Gradient Vector, Tangent Planes and Normal Lines — Question 7

Question 7 – Solution

Strategy. Match the unknown gradient to the plane’s normal; the first gradient component fixes the proportionality factor.

Step 1: Match normals $$\nabla F(P)=\left\langle 2,2a,2b\right\rangle.$$ This must equal $\lambda\left\langle 1,2,3\right\rangle$. The first component gives $\lambda=2$. Therefore $$2a=4,\qquad 2b=6,$$ so $$\boxed{a=2,\qquad b=3}.$$ Since $P$ lies on the surface, $$c=1^2+2(1)^2+3(1)^2=\boxed{6}.$$

Step 2: Regularity For the resulting surface, $$\nabla F(P)=\left\langle 2,4,6\right\rangle\ne\left\langle 0,0,0\right\rangle,$$ so $P$ is regular.

Step 3: Line and plane verification The normal line is $$\boxed{(x,y,z)=(1,1,1)+t(1,2,3)}.$$ The plane obtained from the gradient is $$2(x-1)+4(y-1)+6(z-1)=0.$$ Dividing by $2$ produces $x+2y+3z=6$, exactly the prescribed plane.