Question

(a). Find the equation of the plane through Po = (1,2,1) with normal vector i = (3,1,2) (b). Find the equation of a plane through Po = (2,3,1) and parallel to the plane P:3x + 2y -- z = 4

Answer 1

02 a) An equation of (Mo, y, z) with the plane containing the point normal Vector N = CA, B, C is - A(n-No) +B(y-7) + C(Z-zo)=0 So, for boint Po=(2, 2, 1) and at = 43,1,2> :- 3(0-1) + 1 (y-2) + 2 (Z-1) = 0 37-3+ y2 + 2 Z-2=0 30ty + 2z=7 b The given plane is in P: 3x+2y-Z=4 So, noemal to this plane is :- ZA,6,>= <3,2,-1) Thus, equation of plane passing through Po= (2,3,1) and parallel to LA, B, C) = 23, 2, 1) is 3(1-2) + 2(4-3) -1 (2-) = 0 > 31 -6 + 2y -6 -Z +1=0 3n+ 2y - Z = 11 66

Oy are! 1 The lines L: N-1 2 t 4-2 3. Z-3 4 Z = 3+4+ >> >= 1+ 2+ , y = 2+3+, 1 So, L, y, z)= (1+2+, 2+ 3+ 3+4+ 1 <1,2,3) + + < 2,3,4> and 2+1 Lg : Z+5 s y 3 -1 6 2 > x= -1+65, y = 3-8, Z = -5+28 (ny,z) 2-1,3,-5)+ & 26,-1,2) <-1,3,-5) + s <6,-1,2> v = {2,3,4) A To = <6,-1,2> к X 3 2 6 -1 2 î (6 + 4) - 3 (4-24) + R (-2-18) 10 i + 20 7 - 20 - <10,20,- 207

(v) AS X to So they are not parallel because two parallel lines have barallel direction Vectors and this Veoss broduct should be zero. (1) Let's say they intersect for t- to and so so. then for H, =S, Y = 9 and 3 = 7, : - 1 + 2t, = -1+686 , 2+3+, - 3-6 , 3+4+, = -5+2%. > 211+to) 1 – 3to = S. and 4(t+2) so a - so 6 2 so (1+to) 1-340 = 2(++2) 3 fors 1+4 = 1-34=) 1+ to = 3-94. = = 0.2 1-34, = 2 (4+2) -) 1-340=24+4 => to = -0.6 fore for and 1t to 3 = 2(t+2) = 1+ to = 6 to +12 3) to = - 2.2 As all values of to are different, So lines are Skew.