How to find the linear equation of the plane through the point (−2,4,2) and perpendicular to the line represented by the vector equation r(t)=⟨1+4t,−2−t,6+2t⟩ ?

1 Answer
Cesareo R.
May 15, 2016

#8 + 4 x - y + 2 z=0#

Explanation:

Given a normal vector #v# to the plane #Pi#, a point #Q# pertaining to the plane and #P=(x,y,z)# a generic plane point, the resulting equation is
#Pi->(P-Q)*v=0#
The straight is given by
#r->P_r+t v# with #P_r = (1,-2,6)# and #v = (4,-1,2)#
so #(x+2,y-4,z-2)*(4,-1,2)=0# is the plane equation

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