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anonymous

  • 4 years ago

find the equation of the plane through point (-2,8,10) and perpendicular to line x=1+T, y=2T, z=4-3T

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  1. anonymous
    • 4 years ago
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    is the answer: x+2y-3z=16?

  2. amistre64
    • 4 years ago
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    our perp line gives us our normal to the plane so strip out the vector in it

  3. amistre64
    • 4 years ago
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    <1,2,-3> now apply this to any random point (x,y,z) - (-2,8,10) 1(x--2) +2(y-8) -3(z-10) = 0 x+2 +2y -16 -3z +30 = 0 x +2y -3z +16 = 0 x +2y -3z = -16 dbl chk the math to be sure, i tend to forget how to add and subtract lol

  4. anonymous
    • 4 years ago
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    yay i'm figuring out how to do this!!

  5. amistre64
    • 4 years ago
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    pretty soon youll be like; I cant believe I didnt know this stuff !!! :)

  6. amistre64
    • 4 years ago
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    i figure in a few weeks we will be going over this in calc 3

  7. anonymous
    • 4 years ago
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    okay so for the plane through the origin and points (x,y,z) and (a,b,c,) i would do do (a,b,c)x(x,y,z) to find the normal right?

  8. anonymous
    • 4 years ago
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    and then use the normal to write an equation

  9. amistre64
    • 4 years ago
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    yes, since the origin is one of our points; the other 2 points are also the vectors; so cross them and get the normal

  10. amistre64
    • 4 years ago
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    then use the normal attached to any of the 3 points (0,0,0) being the most useful id assume

  11. anonymous
    • 4 years ago
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    will my answer be a dot product?

  12. amistre64
    • 4 years ago
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    the cross of 2 vectors is a vector

  13. anonymous
    • 4 years ago
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    of the normal and (x,y,z)

  14. amistre64
    • 4 years ago
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    your answer will be a dot product of the normal and every vector in the plane: <x-0, y-0, z-0> would define all vector in the plane

  15. anonymous
    • 4 years ago
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    okay awesome thats what i got!

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