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Very useful notes: http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx (with great pics) Nice video also: http://patrickjmt.com/finding-the-scalar-equation-of-a-plane/ Hope this helps. :)
thanks! I'll check them out right now.
oops* the line r(t) should be = <2,-1,0>+t<2,3,0>
I watched the video and went over the pauls notes section (both of which I've used and are great resources) but I am drawing a blank on the line part r(t).
I believe I need to do something to it to get a normal vector and then I cross that with the point given and then sub the values into the equation a(x-x0)+b(y-y0)+c(z-z0)=0
You just need to take two vectors on the plane and cross product that to get the normal.
<3,-1,2> X <2,-1,0> ?
Not quite, you can also think of in terms of points. Let A(3,-1,2). Take another two points on the line (by plug in two values of t, up to you which t) to get B and C. Now you have Vector BA and BC and you can cross product that.
is this on the right track? i got <2+2t,-1+3t,0> would this be correct to sub a number in for t?
correct you substitute two values of t to get two points on the line. Then from the three points you have two vectors.
so I got these three points t(0) B= <2,-1,0> , t(1) C=<4,2,0> now I take that <3,-1,2> X <2,-1,0> => Vect BA <2,-1,0>X<4,2,0> => BC BA X BC => a point take that point and sub the x,y,z values into the equation a(x-x0)+b(y-y0)+c(z-z0)=0
Remember to get the vector say BC you do <4-2,2-(-1),0-0>=<2,3,0>
hmm, For BC i got 0,0,8 when I <2,-1,0>X<4,2,0> I believe you used -1 for the sub while i used 0
I went through and got <32,-16,0> for the abc values and ended with: 32(x-3)-16(y+1)+0
Hod on .... I used the points that you obtained earlier. Remember that if A(a,b,c) and B(d,e,f) Then the vector AB is given by
ohhh, I was doing a+d. Let me retry
<2t-2,3t+1,2t-2> does this look better?
BA X BC