Question

Consider the following geometrical objects in R3: li: x=2 -21, y = -1, z = 2, ER 12: (x, y, z) = (0,5, -1) + (2, -1, 1), 1ER II : 3x + 4y - 2z = 1 II2 : 3x + 3y + z = -1 (a) Find the intersection point of , and 2. (b) Find the line 63 containing A(2,3,4) and is parallel to both II, and II. (c) () Determine whether II, and 2 are intersecting, parallel or 2 is a line on IT. (11) Find the shortest distance between II, and C 2. (d) Find the general equation of a plane II, that is parallel to II, and a distance 319 units from II

Answer 1

Ο Mon li: 2 = 2-27 9 y=-t, = 2 HER by: <x,y,z) (0,5,-1) ++,(2,1,1) ter TT, : 320 + 47-22 12 3x + 3y + 2 ܝܙ (а @ Let the pb. of intersection of l, and l is (6,%92) sob. (26.9%, 2.) = (2-27, ,-, , 2) = (0,5,-1) + Łą (2,7,1) (2-24,94,92) = (2+,, 5-tz , t -1) or, en 2-2t, 252 -ti = 22-1 5-t2 (2) & 2 3 3 = 2+1 3 Tue From t₂ So, using it from 0 g 2-27, 2-24 ti 2tz 2x 3 = 6 → -2 Now 2 -t, = 5-t, is satisfied for t= -2, t₂=3. . the pt of intersection is (2-2t, s-t,,2)=-2 it (6,2,2) (Ans)

o Let the equation of the reguired line Y-3 whouse is x-2 l 7-4 N and it is passing so 5 are de's are l, m,n l, m,n and it is through A ( 2,3,4). Since the st. line Ch is parallel to 71 3l +4 m -2n = = 0 For's of the normal to the plome 7, ou be 3,4,-2 Similarly since 4 is I to y and drs of the normal to the plane z 31 +3m +n solving 64 © by cross multiplication weger l so drs of l are -9 10, -9,-3 or -10,9,3 So, the require st. line is le Y-3 x-2 -9 3,3,1 ز So 0 6 m n Wola 10 Z-4 -3 ice la ۔ February 2007 lz: x= h = 2+ 10 t 3-97 I > 4-34 Y ܟܢ 17 18 24 25 TER,

16 Wed Take any point on la as (0,5,-1) + 6(2,-1,1) (2+, 5-7,7-1) Let us put a= 2t , y = 5-t, 2=t-1 in the ect of the plane the and hence We get 3 (24) + 3(5-1) + (t-1) = -1 on, (6-3+1) 46 = -15 ~ t= - Hence the line la intersects Iz at (21-44),5-4-4), -1) ie. (-2,,-) 1-15 + 1 or, 15 4 Thu 4 4 © (i) Let us consider the st. line la intersects II, at (0,5,-1) + t (2,-1, 1) = (2t, 5-tot-1). If we get some real value of t, then ous assumption will be true', Since (26,5-6, 6-1) lies (by our assumpt.) 2 Tebruary 200 on ti so } OT, 3(26)+ 4(5-1)-2(t-1) = 1 © 16-4-2)t +20 +2 = 1 Ot + 22 = 1 22=1 OT !!!)

So, our assumption was not intersect wrong, so le does r, ri Now dris of la are 2,-1, 1 and dris of the normal to the plane II, are 3, 4,-2 2x3 + (-1)*4 + 1*(-2) 6 -4 -2 = 0 → la is parallel to the plane Tz. Now = 22 71 10 sat Now since I is passing through (0,5 ,-1) and this point does not lying on the plane Thi 3x 0 + 4*5-2x(-1) Therefore le does nota line on a l is parallel to the plane (Ang) ©Gi) The shortest distance between TI, mely | 3x0 +485-2*(-1) - 11 units (since (0,5,-1) √3+ 4+ (-292 on la .. b is 77, is lies on 30 2 April 2007 ار 10 11 13 14 1 21 129 units 21.29 29 units.

d e is some и1 9 Equation of of of a plane parallel to Thy is 3*+.3.4.-4.Z constant, Here TT, om.d. They are not parallel , so Ty does not exist, since any parallel plane of..IT must meet I in a line 2 We will & find out the intersecting line of 3x+3y + z = 11, ! 3x+47-27 = 1, and O 2

Let z=0 then os f ① o 3x + 3y -C=0 3x + 4y - 1 2 -3+40 У -3+30 ? 129 15 2 = -3+40 3 y=e-1 , Z=0 2

Tue 2 Also let mo & a point on the line of intersection is '(-394, e-1,0) drs of that line is lo me Then 3 l + 3 m+ n=0 l 3 l + 4 me 2n=o. So, the egn. of line of intersection is a Oso a-3 the Y-(1-1) } * -10 slm is Z-o V -10 9 3