Gradient Vector, Tangent Planes and Normal Lines — Question 8

Question 8 – Solution

Strategy. Use the gradient for the plane and line, then compare the point-to-plane formula with the length of the normal segment.

Step 1: Plane and line For $F=xyz$, $$\nabla F=\left\langle yz,xz,xy\right\rangle, \qquad \nabla F(P)=\left\langle 4,4,4\right\rangle.$$ Thus the tangent plane is $$4(x-2)+4(y-2)+4(z-2)=0,$$ or $$\boxed{x+y+z=6}.$$ A normal line is $$\boxed{(x,y,z)=(2,2,2)+t(1,1,1)}.$$

Step 2: Origin on the line At $t=-2$, all three coordinates are zero. Thus the segment from the origin to $P$ follows the plane’s normal direction, making $P$ the perpendicular foot.

Step 3: Distance checks The normal-segment length is $$\|\left\langle 2,2,2\right\rangle\|=\boxed{2\sqrt 3}.$$ The point-to-plane formula independently gives $$\frac{|0+0+0-6|}{\sqrt{1^2+1^2+1^2}} =\frac 6{\sqrt 3}=\boxed{2\sqrt 3}.$$ The agreement verifies both the geometry and the scale of the normal vector.