Gradient Vector, Tangent Planes and Normal Lines — Question 1

Question 1 – Solution

Strategy. For a regular level surface, the gradient at the point is normal to the tangent plane.

Step 1: Gradient and regularity $$\nabla F=\left\langle 2x,4y,6z\right\rangle, \qquad \nabla F(1,1,1)=\left\langle 2,4,6\right\rangle.$$ This vector is nonzero, so $P$ is regular. Also $F(1,1,1)=1+2+3=6$, confirming that $P$ lies on the surface.

Step 2: Tangent plane Using $\nabla F(P)$ as a normal, $$2(x-1)+4(y-1)+6(z-1)=0.$$ After division by $2$, $$\boxed{x+2y+3z=6}.$$

Step 3: Normal line A simplified direction parallel to the gradient is $\left\langle 1,2,3\right\rangle$. Hence $$\boxed{(x,y,z)=(1,1,1)+t(1,2,3)},$$ or $x=1+t$, $y=1+2t$, $z=1+3t$.

Verification The line passes through $P$ at $t=0$, and $\left\langle 1,2,3\right\rangle$ is perpendicular to every displacement $\left\langle u,v,w\right\rangle$ in the plane because the plane equation requires $u+2v+3w=0$.