Angle Between a Line and a Surface: 04-BS-1, May 2018, Q5

Question 5: Angle Between a Line and a Surface (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 of intersection between the line and the surface.

Approach. First locate the intersection point by substituting the line into the surface equation. Then use $\sin\varphi=\dfrac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}$ — the complement of the angle between the line's direction $\mathbf d$ and the surface normal $\mathbf n$ — since the angle between a line and a surface is measured from the tangent plane, not from the normal.

Find the intersection point. Substitute $x=1-t,\ y=t,\ z=2+3t$ into $z=4-x^2+y^2$: $$2+3t=4-(1-t)^2+t^2=4-(1-2t+t^2)+t^2=3+2t\ \Longrightarrow\ t=1.$$ At $t=1$: $(x,y,z)=(0,1,5)$ (check: $4-0+1=5$✓).
Line direction and surface normal. $\mathbf d=(-1,1,3)$. Writing the surface implicitly as $F(x,y,z)=z+x^2-y^2-4=0$, $\nabla F=(2x,-2y,1)$, so at $(0,1,5)$: $\mathbf n=(0,-2,1)$.
Apply the line–surface angle formula. $$\mathbf d\cdot\mathbf n=(-1)(0)+(1)(-2)+(3)(1)=1,\qquad|\mathbf d|=\sqrt{11},\qquad|\mathbf n|=\sqrt5.$$ $$\sin\varphi=\frac{|1|}{\sqrt{11}\sqrt5}=\frac1{\sqrt{55}}.$$

$$\varphi=\boxed{\arcsin\!\left(\frac1{\sqrt{55}}\right)}\approx7.7^\circ$$

Quantity	Result
Intersection point	$(0,1,5)$ at $t=1$
$\mathbf d\cdot\mathbf n$	$1$
Angle $\varphi$	$\arcsin(1/\sqrt{55})\approx7.7^\circ$

Concept & Theory

Topic. Angle between a line and a surface at their point of intersection.

Key relations. The angle $\theta$ between the line direction $\mathbf d$ and the surface normal $\mathbf n$ satisfies $\cos\theta=|\mathbf d\cdot\mathbf n|/(|\mathbf d||\mathbf n|)$; the angle between the line and the surface itself is the complement, $\varphi=90^\circ-\theta$, so $\sin\varphi=\cos\theta$.

Why this works. A line lying exactly in the tangent plane makes a $0^\circ$ angle with the surface but a $90^\circ$ angle with the normal, which is precisely the sine/cosine complementary relationship; reporting $\arccos$ of the dot-product ratio directly (instead of taking the complement) is the single most common error in this problem type.

Common pitfall. Reporting $\arccos(|\mathbf d\cdot\mathbf n|/|\mathbf d||\mathbf n|)$ as the final answer instead of its complement $\arcsin$ of the same ratio; substituting the parameter into the wrong branch (forgetting to solve for $t$ first and instead guessing a point).

Source. Stewart, Calculus: Early Transcendentals, Ch. 14 (§14.6 Gradient Vectors and Normal Lines); Kreyszig, Advanced Engineering Mathematics, Ch. 9 (§9.7 Gradient of a Scalar Field).

Question 5 of 8: Angle Between a Line and a Surface

Question 5: Angle Between a Line and a Surface (20 marks)