Two Lines in Space — Intersection, Orthogonal Line, Common Plane: 04-BS-1, December 2018, Q4

Question 4: Two Lines in Space — Intersection, Orthogonal Line, Common Plane (a) 7, (b) 7, (c) 6 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. $L_1:(3+2t,\,3,\,1-t)$, direction $\mathbf d_1=(2,0,-1)$; $L_2:(s,\,1-2s,\,2+s)$, direction $\mathbf d_2=(1,-2,1)$.

Find. (a) Whether $L_1,L_2$ intersect. (b) A line orthogonal to both. (c) Whether a common plane exists.

Approach. Match $x$- and $y$-coordinates to solve for $t,s$, then check whether the resulting $z$-coordinates agree. The direction of a line orthogonal to both is $\mathbf d_1\times\mathbf d_2$. Two lines admit a common plane only if they are parallel or intersecting; skew lines admit none.

$L_1$ (blue) and $L_2$ (purple) pass near each other — at the $t=-2$, $s=-1$ solution of the $x,y$-equations, $L_1$ reaches $z=3$ while $L_2$ reaches $z=1$ — so the lines miss each other (skew). $L_3$ (green, dashed) is the direction $\mathbf d_1\times\mathbf d_2$ orthogonal to both, anchored here at $(3,3,1)$.

(a) Solve the $x,y$-equations. $y$: $3=1-2s\Rightarrow s=-1$. $x$: $3+2t=s=-1\Rightarrow t=-2$.
(a) Check the $z$-equation. $z_1=1-t=1-(-2)=3$; $z_2=2+s=2+(-1)=1$. Since $3\ne1$, the unique $(t,s)$ that satisfies the $x$- and $y$-equations fails the $z$-equation, so $$\boxed{\text{the lines do NOT intersect.}}$$ Since $\mathbf d_1=(2,0,-1)$ and $\mathbf d_2=(1,-2,1)$ are not parallel (not scalar multiples), the two lines are skew.
(b) Orthogonal direction. $$\mathbf d_1\times\mathbf d_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&0&-1\\1&-2&1\end{vmatrix}=\bigl(0(1)-(-1)(-2),\ -[2(1)-(-1)(1)],\ 2(-2)-0(1)\bigr)=(-2,-3,-4).$$ Any point may serve as the anchor since only the direction is asked for; anchoring at $(3,3,1)$ (the $t=0$ point of $L_1$) gives $$L_3:(x,y,z)=(3,3,1)+r(-2,-3,-4),\quad r\in\mathbb R.$$
(c) Common plane? A plane can contain two lines only if they are parallel or intersecting (coplanar). Step 2 showed $L_1,L_2$ are neither — they are skew. Therefore $$\boxed{\text{no plane contains both } L_1 \text{ and } L_2.}$$

Quantity	Result
(a) Intersect?	No ($z_1=3\ne z_2=1$ at the only $x,y$-consistent $t,s$)
(b) Orthogonal direction $\mathbf d_1\times\mathbf d_2$	$(-2,-3,-4)$
(b) $L_3$	$(3,3,1)+r(-2,-3,-4)$
(c) Common plane?	No (lines are skew)

Concept & Theory

Topic. Relative position of two lines in 3-space: intersecting, parallel, or skew.

Key relations. Two lines with non-parallel direction vectors either intersect at a unique point or are skew. Solving any two of the three coordinate equations pins down candidate parameters; the third equation is the consistency check. A line orthogonal to both given lines has direction $\mathbf d_1\times\mathbf d_2$. Two lines determine a common plane iff they are coplanar (parallel or intersecting).

Why this works. Three scalar equations in two unknowns ($t,s$) is an over-determined system; generically it has no solution unless the lines happen to cross, which is exactly the geometric condition for intersection. Skew lines are neither parallel nor intersecting, so by definition no plane can contain both simultaneously.

Common pitfall. Stopping after matching only two coordinate equations and declaring "intersection" without checking the third; assuming any two non-parallel lines in 3-space must intersect (false in three dimensions — skew lines are the generic case).

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

Question 4 of 8: Two Lines in Space — Intersection, Orthogonal Line, Common Plane

Question 4: Two Lines in Space — Intersection, Orthogonal Line, Common Plane (a) 7, (b) 7, (c) 6 marks