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are those equa;l signs spoose to be (+) signs?
the normal vector appeart to be <3,2,6> assuming those are sposed to be (+) signs
the equation appears to be: 3x-6 +2y-4 +6z+18 = 0 3x + 2y +6z +8 = 0 would be the eqaution of the plane
no i wrote it as it is
that isn't in the answer choices
what are the answer choices :)
ok give me a secon it's kind of a lot to type out
its looks line parametric representation of the line... what is the topic of your study?
a) x=2+2t, y=2+3t, z=-3+t b)x=2+3t, y=2+2t, z=-3+6t c)x=2+2t, y=2-3t, z=-3-t d)x=2+3t, y=2-2t, z=-3-6t
this is cal 3...can you help?
a) x=2+2t, y=2+3t, z=-3+t b)x=2+3t, y=2+2t, z=-3+6t c)x=2+2t, y=2-3t, z=-3-t d)x=2+3t, y=2-2t, z=-3-6t my guess is (b) but i have no proof to back that up......its been awhile :)
ok thanks for looking
To find the parametric equations of the line passing through the point (2,2,-3) and parallel to the vector <3,2,6>. We first find the vector equation of the line. Here, r_0=<2,2,-3> and v=<3,2,6>. Thus, the line has vector equation r=<2,2,-3>+t<3,2,6>. The parametric equations of the line are: x=2+3t, y=2+2t, and z=-3+6t. [B]
copied that off the web and included your numbers :) http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/lineplane/lineplane.html