help me find this normal vector!!! 1. The problem statement, all variables and given/known data Find a normal vector n and the equation of the tangent plane to the surface S at the point x0 S: z=ln(x2+3y2) x0=(2,-1,ln7) 2. Relevant equations 3. The attempt at a solution ok so I have this general formula from the textbook. z= z0+Fx(x0,y0)(x-x0)+Fy(x0,y0)(y-y0) and n=(Fx(x0,y0) , Fy(X0,Y0) , -1) respectively. so using these S: z= ln(x2+3y2) let u=x2+3y2 z=ln(u) dz/du=(1/u) (u') Fx(x,y) = d/dx(u)/u = 2x/(x2+3y2) = 4 Fy = d/dy (u)/u = 6y/ (x2+3y2) = -6 z= ln7 + 4(x-2)-6(y+1) z= ln7 +4x -6y -14 ln7 +4x -6y -14 -z = 0 so n =(4,-6,-1)T but they have 4x-6y-7z-14+7ln7=0 and n= (4, -6, -7)T ?? Im not sure what I have done wrong but it looks very close to what they have. any help is appreciated.