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anonymous

  • 5 years ago

A plane that contains (3,-1,2) point and the line r(t)=<2,-1,0>. What is the equation of that plane?

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  1. mathteacher1729
    • 5 years ago
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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. :)

  2. anonymous
    • 5 years ago
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    thanks! I'll check them out right now.

  3. anonymous
    • 5 years ago
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    oops* the line r(t) should be = <2,-1,0>+t<2,3,0>

  4. anonymous
    • 5 years ago
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    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).

  5. anonymous
    • 5 years ago
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    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

  6. watchmath
    • 5 years ago
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    You just need to take two vectors on the plane and cross product that to get the normal.

  7. anonymous
    • 5 years ago
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    <3,-1,2> X <2,-1,0> ?

  8. watchmath
    • 5 years ago
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    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.

  9. anonymous
    • 5 years ago
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    is this on the right track? i got <2+2t,-1+3t,0> would this be correct to sub a number in for t?

  10. watchmath
    • 5 years ago
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    correct you substitute two values of t to get two points on the line. Then from the three points you have two vectors.

  11. anonymous
    • 5 years ago
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    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

  12. watchmath
    • 5 years ago
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    Remember to get the vector say BC you do <4-2,2-(-1),0-0>=<2,3,0>

  13. anonymous
    • 5 years ago
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    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

  14. anonymous
    • 5 years ago
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    I went through and got <32,-16,0> for the abc values and ended with: 32(x-3)-16(y+1)+0

  15. watchmath
    • 5 years ago
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    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 <d-a,e-b,f-c>

  16. anonymous
    • 5 years ago
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    ohhh, I was doing a+d. Let me retry

  17. anonymous
    • 5 years ago
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    <2t-2,3t+1,2t-2> does this look better?

  18. watchmath
    • 5 years ago
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    Ok, lets use (a,b,c) for point and <a,b,c> for vector to avoid some confusion. Using the points A(3,-1,2) and B(2,-1,0) , C(4,2,0) that you obtained by plugin t=0,t=1 we have two vectors BA=<3-2,-1-(-1),2-0>=<1,0,2> BC=<4-2,2-(-1),0-0>=<2,3,0> Then you cross product BA and BC to get the normal vector.

  19. anonymous
    • 5 years ago
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    <-6,4,3> ?

  20. anonymous
    • 5 years ago
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    BA X BC

  21. watchmath
    • 5 years ago
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    yes

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