Gradient Vector, Tangent Planes and Normal Lines — Question 3

Question 3 – Solution

Strategy. A curve lying on both surfaces has a tangent perpendicular to both surface normals, so use their cross product.

Step 1: Surface normals Let $$F=x^2+y^2+z^2,\qquad G=x+y+z.$$ At $P$, $$\nabla F(P)=\left\langle 2,4,4\right\rangle,\qquad \nabla G(P)=\left\langle 1,1,1\right\rangle.$$ Their cross product is $$\nabla F(P)\times\nabla G(P) =\left\langle 0,2,-2\right\rangle,$$ so a simplified tangent direction is $\left\langle 0,1,-1\right\rangle$.

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

Step 3: Verification First, $1^2+2^2+2^2=9$ and $1+2+2=5$, so $P$ lies on both surfaces. Next, $$\left\langle 2,4,4\right\rangle\cdot\left\langle 0,1,-1\right\rangle=0, \qquad \left\langle 1,1,1\right\rangle\cdot\left\langle 0,1,-1\right\rangle=0.$$ Thus the direction lies in both tangent planes, as required.