Question

Find the general equation and a vector equation of the plane that passes through the points p(1,2,4) Q(1,-1,6) R(1,4,8)

Answer 1

a) find the General & vector aequation that passes Hulpligt the points pl},2,4), (1,-1, 6), RC1,4,8). Gol : we rewrite the points as: P(x0,41, 2,) = P(1, 2, 4) Q(X2, 42,22) = Q(1,-1,6) R (x3, 43, 23) = R(1,4,8) Nome, P Q =<x2-x, y2-4, 2₂- 27 =<1-1,-1-2,6-47 P Q =<0,-3,27 'R =<x3-, 43-y, zz-z, > = <1-1,4-2, 8-47 PP =co,2,47 < Tv To-3 2 0 2 4 = î (-3x 4 - 282) - î (0-0)+(0-0) på XPR = -16 I toog too R from here, we Got a=-16, 6=0 ,c=0 het (1, 2, 4) = p (no, yo, 20) Plugging all values in the equation: - a( x- Ho) + bly-yo) +C (2-20) = 0 7 - 16(x-1) toly-2) to (2-4)=0 - 16 a +16 totoo -16 x + 16 = 0 & general form -16% toy +0.2 +16=D

do vector lequation of a plane passing througe three points is : (_ă”). [15-å) x (2-a)] =0 Nove, the plane passes through the points :- AC1,2,4) B(1,-1, 6) C(1,4,8) a = 1 î+29+ 4 ^ 15 = 1ê-lŷ to Å I a=lî+47 +8 ² so, (5-a)=(11-17 + 6 Å)-(17+ 2q + 4 ) =(1-1) i +(-4-2) +(6-4) Ŕ 5-a ) = oî - 39 +2 Ŕ & (2-a)=(lît49 +8 Å)-(lêt2h + 4 ") =(1-1) î x(4-2) 9+(8-4)  ca_as= oê t2 ŷ +4 ľ Nowe, (5-a)x(è-a) = 10 -3 2 1 10 24 - î (-12-4)-ġ (0-0) + (0-0) =-16 î to ý to  - vector regn of a plane is :- [e - (lê t2j+ 4 %)]. (-16 ê) = 0 fa-l lê +2 ġ +4213.4-16†) = 0
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