Question

find the vector eq of a plane pasing thru points (2,1,-1) and (-1,3,4) and perpendicular to the plane x-2y+4z=10.

Answer 1

@ExperimentX

Answer 2

let the equation of plane be
ax+by+cz+d=0
now since the plane passes through (2,1,-1) and (-1,3,4) ... put the values of (x,y,z) in above.
2a+b-c+d=0
-1a+3b+4c+d=0
----------------------
and finally since it is perpendicular to x-2y+4z=0
a*(1) + b*(-2) + 4*(c) = 0

when you solve these you get,
http://www.wolframalpha.com/input/?i=2a%2Bb-c%2Bd%3D0%2C+-1a%2B3b%2B4c%2Bd%3D0%2C+a*%281%29+%2B+b*%28-2%29+%2B+4*%28c%29+%3D+0
put the values,,, get rid of 'n' by taking common and equate it to zero. you have your plane.