Question 4 of 7: Tangent Line to the Intersection of Two Quadric Surfaces

Question 4: Tangent Line to the Intersection of Two Quadric 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 quadric surfaces $F_1=3x^2+2y^2-z^2-1=0$ and $F_2=x^2+y^2+z^2-6y=0$, intersecting at $(1,1,2)$.

Find. Parametric equations of the line tangent to the curve of intersection at $(1,1,2)$.

Approach. The tangent line to the intersection curve is orthogonal to both surface gradients at the point, so its direction is $\nabla F_1\times\nabla F_2$.

1. Confirm the point lies on both surfaces. $F_1(1,1,2)=3+2-4-1=0$ ✓. $F_2(1,1,2)=1+1+4-6=0$ ✓.
2. Compute the gradients at the point. $$\nabla F_1=(6x,4y,-2z)\Big|_{(1,1,2)}=(6,4,-4),\qquad \nabla F_2=(2x,2y-6,2z)\Big|_{(1,1,2)}=(2,-4,4).$$
3. Cross the gradients for the tangent direction. $$\nabla F_1\times\nabla F_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\6&4&-4\\2&-4&4\end{vmatrix}=(4\cdot4-(-4)(-4))\mathbf i-(6\cdot4-(-4)\cdot2)\mathbf j+(6(-4)-4\cdot2)\mathbf k=(0,-32,-32).$$ This simplifies to the direction $(0,1,1)$.

$$\boxed{x=1,\quad y=1+t,\quad z=2+t}$$