Gradient Vector, Tangent Planes and Normal Lines — Question 2

Question 2 – Solution

Strategy. Differentiate the defining level function; its gradient supplies both the plane normal and the normal-line direction.

Step 1: Gradient $$\nabla F=\left\langle ye^z,xe^z,xye^z\right\rangle.$$ At $P=(1,2,0)$, $$\nabla F(P)=\left\langle 2,1,2\right\rangle,\qquad \|\nabla F(P)\|=3.$$ Therefore the two unit normals are $$\boxed{\pm\frac 13\left\langle 2,1,2\right\rangle}.$$

Step 2: Tangent plane $$2(x-1)+(y-2)+2(z-0)=0,$$ so $$\boxed{2x+y+2z=4}.$$

Step 3: Normal line Using the unsimplified integer direction, $$\boxed{(x,y,z)=(1,2,0)+t(2,1,2)}.$$ Equivalently, $$\boxed{\frac{x-1}{2}=y-2=\frac z2}.$$

Verification The point satisfies $1\cdot 2\cdot e^0=2$. The line direction equals the plane’s coefficient vector, so it is perpendicular to the plane.