anonymous
  • anonymous
find a vector equation and parametric equations through the point (1,0,6) and perpendicular to the plane x + 3y + z =5
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
is the point on a line? thru the plane?
anonymous
  • anonymous
ummm what?
amistre64
  • amistre64
if so, the normal of the plance is the perp to it; strip off your coeefs for the line vector: <1,3,1> would be the vector that is perpenducular to the plane

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amistre64
  • amistre64
since 1,0,6 is not on the plane im going to assume its on the line perp to it (the perp line is called a normal)
anonymous
  • anonymous
okay I'm with you so far
amistre64
  • amistre64
x = Px + 1t y = Py + 3t z = Pz + 1t given the point (1,0,6) this makes x = 1 + 1t y = 0 + 3t z = 6+ 1t
anonymous
  • anonymous
okay but why are the points (1,3,1) perpendicular to the plane?
amistre64
  • amistre64
its a result of how the equation of a plane is defined
amistre64
  • amistre64
the perp to the plane is a vector that dot producted to every line in the plane
amistre64
  • amistre64
so given a point (Px,Py,Pz) in space: we can construct any vector from it by taking any random point (x,y,z) and subtracting the 2 of them: (x-Px, y-Py, z-Pz) is every vector that comes out from our point into space
amistre64
  • amistre64
to elimante all the vectors that are not in a plane; we need to establish a perp vector to it so that when we dot it to these vectors we get 0, so all vectors of the form: (x-Px,y-Py,z-Pz) dot (Nx,Ny,Nz) = 0 Nx(x-Px) +Ny(y-Py) +Nz(z-Pz) = 0 is the equation of the plane
anonymous
  • anonymous
okay makes sense!
amistre64
  • amistre64
good :)
anonymous
  • anonymous
you're way better at explaining things than my book...
amistre64
  • amistre64
thats becasue im crazy lol
anonymous
  • anonymous
lol, does normal mean perpendicular?
mathmate
  • mathmate
yes.
mathmate
  • mathmate
Actually, yes in this context. The word normal is used for other things in maths elsewhere.

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