Angle of Intersection Between a Line and a Hyperboloid: 04-BS-1, December 2013, Q7

Question 7: Angle of Intersection Between a Line and a Hyperboloid (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. Line $\mathbf r(t)=(2-t,\ t,\ 2+2t)$; surface $z=8-x^2+y^2$.

Find. The angle $\varphi$ between the line and the surface at their intersection point.

Approach. First find the parameter $t$ (hence the point) where the line satisfies the surface equation. Then find the surface's normal there (gradient of $F=x^2-y^2+z-8$), and use the fact that the angle between a line and a surface is the complement of the angle between the line's direction and the surface normal: $\sin\varphi=\dfrac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}$.

Find the intersection point. Substitute the line into $z=8-x^2+y^2$: $$2+2t=8-(2-t)^2+t^2=8-(4-4t+t^2)+t^2=4+4t.$$ $$\Rightarrow\quad 2+2t=4+4t\;\Rightarrow\;-2=2t\;\Rightarrow\;t=-1.$$ Point: $(x,y,z)=(2-(-1),\,-1,\,2+2(-1))=(3,-1,0)$.
Surface normal at the point. Writing the surface as $F(x,y,z)=x^2-y^2+z-8=0$, $\nabla F=(2x,-2y,1)$. At $(3,-1,0)$: $$\mathbf n=(6,\,2,\,1).$$
Line direction and the angle formula. The line's direction is $\mathbf d=(-1,1,2)$ (coefficients of $t$). The angle between the line and the surface (not its normal) satisfies $$\sin\varphi=\frac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}.$$ $$\mathbf d\cdot\mathbf n=(-1)(6)+(1)(2)+(2)(1)=-6+2+2=-2,\qquad |\mathbf d|=\sqrt{1+1+4}=\sqrt6,\qquad |\mathbf n|=\sqrt{36+4+1}=\sqrt{41}.$$ $$\sin\varphi=\frac{2}{\sqrt6\sqrt{41}}=\frac{2}{\sqrt{246}}=\frac{\sqrt{246}}{123}.$$

$$\varphi=\boxed{\arcsin\!\left(\dfrac{2}{\sqrt{246}}\right)}\approx7.33^\circ$$

Quantity	Result
Intersection point	$(3,-1,0)$ at $t=-1$
Line direction $\mathbf d$	$(-1,1,2)$
Surface normal $\mathbf n$	$(6,2,1)$
Angle $\varphi$	$\arcsin(2/\sqrt{246})\approx7.33^\circ$

Concept & Theory

Topic. Angle between a curve and a surface at a point of intersection.

Key relations. A surface $F(x,y,z)=0$ has normal $\nabla F$ at each point; the angle $\theta$ between a direction vector $\mathbf d$ and the normal satisfies $\cos\theta=\dfrac{\mathbf d\cdot\mathbf n}{|\mathbf d||\mathbf n|}$; the angle between the line and the *surface itself* (i.e. its tangent plane) is the complement, $\varphi=90^\circ-\theta$, so $\sin\varphi=\cos\theta=\dfrac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}$.

Why this works. The tangent plane at the intersection point is perpendicular to $\mathbf n$; the angle between a line and a plane is measured from the line to its projection onto the plane, which is precisely $90^\circ$ minus the angle from the line to the plane's normal.

Common pitfall. Reporting $\arccos$ of the dot-product ratio directly as "the angle" — that gives the angle to the *normal*, not to the surface; the two differ by $90^\circ$ and the sine/cosine values are not interchangeable here. Also a sign slip expanding $(2-t)^2$.

Source. Stewart, Calculus: Early Transcendentals, Ch. 14 (§14.6 Directional Derivatives and the Gradient Vector — normal lines and tangent planes).

Question 7 of 8: Angle of Intersection Between a Line and a Hyperboloid

Question 7: Angle of Intersection Between a Line and a Hyperboloid (20 marks)