find a vector equation and parametric equations through the point (1,0,6) and perpendicular to the plane x + 3y + z =5
Stacey Warren - Expert brainly.com
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is the point on a line? thru the plane?
if so, the normal of the plance is the perp to it; strip off your coeefs for the line vector:
would be the vector that is perpenducular to the plane
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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)
okay I'm with you so far
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
okay but why are the points (1,3,1) perpendicular to the plane?
its a result of how the equation of a plane is defined
the perp to the plane is a vector that dot producted to every line in the plane
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
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
okay makes sense!
you're way better at explaining things than my book...
thats becasue im crazy lol
lol, does normal mean perpendicular?
Actually, yes in this context.
The word normal is used for other things in maths elsewhere.