Tangent Line to the Intersection of Two Surfaces: 04-BS-1, May 2014, Q3

Question 3: Tangent Line to the Intersection of Two Surfaces (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two implicit surfaces $F_1=3x^2+2y^2-2z-1=0$ and $F_2=x^2+y^2+z^2-4y-2z+2=0$, meeting at $(1,1,2)$ (confirmed: $F_1(1,1,2)=3+2-4-1=0$; $F_2(1,1,2)=1+1+4-4-4+2=0$).

Find. The tangent line to the intersection curve at $(1,1,2)$.

Approach. The intersection curve lies on both surfaces, so its tangent direction is perpendicular to both surfaces' gradients at the point — i.e. it is $\nabla F_1\times\nabla F_2$.

Gradients at the point. $\nabla F_1=(6x,4y,-2)\big|_{(1,1,2)}=(6,4,-2)$. $\nabla F_2=(2x,2y-4,2z-2)\big|_{(1,1,2)}=(2,-2,2)$.
Cross product (tangent direction). $$\vec d=\nabla F_1\times\nabla F_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\6&4&-2\\2&-2&2\end{vmatrix}=(4(2)-(-2)(-2))\mathbf i-(6(2)-(-2)(2))\mathbf j+(6(-2)-4(2))\mathbf k=(4,-16,-20).$$ Simplifying by a factor of 4: $\vec d\parallel(1,-4,-5)$.
Write the parametric line. Through $(1,1,2)$ with direction $(1,-4,-5)$: $$\boxed{(x,y,z)=(1,1,2)+t(1,-4,-5),\quad t\in\mathbb R}$$

Quantity	Result
$\nabla F_1$ at $(1,1,2)$	$(6,4,-2)$
$\nabla F_2$ at $(1,1,2)$	$(2,-2,2)$
Tangent line	$(x,y,z)=(1,1,2)+t(1,-4,-5)$

Concept & Theory

Topic. Tangent line to the intersection curve of two implicit surfaces.

Key relations. A surface $F(x,y,z)=0$ has normal $\nabla F$ at each point; a curve lying in both surfaces $F_1=0$ and $F_2=0$ must, at each point, have a tangent vector orthogonal to both normals, so its direction is $\nabla F_1\times\nabla F_2$ (assuming the gradients are not parallel there).

Why this works. Moving along the intersection curve keeps both $F_1$ and $F_2$ identically zero, so the curve's velocity vector is orthogonal to $\nabla F_1$ and to $\nabla F_2$ by the chain rule ($\tfrac{d}{dt}F_i(\mathbf r(t))=\nabla F_i\cdot\mathbf r'(t)=0$); the cross product is exactly the (unique, up to scale) vector orthogonal to two given vectors.

Common pitfall. Sign errors expanding the $3\times3$ cross-product determinant, and forgetting to confirm the given point actually lies on both surfaces before differentiating.

Source. Stewart, Calculus: Early Transcendentals, Ch. 14 (§14.6 Tangent Planes and Normal Lines).

Question 3 of 8: Tangent Line to the Intersection of Two Surfaces

Question 3: Tangent Line to the Intersection of Two Surfaces (20 marks)