Two Lines in Space — Intersection, Common Perpendicular, and Containing Plane: 04-BS-1, December 2015, Q3

Question 3: Two Lines in Space — Intersection, Common Perpendicular, and Containing Plane (a) 10, (b) 5, (c) 5 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. $L_1$ through $P_1=(3,3,1)$ with direction $\mathbf d_1=(2,0,-1)$; $L_2$ through $P_2=(0,1,2)$ with direction $\mathbf d_2=(1,-2,1)$.

Find. (a) Whether $L_1,L_2$ meet. (b) A line orthogonal to both. (c) A plane containing both, if one exists.

$L_1$ and $L_2$: their $xy$-shadows cross, but the two lines pass at different heights there — they are skew.

Approach. (a) Set the $x$- and $y$-components equal to solve for the two parameters, then check whether the $z$-components agree at that parameter pair. (b) The direction orthogonal to both lines is $\mathbf d_1\times\mathbf d_2$; any line with that direction qualifies. (c) A single plane contains both lines only if they are coplanar (intersecting or parallel); skew lines admit no such plane.

(a) Match $x$ and $y$. $L_1$ has $y=3$ always; $L_2$ has $y=1-2s$, so $3=1-2s\Rightarrow s=-1$. Matching $x$: $3+2t=s=-1\Rightarrow t=-2$.
(a) Check $z$ at this parameter pair. $L_1$: $z=1-t=1-(-2)=3$. $L_2$: $z=2+s=2+(-1)=1$. Since $3\ne1$, the two lines do not intersect — and since $\mathbf d_1=(2,0,-1)$ is not a scalar multiple of $\mathbf d_2=(1,-2,1)$, they are not parallel either, so $$\boxed{L_1\text{ and }L_2\text{ are skew (no intersection).}}$$
(b) Cross product of the two directions. $$\mathbf d_1\times\mathbf d_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&0&-1\\1&-2&1\end{vmatrix}=\big(0(1)-(-1)(-2)\big)\mathbf i-\big(2(1)-(-1)(1)\big)\mathbf j+\big(2(-2)-0(1)\big)\mathbf k=(-2,-3,-4).$$ This vector is perpendicular to both $\mathbf d_1$ and $\mathbf d_2$ by construction, so any line with this direction is orthogonal to both $L_1$ and $L_2$ — the problem does not require it to meet either line. Taking, for concreteness, the line through the origin: $$\boxed{(x,y,z)=u(-2,-3,-4),\quad u\in\mathbb R}$$
(c) Coplanarity test. Two lines lie in a common plane exactly when they intersect or are parallel. Part (a) showed $L_1,L_2$ are neither (they are skew), so $$\boxed{\text{No plane contains both }L_1\text{ and }L_2.}$$

Part	Result
(a) Intersection	None — lines are skew ($z=3$ vs. $z=1$ at the only $x,y$-matching parameters)
(b) Orthogonal direction	$\mathbf d_1\times\mathbf d_2=(-2,-3,-4)$
(c) Containing plane	Does not exist ($L_1,L_2$ skew)

Concept & Theory

Topic. Classifying two lines in $\mathbb R^3$: intersecting, parallel, or skew.

Key relations. Two lines are parallel iff their direction vectors are scalar multiples; if not parallel, they intersect iff the systems obtained from matching all three coordinate equations are consistent for a single $(t,s)$ pair. A common perpendicular direction is always $\mathbf d_1\times\mathbf d_2$. Two lines are coplanar (lie in one plane) iff they intersect or are parallel — skew lines are, by definition, the case where neither holds.

Why this works. With only two equations ($x$ and $y$) available to fix two unknowns ($t,s$), the system is already fully determined before $z$ is checked; the $z$-equation is then an independent consistency check. If it fails, no single $(t,s)$ satisfies all three coordinate equations simultaneously, so the lines never meet.

Common pitfall. Stopping after matching only $x$ and $y$ and declaring an intersection without checking $z$ — this is the single most common error in this problem type. Also, assuming any two lines admit a containing plane; skew lines fundamentally do not.

Source. Stewart, Calculus: Early Transcendentals, Ch. 12 (§12.5 Equations of Lines and Planes in Space).

Question 3 of 8: Two Lines in Space — Intersection, Common Perpendicular, and Containing Plane

Question 3: Two Lines in Space — Intersection, Common Perpendicular, and Containing Plane (a) 10, (b) 5, (c) 5 marks