Wize University Linear Algebra Textbook > Equations of Lines and Planes

Intersection of Two Lines

Intersection of a Line and a PlaneNext Section

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Intersection of Two Lines
Possible Intersection Scenarios
Lines are coincident (same line) 
They are parallel and intersect (share one point)   ⟹  \implies⟹ infinitely many points of intersection
Lines are parallel but do not intersect   ⟹  \implies⟹ zero points of intersection
Lines are skew[R3 and higher ONLY]\left[ \colorTwo{\reals^3 \text{ and higher ONLY}} \right][R3 and higher ONLY]
They are not parallel but they do not intersect   ⟹  \implies⟹ zero points of intersection
Lines are not parallel and they intersect   ⟹  \implies⟹ one point of intersection
Table Summary
ParallelIntersect# Intersection PointsCoincident✓✓∞Parallel, Non-Coincident✓×0Skew [R3+]××0Intersecting×✓1\def\arraystretch{1.5}
\begin{array}{|lc c c|}
\hline
 & \text{Parallel} & \text{Intersect} & \text{\# Intersection Points}\\
\hline
\text{Coincident}&\colorThree{\bm\checkmark} &  \colorThree{\bm\checkmark}  & \infty\\
\hline
\text{Parallel, Non-Coincident}&\colorThree{\bm\checkmark} &\red{\bm{\times}}  & 0\\
\hline
\text{Skew}\ [\reals^{3+}]&\red{\bm{\times}} &\red{\bm{\times}}  & 0\\
\hline
\text{Intersecting}&\red{\bm{\times}} & \colorThree{\bm\checkmark}   & 1\\
\hline
\end{array}CoincidentParallel, Non-CoincidentSkew [R3+]Intersecting​Parallel✓✓××​Intersect✓××✓​# Intersection Points∞001​​
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Determining Collinearity
Two lines are parallel if they have:
Collinear direction vectors
Collinear normal vectors
The direction vector of one line is orthogonal to the normal vector of the other 
Determining Intersections
To determine whether two lines intersect:
Make sure at least one line is in parametric form (convert if needed)
Substitute the parametric equations into the equation of the other line
Solve for the parameter
Determine the number of intersections: 
If you don't get a consistent result when solving for the parameter (e.g. 2=52=52=5)
  ⟹  \implies⟹no intersection (or skewed in R3+\reals^{3+}R3+)
If you get a single valid solution set (e.g. t=1, s=0t=1, \ s=0t=1, s=0)
  ⟹  \implies⟹one point of intersection
If you only get redundant equations (e.g. 0=00=00=0)
  ⟹  \implies⟹infinitely many points of intersection

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Example: Intersection of Two Lines (One POI)
Find the point(s) of intersection between the lines:
l1: ⟨x,y⟩=⟨2,1⟩+t⟨−1,3⟩l_1:\ \lang x,y\rang = \lang 2,1\rang +t\lang -1,3\rangl1​: ⟨x,y⟩=⟨2,1⟩+t⟨−1,3⟩
l2: x+2y=14l_2: \ x+2y=14l2​: x+2y=14

The direction vector of l1l_1l1​ is ⟨−1,3⟩\lang -1,3\rang⟨−1,3⟩.
The normal vector of l2l_2l2​ is ⟨1,2⟩\lang 1,2\rang⟨1,2⟩.
These are not orthogonal (dot product is not 0), meaning the lines are not parallel. 
Since these lines are in R2\reals^2R2, they must intersect at exactly one point (skew lines are only possible in R3+\reals^{3+}R3+).

Writing l1l_1l1​ in parametric form:   l1: {x=2−ty=1+3tl_1:\ 
\begin{cases}
x=2-t\\
y=1+3t
\end{cases}l1​: {x=2−ty=1+3t​ 
  
We can substitute these expressions into the standard form of the second equation:
 x+2y=14(2−t)+2(1+3t)=144+5t=145t=10t=2\begin{aligned}
x + 2y &=14\\
(2−t)+2(1+3t)&=14\\
4+5t &= 14\\
5t &= 10\\
t&=\colorOne{2}
\end{aligned}x+2y(2−t)+2(1+3t)4+5t5tt​=14=14=14=10=2​
Now we can substitute t=2t=2t=2  into the parametric equations of l1l_1l1​ to find the point of intersection:
⟨2,1⟩+2⟨−1,3⟩=⟨0,7⟩\lang2,1\rang+\colorOne{2}\lang −1,3 \rang=\lang 0,7\rang⟨2,1⟩+2⟨−1,3⟩=⟨0,7⟩

This position vector corresponds to the point of intersection (0,7)\left(0,7\right)(0,7).

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Example: Intersection of Two Lines (no POI)
Find the point(s) of intersection of the lines:
l1: x⃗=⟨0,−1,5⟩+t⟨2,1,−3⟩l_1:\ \vec x = \lang 0,−1,5\rang + t\lang 2,1,-3\rangl1​: x=⟨0,−1,5⟩+t⟨2,1,−3⟩ 
l2: x⃗=⟨2,0,5⟩+s⟨−4,−2,6⟩l_2:\ \vec x = \lang 2,0,5\rang +s\lang -4,-2,6\rangl2​: x=⟨2,0,5⟩+s⟨−4,−2,6⟩

The direction vectors are d⃗1=⟨2,1,−3⟩\vec d_1 = \lang 2,1,-3\rangd1​=⟨2,1,−3⟩  and  d⃗2=⟨−4,−2,6⟩\vec d_2 = \lang -4,-2,6\rangd2​=⟨−4,−2,6⟩.
Since d⃗2=−2d⃗1\vec d_2 = -2\vec d_1d2​=−2d1​, the lines are parallel.

Now check if the lines are coincident by seeing if they share a point.

(2,0,5)(\colorOne{2},
\colorTwo{0},
\colorThree{5})(2,0,5) is a point on l2l_2l2​, so let's see if it's also on l1l_1l1​ (which we convert to parametric form):
l1:{x=2ty=−1+tz=5−3t  →  {2=2t0=−1+t5=5−3tl_1:\begin{cases}
x=2t\\
y=-1+t\\
z=5-3t
\end{cases}
\space \space \to \space\space 
\begin{cases}
\colorOne{2}=2t\\
\colorTwo{0}=-1+t\\
\colorThree{5}=5-3t
\end{cases}
l1​:⎩⎨⎧​x=2ty=−1+tz=5−3t​  →  ⎩⎨⎧​2=2t0=−1+t5=5−3t​
By solving this set of equations, we find that t=1t=1t=1 for the first two equations but not for the third. Inconsistent.
Thus, (2,0,5)\left(2,0,5\right)(2,0,5) is a point on l2l_2l2​ but not on l1l_1l1​, so they are not the same line.
Therefore, the lines are parallel and non-coincident, so there are no points of intersection.

Alternatively
We can jump straight into solving the set of parametric equations:
l1:{x=2ty=−1+tz=5−3tl2:{x=2−4sy=−2sz=5+6sl_1:\begin{cases}
x=2t\\y=-1+t\\z=5-3t
\end{cases} \qquad

l_2:\begin{cases}
x=2-4s\\y=-2s\\z=5+6s
\end{cases}l1​:⎩⎨⎧​x=2ty=−1+tz=5−3t​l2​:⎩⎨⎧​x=2−4sy=−2sz=5+6s​
Setting them to be equal to each other:
{2t=2−4s−1+t=−2s5−3t=5+6t\begin{cases}
2t=2-4s\\-1+t=-2s\\5-3t=5+6t
\end{cases}⎩⎨⎧​2t=2−4s−1+t=−2s5−3t=5+6t​
Solving the first equation for ttt yields: 
t=1−2s(∗)t=1-2s \quad (*)t=1−2s(∗) 
Substituting (∗)(*)(∗) into the second equation:
−1+(1−2s)=−2s→−2s=−2s-1+\left(1-2s\right)=-2s\quad \to \quad -2s=-2s−1+(1−2s)=−2s→−2s=−2s
This is a redundant equation (always true), so it does not provide any information.
Substituting (∗)(*)(∗) into the third equation:
5−3(1−2s)=5+6s2+6s=5+6s2=5\begin{aligned}
5-3(1-2s) &= 5+6s\\
2+6s&=5+6s\\
2&=5
\end{aligned}5−3(1−2s)2+6s2​=5+6s=5+6s=5​
We get an inconsistent result because there is no point where the two lines intersect.

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Example: Intersection of Two Lines (Infinite POIs)
Find the point(s) of intersection of the lines:
l1: x⃗=⟨1,−3⟩+t⟨−1,2⟩l_1:\ \vec x = \lang 1,-3\rang +t\lang -1,2\rangl1​: x=⟨1,−3⟩+t⟨−1,2⟩
l2: x⃗=⟨−1,1⟩+s⟨3,−6⟩l_2:\ \vec x = \lang -1,1\rang +s\lang 3,-6 \rangl2​: x=⟨−1,1⟩+s⟨3,−6⟩

The direction vectors are d⃗1=⟨−1,2⟩\vec d_1 = \lang -1,2\rangd1​=⟨−1,2⟩ and  d⃗2=⟨3,−6⟩\vec d_2 = \lang 3,-6\rangd2​=⟨3,−6⟩.
Since d⃗2=−3d⃗1\vec d_2 = -3\vec d _1d2​=−3d1​, the lines are parallel.
Now check if the lines are coincident by seeing if they share a point.
(−1,1)(\colorOne{-1},
\colorTwo{1})(−1,1) is a point on l2l_2l2​, so let's see if it's also on l1l_1l1​ (which we convert to parametric form):
l1:{x=1−ty=−3+2t→{−1=1−t1=−3+2tl_1:\begin{cases}
x=1-t\\
y=-3+2t
\end{cases}

\quad \rightarrow \quad

\begin{cases}
\colorOne{-1}=1-t\\
\colorTwo{1}=-3+2t
\end{cases}l1​:{x=1−ty=−3+2t​→{−1=1−t1=−3+2t​

We see that t=2t=2t=2 is the solution to both equations, so the point (−1,1)\left(-1,1\right)(−1,1) is also on l1l_1l1​.
Since the lines are parallel and share a point, they coincide (same line) and have infinitely many points of intersection.
Alternatively
We can jump straight into solving the parametric equations:
l1:{x=1−ty=−3+2tl2:{x=−1+3sy=1−6sl_1:\begin{cases}
x=1-t\\
y=-3+2t
\end{cases}
\qquad 
l_2:\begin{cases}
x=-1+3s\\
y=1-6s
\end{cases}l1​:{x=1−ty=−3+2t​l2​:{x=−1+3sy=1−6s​
Setting these equations equal, we get
1−t=−1+3s−3+2t=1−6s\begin{aligned}
1-t&=-1+3s\\
-3+2t&=1-6s
\end{aligned}1−t−3+2t​=−1+3s=1−6s​
Solving for ttt in the first equation:
 t=2−3s(∗)t=2-3s \quad (*)t=2−3s(∗)
Substitute (∗)(*)(∗) into the second equation: 
−3+2(2−3s)=1−6s1−6s=1−6s0=0\begin{aligned}
-3+2\left(2-3s\right)&=1-6s\\
1-6s&=1-6s\\
0&=0
\end{aligned}−3+2(2−3s)1−6s0​=1−6s=1−6s=0​
This is a redundant equation.
Since any value of sss satisfies the equations, there are infinitely many solutions.
Therefore, the lines are the coincident with infinitely many points of intersection.

Practice: Intersection of Two Lines
Select all of the following statements that are true given the lines: 

𝑙1:x⃗=⟨1,1,2⟩+t⟨−1,2,3⟩𝑙_1: \vec x = \lang 1,1,2\rang + t\lang −1,2,3\rang l1​:x=⟨1,1,2⟩+t⟨−1,2,3⟩
l2:x⃗=⟨0,3,5⟩+s⟨2,−4,−6⟩l_2: \vec x = \lang 0,3,5\rang  + s\lang 2, −4, −6\rangl2​:x=⟨0,3,5⟩+s⟨2,−4,−6⟩

A) l1l_1l1​ and l2l_2l2​ are parallel

B) l1l_1l1​ and l2l_2l2​ coincide

C) 𝑙1𝑙_1l1​ and 𝑙2𝑙_2l2​ are skew

D) 𝑙1𝑙_1l1​ and 𝑙2𝑙_2l2​ intersect at exactly one point

I don't know

Practice: Intersection of Two Lines
Determine the point(s) of intersection, if any, of the followings lines, or state whether the lines are parallel or skewed.

l1: x⃗=⟨1,3,4⟩+t⟨−8,16,24⟩l_1:\ \vec x = \lang 1,3,4\rang+t\lang −8,16,24\rangl1​: x=⟨1,3,4⟩+t⟨−8,16,24⟩
l2: x⃗=⟨−1,7,10⟩+s⟨2,−4,−6⟩l_2:\ \vec x = \lang −1,7,10\rang +s\lang 2,−4,−6\rang l2​: x=⟨−1,7,10⟩+s⟨2,−4,−6⟩

The lines intersect at exactly one point: (1,−2,−3)(1,-2,-3)(1,−2,−3)

The lines intersect at exactly one point: (3,−6,−9)(3,-6,-9)(3,−6,−9)

The lines coincide (infinitely many points of intersection)

The lines are parallel and non-coincident (no intersection)

The lines are skewed (no intersection but not parallel)

I don't know

Extra Practice

Practice: Intersection of lines

Practice: Intersection of lines
Determine the point(s) of intersection, if any, of the followings lines, or state whether the lines are parallel or skewed.

Practice: Intersection of lines

Practice: Intersection of lines
Determine the point(s) of intersection, if any, of the followings lines, or state whether the lines are parallel or skewed.

Consider the two lines, L_1:(1,0,0)+s(1,1,1) and L_2:(-1,2,3)+t(4,0,c).
a) For what value of c two lines intersects.
b) Find the equation of the plane containing both lines.

Consider two lines: L1:(1,0,0)+s(1,1,1) and L2:(-1,2,3)+t(4,0,c).
a) For what value of c two lines intersects.
b) Find the equation of the plane containing both lines.

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.6

 Given the planes
π1 : 4x−3y+z=2\bcb{\pi_1 \, : \, 4x - 3y + z = 2}π1​:4x−3y+z=2 and π2 : x+2y+2z=0\bcb{\pi_2 \, : \, x + 2y + 2z = 0}π2​:x+2y+2z=0, and the lines
L1 : {x=3t+2y=−2t+1z=t, L2 : {x=−2t+1y=t+1z=t+3, L3 : {x=t+3y=−tz=2t+1\bcb{L_1 \, : \, \left\{ \begin{matrix} x & = & 3t + 2 \\ y & = & -2t + 1 \\ z & = & t \end{matrix} \right., ~ 
L_2 \, : \, \left\{ \begin{matrix} x & = & -2t + 1 \\ y & = & t + 1 \\ z & = & t + 3 \end{matrix} \right., ~ 
L_3 \, : \, \left\{ \begin{matrix} x & = & t + 3 \\ y & = & -t \\ z & = & 2t +1 \end{matrix} \right.}L1​:⎩⎨⎧​xyz​===​3t+2−2t+1t​, L2​:⎩⎨⎧​xyz​===​−2t+1t+1t+3​, L3​:⎩⎨⎧​xyz​===​t+3−t2t+1​,

Practice: Intersection of lines

Practice: Intersection of lines
Which of the following statements is/are true about these lines?
L1: [1,1,2]+𝑡[−1,2,3]L_1:\ \left[1,1,2\right]+𝑡\left[−1,2,3\right]L1​: [1,1,2]+t[−1,2,3]

Which of the following statements is true about the lines
L1:x=4+ty=−1+tz=−3−2tL_1:\begin{array}{l}x=4+t\\y=-1+t\\z=-3-2t\end{array}L1​:x=4+ty=−1+tz=−3−2t​and L2:x=1+ry=2−rz=3−2rL_2:\begin{array}{l}x=1+r\\y=2-r\\z=3-2r\end{array}L2​:x=1+ry=2−rz=3−2r​

Practice: Intersection of lines

Practice: Intersection of lines
Determine the point(s) of intersection, if any, of the followings lines, or state whether the lines are parallel or skewed.

Find the value of cccif the point (a,b,c)\left(a,b,c\right)(a,b,c)is a point of intersection between the lines
L1:x=1+ty=t+2z=3−2tL_1:\begin{array}{l}x=1+t\\y=t+2\\z=3-2t\end{array}L1​:x=1+ty=t+2z=3−2t​and L2:x=−1+2ry=1+rz=4−rL_2:\begin{array}{l}x=-1+2r\\y=1+r\\z=4-r\end{array}L2​:x=−1+2ry=1+rz=4−r​

Line Intersections

Consider the lines:
l1:(x1,x2,x3,x4)=(0,−3,−1,2)+s(1,0,2,−1)l_1:(x_1,x_2,x_3,x_4)=(0,-3,-1,2)+s(1,0,2,-1)l1​:(x1​,x2​,x3​,x4​)=(0,−3,−1,2)+s(1,0,2,−1)
l2:(x1,x2,x3,x4)=(6,−4,2,0)+t(−4,1,1,0)l_2:(x_1,x_2,x_3,x_4)=(6,-4,2,0)+t(-4,1,1,0)l2​:(x1​,x2​,x3​,x4​)=(6,−4,2,0)+t(−4,1,1,0)

Wize University Linear Algebra Textbook > Equations of Lines and Planes

Intersection of Two Lines

Popular Courses

Intersection of Two Lines

Possible Intersection Scenarios

Determining Collinearity

Determining Intersections

Example: Intersection of Two Lines (One POI)

Example: Intersection of Two Lines (no POI)

Alternatively

Example: Intersection of Two Lines (Infinite POIs)

Alternatively

Practice: Intersection of Two Lines

Practice: Intersection of Two Lines

Extra Practice

Practice: Intersection of lines

Practice: Intersection of lines

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.6

Practice: Intersection of lines

Practice: Intersection of lines

Line Intersections