Question

9. (a) Let P1(21, 91, zı), P2 (22, 42, z2),P3 (13, 93, 23) be three non-collinear points in R, that is, three points which do not all lie on a straight line. Then the equations of the plane through these three points is: 2 y 21 1 = 0 22 Y2 22 1 2 1 2141 I3 Y3 23 1 Page 1 (b) Find the equation of the plane through P1(1,2,2), P2(1, 2, -1), P3(0,1,2)

Answer 1

a.

Z 0 The general form of the equation of a planeis ax+by+cz +d=0(i), where (x,y,z) be any point on therequired plane. The required plane passes through the non-collinear points ( 11,91, 2), (22,73,22), (13,Y3, 23). Thus these points satisfy (). :: ax: +by; +cz, +d = 0) (ii) ax, +by, +cz, +d = 0) (1) ax: +byz +cz3 +d = 0) (iv) In matrix form, AX = 0, we get V * 2 db b where X * 22 d [*3 Y3 Zz d The above system has a non zero solution because of non zero element 1 in X. This happens only if det(A)= 0. al *1 i 2 d = 0 11 » 21 = 0 12 y 22 12 V222 [X3 Y3 Zz |73 ; 23 II c 0 N > Z 1 1 d X V Z 1 0. 11 Vi 21 12 V 22 13 Y3 23

X Z 1 ܢܝܕ b). The equation of a plane pas sin g through the points (1,2,2), (1,2,-1),(0,1,2) is (1). Elementary row operations of the form R → R + KR;, does not change the value of determinant. 1 2 2 = 0 (ii). R + Ry, interching ;* anf j* rows changes the sign of the value 1 2 -1 1 of determinant 0 1 2 (iii). R → KR;, changes |A to k|A|. Applying R+R-RA, RR-R, R + R3 - Ry, we get x y-1 2 -2 01 0 0 = 0 Applying RR-R, we get -3 0 0 2 x y-1 2 -2 0 0 3 = 0, On evaluating the determinant along the fourth column, we get 1 2 x y-1 -21 3 = 0, On evaluating the determinant along thesecon row, we get -3 1 1 1 1 11 0 -3 1 1 Z- 10 0 1 X -3 y-11 = 0 X 11- >x-y+1=0