Two Lines in Space — Intersection, Orthogonal Line, and Common Plane: 04-BS-1, December 2014, Q3

Question 3: Two Lines in Space — Intersection, Orthogonal Line, and 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. $L_1:\ (x,y,z)=(2-t,\,3t,\,1+t)$, direction $\mathbf d_1=(-1,3,1)$, point $(2,0,1)$. $L_2:\ (x,y,z)=(1+s,\,3-2s,\,2+4s)$, direction $\mathbf d_2=(1,-2,4)$, point $(1,3,2)$.

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

Two lines meeting at $(1,3,2)$; the third orthogonal line (and the common plane's normal) runs along $\mathbf d_1\times\mathbf d_2$.

Approach. (a) Set the two parametrizations equal component-wise and solve for $t,s$; if a common $(t,s)$ satisfies all three equations, the lines intersect there. (b) Any line orthogonal to both directions runs along $\mathbf d_1\times\mathbf d_2$. (c) Two lines that intersect and are not parallel determine a unique plane, with the same normal $\mathbf d_1\times\mathbf d_2$.

(a) Solve for $t,s$. Equating $x$: $2-t=1+s\Rightarrow s=1-t$. Equating $y$: $3t=3-2s$. Substituting $s=1-t$: $3t=3-2(1-t)=1+2t\Rightarrow t=1$, so $s=0$.
(a) Check the third equation and report the point. $z$: $1+t=1+1=2$ from $L_1$; $2+4s=2+0=2$ from $L_2$ — consistent. So the lines intersect at $$\boxed{(x,y,z)=(1,3,2)}.$$
(b) Orthogonal direction via cross product. $$\mathbf d_1\times\mathbf d_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\-1&3&1\\1&-2&4\end{vmatrix}=(3(4)-1(-2))\mathbf i-((-1)(4)-1(1))\mathbf j+((-1)(-2)-3(1))\mathbf k=(14,\,5,\,-1).$$ A third line orthogonal to both, through the intersection point: $$\boxed{(x,y,z)=(1,3,2)+u(14,5,-1)}.$$
(c) Plane through both lines. Since $L_1,L_2$ intersect (are coplanar) and are not parallel, they determine a unique plane with normal $\mathbf d_1\times\mathbf d_2=(14,5,-1)$ through $(1,3,2)$: $$14(x-1)+5(y-3)-1(z-2)=0\ \Rightarrow\ \boxed{14x+5y-z=27}.$$

Part	Result
(a) Intersection point	$(1,3,2)$, at $t=1,\ s=0$
(b) Orthogonal line	$(1,3,2)+u(14,5,-1)$
(c) Common plane	$14x+5y-z=27$

Concept & Theory

Topic. Intersection of two lines in space, orthogonal directions via cross product, and the plane determined by two intersecting lines.

Key relations. Two lines with directions $\mathbf d_1,\mathbf d_2$ intersect iff a common parameter pair $(t,s)$ satisfies all three coordinate equations simultaneously (two equations determine candidates for $t,s$; the third must be checked for consistency). Any vector orthogonal to both directions is parallel to $\mathbf d_1\times\mathbf d_2$, which also serves as the normal of the (unique) plane containing two coplanar, non-parallel lines.

Why this works. Three unknowns ($t,s$, and the requirement of equal $z$) but only two free parameters means intersection is a codimension-1 condition — solving two of the three coordinate equations pins down candidate $(t,s)$, and the third equation is the actual test of intersection (as opposed to skew lines, where it fails).

Common pitfall. Solving only two of the three coordinate equations and declaring intersection without checking the third — two lines in 3-space are skew far more often than they intersect, so the consistency check is not optional.

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

Question 3 of 8: Two Lines in Space — Intersection, Orthogonal Line, and Common Plane

Question 3: Two Lines in Space — Intersection, Orthogonal Line, and Common Plane (a) 7, (b) 7, (c) 6 marks