Angle of Intersection Between a Line and a Hyperboloid: 04-BS-1, December 2015, Q6

Question 6: 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 $\mathbf r(t)=(1-t,\ t,\ 2+3t)$; surface $z=4-x^2+y^2$.

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

Approach. First find the parameter $t$ where the line satisfies the surface equation. Then find the surface's normal there (gradient of $F=x^2-y^2+z-4$), and use $\sin\varphi=\dfrac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}$ — the complement of the line-to-normal angle.

Find the intersection point. Substitute the line into $z=4-x^2+y^2$: $$2+3t=4-(1-t)^2+t^2=4-(1-2t+t^2)+t^2=3+2t.$$ $$\Rightarrow\quad 2+3t=3+2t \;\Rightarrow\; t=1.$$ Point: $(x,y,z)=(1-1,\,1,\,2+3)=(0,1,5)$.
Surface normal at the point. Writing the surface as $F(x,y,z)=x^2-y^2+z-4=0$, $\nabla F=(2x,-2y,1)$. At $(0,1,5)$: $$\mathbf n=(0,\,-2,\,1).$$
Line direction and the angle formula. The line's direction is $\mathbf d=(-1,1,3)$. The angle between the line and the surface (not its normal) satisfies $$\sin\varphi=\frac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}.$$ $$\mathbf d\cdot\mathbf n=(-1)(0)+(1)(-2)+(3)(1)=1,\qquad |\mathbf d|=\sqrt{1+1+9}=\sqrt{11},\qquad |\mathbf n|=\sqrt{0+4+1}=\sqrt5.$$ $$\sin\varphi=\frac{1}{\sqrt{11}\sqrt5}=\frac{1}{\sqrt{55}}.$$

$$\varphi=\boxed{\arcsin\!\left(\dfrac{1}{\sqrt{55}}\right)}\approx7.75^\circ$$

Quantity	Result
Intersection point	$(0,1,5)$ at $t=1$
Line direction $\mathbf d$	$(-1,1,3)$
Surface normal $\mathbf n$	$(0,-2,1)$
Angle $\varphi$	$\arcsin(1/\sqrt{55})\approx7.75^\circ$

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 $\mathbf d$ 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 (its tangent plane) is the complement, 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 that plane, which is exactly $90^\circ$ 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 is the angle to the normal, not to the surface; the two differ by $90^\circ$. Also a sign slip expanding $(1-t)^2$.

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

Question 6 of 8: Angle of Intersection Between a Line and a Hyperboloid

Question 6: Angle of Intersection Between a Line and a Hyperboloid (20 marks)