A normal to 3x-y+4z=12 passes through A(5,1,6). Find the point of intersection of the normal and the plane.
So from what I know the normal would be:
n = (3,-1,4)
But I don't know how to go about answering this....Any help is appreciated...
Hello : the nomal is definid by : x-5=3t y-1=-t z-6=4t x=5+3t and y=1-t and z=6+4t Remplace in plan equation : 3(5+3t) -(1-t)+4(6+4t)=12 solve : I'have t= -1 remplace in system : x=5+3( -1 ) and y =1-(-1 ) and z= 6+4( -1) coclusion: the point M( 2 , 2 , 2 )