Question 3 of 8: Tangent Plane and Tangent Line to a Surface Intersection

Question 3: Tangent Plane and Tangent Line to a Surface Intersection (a) 10, (b) 10 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. $f(x,y,z)=x^2+y^2+z^2+2y-3x$, $g(x,y,z)=3x+y^2-z^2$, common point $P=(3,-1,1)$ (check: $f(P)=9+1+1-2-9=0$ and $g(P)=9+1-1=9$, so $P$ lies on both level surfaces).

Find. (a) The tangent plane to $g=9$ at $P$. (b) The tangent line to the curve $f=0\cap g=9$ at $P$.

Approach. A level surface's tangent plane uses its gradient as normal vector. The curve of intersection of two surfaces is tangent, at any common point, to the direction $\nabla f\times\nabla g$ (perpendicular to both surface normals, hence lying in both tangent planes).

1. (a) Gradient of $g$ at $P$. $\nabla g=(3,\,2y,\,-2z)$, so $\nabla g(P)=(3,-2,-2)$.
2. (a) Tangent plane. $$3(x-3)-2(y+1)-2(z-1)=0\ \Longrightarrow\ \boxed{3x-2y-2z=9}.$$
3. (b) Gradient of $f$ at $P$. $\nabla f=(2x-3,\,2y+2,\,2z)$, so $\nabla f(P)=(3,0,2)$.
4. (b) Cross the two gradients. The tangent direction to the intersection curve is $$\nabla f\times\nabla g=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\3&0&2\\3&-2&-2\end{vmatrix}=\bigl(0(-2)-2(-2),\ 2(3)-3(-2),\ 3(-2)-0(3)\bigr)=(4,12,-6).$$ Dividing by 2, a convenient direction vector is $\mathbf d=(2,6,-3)$.
5. (b) Write the line. $$(x,y,z)=(3,-1,1)+s(2,6,-3),\quad s\in\mathbb R.$$

$$\text{(a)}\ \boxed{3x-2y-2z=9}\qquad\text{(b)}\ (x,y,z)=(3,-1,1)+s(2,6,-3)$$