pravin19 Posted April 11, 2015 Share Posted April 11, 2015 (edited) Determine scalar, vector and parametric equations for the plane that passes through the points A(1,-2,0), B(1,-2,2), and C(0,3,2). This is what I got... AB AB=OA-OB =(1,-2,2)-(1,-2,0) =(0,0,2) AC=OA-OC =(0,3,2)-(1,-2,0) =(-1,5,2) AB x AC (0)(2)-(5)(2) =0-10 =-10 (2)(-1)-(2)(0) =-2-0 =-2 (0)(5)-(-1)(0) =0 The normal vector is (-10,-2,0). -10x-2y+0z+d=0 -10(1)-2(-2)+0(0)+d=0 -10+4+d=0 -6+d=0 d=6 The scalar equation is -10x-2y+6=0 The vector equation is (1,-2,0) + s(0,0,2) + t(-1,5,2) Parametric Equations: x=1-t y=-2+5t z=2s+2t Did I do this right? Thanks for the help in advance. Edited April 11, 2015 by pravin19 Link to comment Share on other sites More sharing options...
studiot Posted April 11, 2015 Share Posted April 11, 2015 The scalar equation is -10x-2y+6=0 Is the coefficient of z truly zero? Link to comment Share on other sites More sharing options...
pravin19 Posted April 11, 2015 Author Share Posted April 11, 2015 Yeah, are my answers right? Link to comment Share on other sites More sharing options...
studiot Posted April 11, 2015 Share Posted April 11, 2015 Thar's a bit abrupt, I was wondering if you realised the implications of that statement. However, according to WolframAlpha yes that is the correct plane. To check, input the determinant as follows matrix {{x,y,z,1},{1,-2,0,1},{1,-2,2,1},{0,3,2,1}} Link to comment Share on other sites More sharing options...
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