anonymous
  • anonymous
find the equation of the plane through point (-2,8,10) and perpendicular to line x=1+T, y=2T, z=4-3T
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
is the answer: x+2y-3z=16?
amistre64
  • amistre64
our perp line gives us our normal to the plane so strip out the vector in it
amistre64
  • amistre64
<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

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anonymous
  • anonymous
yay i'm figuring out how to do this!!
amistre64
  • amistre64
pretty soon youll be like; I cant believe I didnt know this stuff !!! :)
amistre64
  • amistre64
i figure in a few weeks we will be going over this in calc 3
anonymous
  • anonymous
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?
anonymous
  • anonymous
and then use the normal to write an equation
amistre64
  • amistre64
yes, since the origin is one of our points; the other 2 points are also the vectors; so cross them and get the normal
amistre64
  • amistre64
then use the normal attached to any of the 3 points (0,0,0) being the most useful id assume
anonymous
  • anonymous
will my answer be a dot product?
amistre64
  • amistre64
the cross of 2 vectors is a vector
anonymous
  • anonymous
of the normal and (x,y,z)
amistre64
  • amistre64
your answer will be a dot product of the normal and every vector in the plane: would define all vector in the plane
anonymous
  • anonymous
okay awesome thats what i got!

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