Two Lines in Space — Intersection, Orthogonal Line, Common Plane: 04-BS-1, May 2013, Q6

Question 6: 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. $L_1:(x,y,z)=(3-2t,\,3,\,3-t)$, direction $\mathbf d_1=(-2,0,-1)$. $L_2:(x,y,z)=(s,\,1-2s,\,-s)$, direction $\mathbf d_2=(1,-2,-1)$.

Find. (a) whether $L_1,L_2$ intersect and, if so, the point. (b) a line orthogonal to both. (c) a plane containing both lines, if one exists.

Figure — lines L1, L2 meet at P(−1, 3, 1); L3 (dashed) along d1×d2 = (−2,−3,4) is orthogonal to both and is normal to the plane 2x + 3y − 4z − 3 = 0 spanned by L1 and L2.

Approach. Equate coordinates of $L_1$ and $L_2$ to solve for $t,s$ (a 2-equation subsystem, then check the third equation); the orthogonal line's direction is $\mathbf d_1\times\mathbf d_2$; since the lines meet at a point, they automatically span a plane whose normal is that same cross product.

(a) Solve for intersection. From $y$: $3=1-2s\Rightarrow s=-1$. From $x$: $3-2t=s=-1\Rightarrow t=2$. Check $z$: $L_1$ gives $3-t=1$; $L_2$ gives $-s=1$ — both equal 1, so the system is consistent. $$\Rightarrow\quad\text{Intersection point } \boxed{P=(-1,3,1)}\ \ (t=2,\,s=-1).$$
(b) Orthogonal line direction. A line orthogonal to both $L_1$ and $L_2$ must have direction $\mathbf d_1\times\mathbf d_2$ (perpendicular to both direction vectors): $$\mathbf d_1\times\mathbf d_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\-2&0&-1\\1&-2&-1\end{vmatrix}=\mathbf i\big(0\cdot(-1)-(-1)(-2)\big)-\mathbf j\big((-2)(-1)-(-1)(1)\big)+\mathbf k\big((-2)(-2)-0\cdot 1\big)=(-2,-3,4).$$ Check: $\mathbf d_1\cdot(-2,-3,4)=4+0-4=0$ and $\mathbf d_2\cdot(-2,-3,4)=-2+6-4=0$, as required. Taking the intersection point $P=(-1,3,1)$ as a convenient point on this line, $$\Rightarrow\quad \boxed{L_3:\ (x,y,z)=(-1-2r,\ 3-3r,\ 1+4r)}.$$
(c) Plane containing both lines. Since $L_1$ and $L_2$ intersect at $P$, they determine a unique plane, whose normal is the same $\mathbf d_1\times\mathbf d_2=(-2,-3,4)$ found in (b). Using point $P=(-1,3,1)$: $$-2(x+1)-3(y-3)+4(z-1)=0 \;\Rightarrow\; -2x-3y+4z+3=0.$$ $$\Rightarrow\quad \boxed{2x+3y-4z-3=0}$$ Check both lines lie in it: $L_1$ gives $2(3-2t)+3(3)-4(3-t)-3=6-4t+9-12+4t-3=0$, and $L_2$ gives $2s+3(1-2s)-4(-s)-3=2s+3-6s+4s-3=0$, for every $t$ and $s$.

Part	Result
(a) Intersection	Yes — $P=(-1,3,1)$
(b) Orthogonal line	$(x,y,z)=(-1-2r,\,3-3r,\,1+4r)$
(c) Common plane	$2x+3y-4z-3=0$

Concept & Theory

Topic. Lines and planes in 3-space: intersection, cross product for orthogonality, plane through two intersecting lines.

Key relations. Two lines intersect iff their parametric coordinate equations have a common consistent solution $(t,s)$; a vector orthogonal to two given directions is $\mathbf d_1\times\mathbf d_2$; a plane through point $P_0$ with normal $\mathbf n$ is $\mathbf n\cdot(\mathbf r-\mathbf r_0)=0$.

Why this works. With three coordinate equations and only two unknowns ($t,s$), the system is overdetermined — solving two of them and checking the third is both necessary and sufficient to confirm intersection (skew lines would fail the check). Because $\mathbf d_1\times\mathbf d_2$ is perpendicular to both direction vectors by definition of the cross product, it simultaneously answers (b) (a line along it) and (c) (a plane normal to it) — the same computation serves both parts once the lines are known to intersect.

Common pitfall. Solving only two of the three coordinate equations and assuming intersection without checking the third (this misses the very common case of two skew lines that only *appear* to meet in the $x,y$ or $x,z$ projection). Also a sign error in the $3\times3$ cross-product expansion (the $\mathbf j$ component carries a minus sign).

Source. Stewart, Calculus: Early Transcendentals, Ch. 12 (Vectors and the Geometry of Space, §12.4–12.5 Cross Product, Lines and Planes).

Question 6 of 8: Two Lines in Space — Intersection, Orthogonal Line, Common Plane

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