Question

QUESTION 1 (15 MARKS) a) Given the line Lj: I = 2 - 2t, y = 5 + 2t, z=t-1 and 1 1 - 2 L2 : =y-3 = 2 4 i. Check whether the lines Lị and L2 parallel, intersect or skewed? (5 marks) ii. Find the shortest distance from the point (1, 2, -1) to the line Li- (3 marks) b) Given two planes 71 : 20 - 4y +z = 5 and T2 : 7x + y - 2z = 14. i. Find the acute angle between the planes. (3 marks) ii. Find the parametric equations of a line that passes through point P(2, -3, 4) and parallel to the line of intersection between planes 11 and 12 (4 marks.)

Answer 1

Solution: (a) Given, L: x = 2-2 t , Z = t-1 =-1+t y=5+24 L2: 2-1 2 y-3 1 I-Z 4 Leti I- Z y-3 1 - 8 (whese s is a parameter) 2 x-1 2 LES 9-3 = 3, 1-2 = 8 x=28+1, y = 8 + 3 Z=48+1 That is, L2: x= 1+2s, y=3ts, Z=1-48. () To check Lis L2 parallel or not: Sinces 2 + } 4 [From O and ©] > -1 # 27-1/4 = L L2 not parallel To check Li, L2 intersect or not: - From O and@, Assume (220790, Zo) be The intersect point of . ① and ② That is, xo= 2-2 t - 28t 1 3 Yo=5+2t = st 3 Zo = t-1 = 1-45 5 From (3) we get, 2-2 t = 2st 1 > 2t=2-28-1 = 1-28 - From @ using 2t - st 3-5 1-28 = 8-2 38=3 8=1

Foom 6 From ② using We get, 2t=1-25 = 1-2(1) 7t=-1 2 The value of s and t we get -2 2-1 = 1-4(1) ³ - 2 2 = -3 which is a contradiction. Hences our assumption is wrong. LI, L2 do not intersect. To check Lu L2 are skewed: Since, L1, L2 two lines are not parallels intersect, Hence, L, L2 Skewed. cü Given, L.: x= 2-269 y = 5+2t 2 Z= -1tt Leto X= α = 12-2t 5+2t L-itt [x is any point on Li] IA (1,2,-1) Z a= T Х Now, X - A = (say) = [1] and Å= (112,-2) AX [] a 5=0 imply. (since, to is perpendicular to a ) -2(1-2t) + 2(3+2+)+1(t) = 0 1-2t 3 + 2t Lott 12-2t 15t2t Eltt Now, - -2+46 + 6 + 4t+t=0 It = - 4 > t = - 4/9

Hence, Ax = 1-21-4/9) 3+21-419) - 419 [using value of t It 3-8 -419 -419 Hence, shortest distance From (1, 2,-1) to Liis. 1AX || - (47)+(12)+(-40) = 289+361 +16 81 666 81 7 2.867 4 unit Hence, shortest distance, 2-8674 unit.