Lines and Planes in Space

12.5 Lines and Planes in Space 3.7 In the plane, a line is determined by a point and slope. In the space, a line is determined by a point and a vector giving the direction of the line. Let L be a line in space: Z passing through 1. (xo, yo, to) and L(x,y,2) panallel to a vector V =vi+vzj+vsR L L is the set of all points P(x,y, 2) s.t 0 y // V . Thus, X P.P =t V for some scalar parameter tEIR. (x-xo) i + (y-yo) j + (2-Zo) K = +(vi+vjj+vk) X , + yo + ZK = (Xo + t V1) i + (yoty) + (2o+tv3) K Hence, the Parametric Equations for a line passing through Lo ( xo, yo, to) and // to V ==vi+vzj+vi K is y = Yo+ tv2, Z = Zo + tV3 Exp: Find parametric equations for the line 1 passing through point (3, -2,1) and // to the vector 2i-3 +3K X = 3+2t, y=-2-t / z=1+3t , - set < 8 2 passing through the points p(1,2,-1) and Q(-1,0,1) PQ = -21-2 - +2K is the direction. Consider £(1,2,-1) x=1-2t , y=2-2t / Z = - 1+2t / - wet

Exp Find parametrization for the line segments joining 38 the points $(1,1,0) and Q(1,1,1), and sketch them. Z PQ = K o Consider 1(1,1,0) Q(1,1,1) \ t y X P(1,1,0) Dis tance from Point to Line in space: The distance from the points to the line L ( L passes through P and // to 3) is I PS / sing / , where I is O is the angle between PS and J Note that IPSX v /= . Hence, the distance is given by: PS*V d= |ps/sing IVI Exp Find the distance from the origin to the line X = 5+3+, y=5+yt, z=3-5t 5(0,0,0) and 1(5,5,-3) and J=3i+4j-51 PSXV = = . 34-5 d= IPSXVI = V 169 +256+25 = - V450 = V9=3 lvl 550 9+16+25

An Equation for a Plane in Space 39 A plane M in space is determined by plane M knowing a point 10 (xo, yo, zo) on the plane n knowing a vector 15 1 to the plane. xf Po Assume M passes through 10 and MIn = Ai+Bj+ck Then M is the set of all points P(x,y, 7) s.t P.P 1 n c Thus n . P.P =0 @=90 cos === ( A i +Bj'+CR). + . [1x- Xo) i + (y-yo) + (2-Zo) R]=0 = 6 A(x- Xo) + B(y-yo) + C (z-zo) =0 That is Ax + By + CZ = Axo + Byo + ( to Exp Find the equation of the plane ) passing through 1.(0,2,-1) and 1 to in = 31-2j-k - 3 2 passing through (1,-1,3) and // to the plane 3x+y+z=7 = fo(1,-1,3) and is = 31+j + k 3 x+y+z = (3)(1) +(1)(-1) + (1)(3) = 5 3 passing through I (2,4,5) Q(1,5,7) and 5(-1,6,8) To Find 1s : PQ = -i+j+2K } 15 = PQ X PS = 12 PS = -31+23+3K 3 2 3 Consider P(2,4,5) and n -3j. K =- -3j+i - X-3y+z=(-1)(2)+ (-3)(4)+(1)(5) = 9

Planes Intersection in Space: 40 Note that. lines in space are parallel iff they have the same directions. two planes are parallel iff their normals are parallel (or n, = Kn2 for Some scalark). two planes that are not parallel intersect in a line. Exp OIFind a vector v 11 to the line of M2, intersection of the planes 3x-6y-27=15 and 2x+y-27=5 Nz n, n==3i-6j-2k n2=2i+j-2K K / M1 v. i K = XN2= 3-6-2 214 i tzj + 15k 0 n 21-2 2 Find the parametric equations for the line intersection We need a vector 11 to the line intersection is 1141+2j+15R We need a point on the line intersection (on the two planes): subs titule "for example" Z=0 Po (3,-1,0) The parametric equations are: X = 3 +14t, y= - 1 +2t, Z = 15t, - ast< 00 not unique point Exp Find the Twhere the line X= 1+2t y=1+5t, Z = 3t meets (intersects) the plane x+y+z = 2. 1+y+z=2 (1+zt)+(1+st)+(3t) =2 (c) lot=0 t=0 Hence, X=1,Y=1,2=0 The point is (1,1,0)

Distance from a Point to a Plane 41 n The distance from the point S to S. L the plane M is d O d = / PS / COSE I Note that PS is = /ps/1n/ woo M The distance is then given by d = 1 ps Cos Q = abs ( PS n . In't = abs ( PS. In' n Exp Find the distance from the point 5(0,0,0) to the plane 3x + 2y + 6Z = 6. The point I is P(2,0,0) which is on the plane. PS = -21 and in = 3i+2 2j + 6K The distance is (3i +2 + 6K d= abs ((-zi). V 9+4 + 36 d= abs V 49 ) = abs ( -6 7 ) = 6 7 Exp Find the angle between the planes 5 X +y- Z = and X- zy + 3Z =-1 The angle between two intersecting planes is the acute angle between their normal vectors (see *) ni=5i+j-R Inil= = 27 n2 = - 2j + 3K => = V14 = cos 1 12/1/201 n, n2 isi) = cos' ( 5-2-3) = cos' (o) = HI V27V14