12.5 Equations of Lines and Planes

Transcription

1 Instructor: Longfei Li Math 43 Lecture Notes.5 Equations of Lines and Planes What do we need to determine a line? D: a point on the line: P 0 (x 0, y 0 ) direction (slope): k 3D: a point on the line: P 0 (x 0, y 0, z 0 ) direction (a vector parallel the line): v Vector Equation of L: Proof: r = r 0 + tv P 0 (x 0, y 0, z 0 ) is a given (known) point on L, P (x, y, z) is an arbitrary point on L. r 0 and r are the position vectors of P 0 and P. a = P 0 P, then by the Triangle Law: Since a is parallel to v, v is a scalar multiple of a: r = r 0 + a a = tv, t R So r = r 0 + tv, t R Remark: r traces out the line: r 0 shifts us onto the line, tv moves us along the line(t > 0 to the same direction as v; t < 0 to the opposite direction to v)

2 Scalar Equations (parametric equations) x = x 0 + at; y = y 0 + bt; z = z 0 + ct, t R Proof: If we write the vectors into their components form and let v =< a, b, c >, then < x, y, z > =< x 0, y 0, z 0 > +t < a, b, c > =< x 0 + at, y 0 + bt, z 0 + ct > Thus, we have the scalar equations: x = x 0 + at; y = y 0 + bt; z = z 0 + ct. Example: Find a vector and parametric equations for the line that passes through the points A(,, 3) and B(4, 6, 8). The vector of the line: v = AB =< 3, 4, 5 > If we pick A(,, 3) as the given point on the line, then vector equation: r =<,, 3 > +t < 3, 4, 5 >=< + 3t, + 4t, 3 + 5t > parametric equations: x = + 3t, y = + 4t, z = 3 + 5t If we pick B(4, 6, 8) as the given point on the line, then vector equation: r =< 4, 6, 8 > +t < 3, 4, 5 >=< 4 + 3t, 6 + 4t, 8 + 5t > parametric equations: x = 4 + 3t, y = 6 + 4t, z = 8 + 5t We could also a scalar multiple of the vector v =< 6, 8, 0 >: vector equation: r =<,, 3 > +t < 6, 8, 0 >=< + 6t, + 8t, 3 + 0t > Remark: Equations are not unique! We can change the given point, choose a different parallel vector. Direction Numbers: If vector v =< a, b, c > is used as the direction of L, then a, b and c are called the direction numbers. Symmetric Equations: x x 0 = y y 0 = z z 0 a b c If one of a, b, c equals 0, we write the symmetric equations as: x x 0 a = z z 0, y = y 0, for instance, if b = 0 c Proof: The symmetric equations are obtained by eliminating t from the parametric equations: Hence, x = x 0 + at t = x x 0 a y = y 0 + bt t = y y 0 b z = z 0 + ct t = z z 0 c x x 0 a = y y 0 b = z z 0 c

3 Example: Find parametric and symmetric equations for the line passes through the point (5,, 3) and is parallel to i + 4j k. parametric equations: x = 5 + t, y = + 4t, z = 3 t symmetric equations: x 5 = y 4 = z 3 Line Segment: The line segment from r 0 to r is given by the vector equations: Proof: r(t) = ( t)r 0 + tr, 0 t If we choose v = r r 0, then the vector equation r(t) = r 0 + t(r r 0 ) r(t) = ( t)r 0 + tr, 0 t Remark: r(t) starts at r 0, ends at r and traces point between. Relations of lines in space: Intersect: pass through the same point. Parallel: have the same direction Skew: not parallel, don t intersect Example: Determine whether the lines L and L are parallel, intersecting, or skew: (a) L : x = + t, y = 3 + t, z = 5 + 3t; L : x = 3s, y = 6s, z = 5 9s Directions: v =<,, 3 >, v =< 3, 6, 9 >. We have v = 3v, scalar multiple parallel 3

4 (b) L : x = + t, y = + t, z = + t; L : x = 3s, y = s, z = 0 Directions: v =<,, >, v =< 3,, 0 >. Not scalar multiple Not parallel. Intersect? If so, for some t and s: + t = 3s + t = s + t = 0 Solve the first equations t =, s = 0. However, the third equation is not satisfied! Not intersecting. So L and L are skew lines. (c) L : x = y 3 = z 3, L : x 3 = y + 4 = z 3 7 : Directions: v =<,, 3 >, v =<, 3, 7 >. Not scalar multiple Not parallel. Intersect? The parametric equations are If intersect, for some t and s, we have L : x = + t, y = 3 t, z = 3t L : x = 3 + s, y = 4 + 3s, z = 7s + t = 3 + s 3 t = 4 + 3s 3t = 7s Solve the first two equations t =, s =. Check the third equation, it s satisfied! So the lines intersect when t =, s =, i.e., at the point (4,, 5). Planes What do we need to determine a plane? a point in the plane: P 0 (x 0, y 0 ) direction (a vector perpendicular to the plane): n Remark: The orthogonal vector n is called a normal vector. 4

5 Let P (x, y, z) be an arbitrary point in the plane, P 0 (x 0, y 0, z 0 ) is a given point in the plane, and n be a normal vector of plane, then n is orthogonal to r r 0, i.e., n (r r 0 ) = 0 Vector Equation of the Plane: n (r r 0 ) = 0 or n r = n r 0 Scalar Equation of the Plane: If n =< a, b, c >, then a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0 Linear Equation of the Plane: where d = (ax 0 + by 0 + cz c ) ax + by + cz + d = 0 Recall: the intercepts are the intersections of the graph with the coordinate axes. Example: Find an equation of the plane through the point (, 4, ) with normal vector n =<, 3, 4 >. Find the intersepts and sketch the plane. (x ) + 3(y 4) + 4(z + ) = 0 x + 3y + 4z = The x intercepts is x = 6, the y intercepts is y = 4, and the z intercepts is z = 3. 5

6 Example: Find an equation of the plane that passes through the points P (3,, ), Q(5,, ) and R(, 4, 7). The vectors in the plane are a = P Q =<, 3, > and b = P R =<, 3, 6 > a b is orthogonal to both a and b, i j n = a b = 3 3 thus is orthogonal to the plane and can be taken as normal. k =< 8 3,, 6 3 >=<, 3, 3 > 6 So the equation of the plane is (x 3) 3(y ) + 3(z ) = 0 Parallel of Planes: Two planes are parallel if their normal vectors are parallel. The Angle between the planes: defined as the acute angle between the normal vectors. Example: Find the angle between the planes x + y + z = 0 and x y + 3z =. The normal vectors of the two planes are n =<,, > and n =<,, 3 > 6

7 If θ is the angle between the two planes, then n n +3 = = n n θ = arccos 7 4 cos θ = Example: Find the parametric equations for the line of intersection L of the planes x + y + z = and x y z =. The normal vectors of the planes are n =<,, > and n =<,, > Since the line L lies in both of the planes, it is to L is i n n = orthogonal to both n and n. Thus, the vector parallel j k =< 0,, > Pick a point on the line of the intersect: ( x+y+z = x y z = set z = 0, we have x + y =, x y = x = 3, y =. So the point ( 3,, 0) lies on L. So far we find the direction of L: < 0,, > and a given point ( 3,, 0) on the line, so the parametric equations are 3 x =, y = + t, z = t Distance: The distance D from a point P (x, y, z ) to the plane ax + by + cz + d = 0 is D= ax + by + cz + d a + b + c Proof: Let P0 (x0, y0, z0 ) be a point in the plane. Then the coordinates of P0 satisfies the plane equation: ax0 + by0 + cz0 + d = 0 d = (ax0 + by0 + cz0 ) 7

8 b = P 0 P =< x x 0, y y 0, z z 0 > From the equation of the plane, the normal vector is n =< a, b, c >. Thus, D = comp n b = n b n = a(x x 0 ) + b(y y 0 ) + c(z z 0 ) a + b + c = ax + by + cz (ax 0 + by 0 + cz 0 ) a + b + c = ax + by + cz + d a + b + c 8