point P 0 (x 0, y 0, z 0 ) on the plane, i.e., (x 0, y 0, z 0 ) solves Ax 0 + By 0 + Cz 0 + D 0, 14

15 P 1 N N P 0 Plane π then the distance is the absolute value of the dot product of! P 0 P 1 and normal direction 1 n p ha, B, Ci A + B + C ³! dist (P 0, π) P 0 P 1 n hx 1 x 0, y 1 y 0, z 1 z 0 i ha, B, Ci p A + B + C A (x 1 x 0 ) + B (y 1 y 0 ) + C (z 1 z 0 ) p A + B + C Ax 1 + By 1 + Cz 1 (Ax 0 + By 0 + Cz 0 ) p A + B + C Ax 1 + By 1 + Cz 1 + D p A + B + C. Example 4.9. Find the distance from P (1,, 3) to the planeπ : x y + 3z 4. Solution: We rewrite plane π in the standard form as x y + 3z

16 So dist (P, π) Ax 1 + By 1 + Cz 1 + D p A + B + C x 1 y 1 + 3z 1 4 p p (3) Distance between two lines. We rst ne the plane π containing line l 1 and being parallel to l. If v 1 and v are directions of l 1 and l,respectively. Then, n v 1 v is a normal vector to the plane π.one may pick any point P 0 on l 1 and this normal vector to obtain the equation of π. Pick any point P 1 on line l,and dist (l 1, l ) dist (P 1, π) (A) Another approach is to use projection. Pick one point from each line, say P 0 l 1, P 1 l. Then, ³ ³!! P 0 P 1 n dist (l 1, l ) Proj n P 0 P 1 (B) j nj Example Consider two skewed lines in Example l 1 : x 1 + t, y + 3t, z 4 t l : x t, y 3 + t, z 3 + 4t. Find their distance. Solution. We rst use method (A). The normal to the plane π containing l 1 and parallel to l is n v 1 v n v 1 v 13 i 6 j 5 k. i j k i j k 16

17 P 1 Line l V dist N Line l 1 V 1 P 0 17

18 Obviously, P 0 (1,, 4) l 1. So equation of π is 13 (x 1) 6 (y + ) 5 (z 4) 0. Since P 1 (0, 3, 3) l, formula (A) and the distance formula lead to dist (l 1, l ) dist (P 1, π) Ax 1 + By 1 + Cz 1 + D p A + B + C 13 (0 1) 6 (3 + ) 5 ( 3 4) p Let now solve the same problem using formula (B). Note that! P 0 P 1 h0, 3, 3i h1,, 4i h 1, 5, 7i n 13 i 6 j 5 k. So dist (l 1, l ) ³! P 0 P 1 n j nj ( 6) + ( 7) ( 5) p (4) Distance between a line l and a plane π. Pick any point P 0 on the line and the distance dist (l, π) between line l and plane π is the distance from P 0 to plane π : Homework: dist (l, π) dist (P 0, π), P 0 l. 1. Determine whether each statement is true or false. (a) Two lines parallel to a third line are parallel. (b) Two lines perpendicular to a third line are parallel. (c) Two planes parallel to a third plane are parallel. (d) Two planes perpendicular to a third plane are parallel. 18

19 (e) Two lines parallel to a plane are parallel. (f) Two lines perpendicular to a plane are parallel. (g) Two planes parallel to a line are parallel. (h) Two planes perpendicular to a line are parallel. (i) Two planes are either intersect or are parallel. (j) Two lines are either intersect or are parallel. (k) A plane and a line are either intersect or are parallel.. Find parametric equation and symmetric equation of line. (a) The line through the point (1, 0, 3) and parallel to h, 4, 5i. (b) The line through the point the origin and parallel to the line x t, y 1 t, z 4 + 3t. (c) The line through (1, 1, 6) and perpendicular to the plane x + 3y + z 5. (d) The line through (, 1, 1) and perpendicular to i+ j + k and i+ k. (e) The line of intersection of the planes x + y +z 1 and x + y Find equation of plane. (a) The plane through (1, 0, 1), (0, 1, 1), and (1, 1, 0). (b) The plane through (5, 1, 0) and parallel to two lines x +t, y t, z 1 and x t, y t, z 3t. (c) The plane through ( 1,, 1) and contains the line of intersection of two planes x + y z and x y + 3z 1. (d) The plane through ( 1,, 1) and perpendicular to the line of intersection of two planes x + y z and x y + 3z 1. (e) The plane that passes through the line of intersection of the planes x z 1 and y + z 3, and is perpendicular to the plane x + y z Determine whether two lines are parallel, skew, or intersecting. If they intersect, nd the point of intersection. 19

20 (a) L 1 : x 1+t, y 3t, z t; L : x 1+ s, y 4+s, z 1 + 3s. (b) L 1 : x y 3 5. (Optional) Find distance. z 1 ; L : x y z + 3 (a) Distance from (1, 0, 1) to the line x 5 t, y 3t, z 1 + t. (b) Distance from (3,, 7) to the plane 4x 6y + z 5. (c) Distance between two planes 3x+6y 9z 4 and x+y 3z. 0

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