Wize High School Grade 12 Calculus Textbook > Intersections of Lines & Planes

Intersection of 2 Lines in R2
Given any 2 lines in R2R^2R2, there are 3 possible scenarios:
1) The lines are parallel but do not coincide → there are no points of intersection
2) The lines coincide (are identical) → there are infinitely many points of intersection
3) The lines intersect → there is exactly one point of intersection (p.o.i.)

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Intersection of 2 Lines in R3
Given any 2 lines in R3R^3R3, there are 4 possible scenarios:
1) The lines are parallel but do not coincide → there are no points of intersection
2) The lines coincide (are identical) → there are infinitely many points of intersection
3) The lines intersect → there is exactly one point of intersection
4) The lines are skewed → the lines are not parallel but do not intersect

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Finding Points of Intersection

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Example: Intersection of Lines
Find the point(s) of intersection (if any) between the lines r1⃗=[1,0,−2]+t[1,−2,−1]\vec {r_1}=[1, 0, -2]+t[1, -2, -1]r1​​=[1,0,−2]+t[1,−2,−1] and r2⃗=[0,1,−5]+s[1,−1,3]\vec{r_2}=[0,1,-5]+s[1, -1, 3]r2​​=[0,1,−5]+s[1,−1,3].

Check direction vectors
m1⃗=[1,−2,1]\vec {m_1}=[1,-2,1]m1​​=[1,−2,1] and m2⃗=[1,−1,3]\vec {m_2}=[1,-1,3]m2​​=[1,−1,3].
The direction vectors are not scalar multiples of one another, the lines either intersect at a point or are skewed.

Find the parametric equations.
L1: x=1+ty=−2tz=−2−tL_1:\ \begin{array}{l}
x=1+t\\
y=-2t\\
z=-2-t
\end{array}L1​: x=1+ty=−2tz=−2−t​ and L2: x=sy=1−sz=−5+3sL_2:\ 
\begin{array}{l}
x=s\\
y=1-s\\
z=-5+3s
\end{array}L2​: x=sy=1−sz=−5+3s​

Set the component equations equal one another
1+t=s−2t=1−s−2−t=−5+3s\begin{array}{c}
1+t=s\\
-2t=1-s\\
-2-t=-5+3s
\end{array}1+t=s−2t=1−s−2−t=−5+3s​

Solve for the parameters
From equation 1: s=1+ts=1+ts=1+t
Sub this into equation 2: −2t=1−(1+t)-2t=1-(1+t)−2t=1−(1+t) → −2t=−t-2t=-t−2t=−t → 0=t0=t0=t
Sub this back into equation 1: s=1+0=1s=1+0=1s=1+0=1
Now, sub t=0, s=1t=0,\ s=1t=0, s=1 into equation 3 to confirm this pair of parameter values: −2−(0)=−5+3(1)-2-(0)=-5+3(1)−2−(0)=−5+3(1) → −2=−2-2=-2−2=−2

Since we get a single value pair for the parameters ttt and sss, the lines intersect at exactly one point.
Sub t=0t=0t=0 into line 1: x=1x=1x=1, y=0y=0y=0, and z=−2z=-2z=−2

Therefore, the lines intersect at the point (1, 0, −2)(1,\ 0,\ -2)(1, 0, −2)

Practice: Intersection of Lines
Find the point(s) of intersection (if any) between the lines r1⃗=[3, −2, −1]+t[1, 1, 1]\vec{r_1}=\left[3,\ -2,\ -1\right]+t\left[1,\ 1,\ 1\right]r1​​=[3, −2, −1]+t[1, 1, 1] and x−22=y+1=z\frac{x-2}{2}=y+1=z2x−2​=y+1=z.

Enter the coordinates of the point of intersection in the form

(x, y, z)

or enter "None" or "Infinitely many"

I don't know

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Example: Intersection of Identical Lines
Find the point(s) of intersection (if any) between the lines r1⃗=[1,0,−2]+t[2,−2,6]\vec {r_1}=[1, 0, -2]+t[2, -2, 6]r1​​=[1,0,−2]+t[2,−2,6] and r2⃗=[0,1,−5]+s[1,−1,3]\vec{r_2}=[0,1,-5]+s[1, -1, 3]r2​​=[0,1,−5]+s[1,−1,3].

Check direction vectors
m1⃗=[2,−2,6]\vec {m_1}=[2,-2,6]m1​​=[2,−2,6] and m2⃗=[1,−1,3]\vec {m_2}=[1, -1, 3]m2​​=[1,−1,3].
Since these are scalar multiples of one another, the lines are either parallel and don't intersect, or they coincide.

Do the lines share a point?
The point (1,0,−2)(1,0,-2)(1,0,−2) is on line 1, sub this into the parametric equations of line 2:
L2: 1=s0=1−s−2=−5+3sL_2:\ 
\begin{array}{l}
1=s\\
0=1-s\\
-2=-5+3s
\end{array}L2​: 1=s0=1−s−2=−5+3s​

Solving these equations, we see that the parameter s=1s=1s=1.

So, the point (1,0,−2)(1,0,-2)(1,0,−2) is also on line 2.

Therefore, the lines coincide and there are infinitely many points of intersection.

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Example: Intersection of Parallel Lines
Find the point(s) of intersection (if any) between the lines r1⃗=[3,1,1]+t[2,−2,6]\vec {r_1}=[3, 1, 1]+t[2, -2, 6]r1​​=[3,1,1]+t[2,−2,6] and r2⃗=[0,1,−5]+s[1,−1,3]\vec{r_2}=[0,1,-5]+s[1, -1, 3]r2​​=[0,1,−5]+s[1,−1,3].

Check direction vectors
m1⃗=[2,−2,6]\vec {m_1}=[2,-2,6]m1​​=[2,−2,6] and m2⃗=[1,−1,3]\vec {m_2}=[1,-1,3]m2​​=[1,−1,3].
Since these are scalar multiples of one another, the lines are either parallel and don't intersection, or the lines coincide.

Do the lines share a point?
The point (3,1,1)(3,1,1)(3,1,1) is on line 1, sub this into the parametric equations of line 2:
L2: 3=s1=1−s1=−5+3sL_2:\ 
\begin{array}{l}
3=s\\
1=1-s\\
1=-5+3s
\end{array}L2​: 3=s1=1−s1=−5+3s​

Solving these equations, we get different values for the parameter sss for each equation.

Therefore, the lines are parallel and do not intersect.

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Example: Intersection of Skewed Lines
Find the point(s) of intersection (if any) between the lines r1⃗=[3,1,1]+t[2,−2,6]\vec {r_1}=[3, 1, 1]+t[2, -2, 6]r1​​=[3,1,1]+t[2,−2,6] and r2⃗=[0,1,−5]+s[1,−2,3]\vec{r_2}=[0,1,-5]+s[1, -2, 3]r2​​=[0,1,−5]+s[1,−2,3].

Check direction vectors
m1⃗=[2,−2,6]\vec {m_1}=[2,-2,6]m1​​=[2,−2,6] and m2⃗=[1,−2,3]\vec {m_2}=[1,-2,3]m2​​=[1,−2,3].
The direction vectors are not scalar multiples of one another, the lines either intersect at a point or are skewed.

Find the parametric equations.
L1: x=3+2ty=1−2tz=1+6tL_1:\ \begin{array}{l}
x=3+2t\\
y=1-2t\\
z=1+6t
\end{array}L1​: x=3+2ty=1−2tz=1+6t​ and L2: x=sy=1−2sz=−5+3sL_2:\ 
\begin{array}{l}
x=s\\
y=1-2s\\
z=-5+3s
\end{array}L2​: x=sy=1−2sz=−5+3s​

Set the component equations equal one another
3+2t=s1−2t=1−2s1+6t=−5+3s\begin{array}{c}
3+2t=s\\
1-2t=1-2s\\
1+6t=-5+3s
\end{array}3+2t=s1−2t=1−2s1+6t=−5+3s​

Solve for the parameters
From equation 1: s=3+2ts=3+2ts=3+2t
Sub this into equation 2: 1−2t=1−2(3+2t)1-2t=1-2(3+2t)1−2t=1−2(3+2t) → 1−2t=−5−4t1-2t=-5-4t1−2t=−5−4t → t=−3t=-3t=−3
Sub this back into equation 1: s=3+2(−3)=−3s=3+2(-3)=-3s=3+2(−3)=−3
Sub this value of t=−3t=-3t=−3 and s=−3s=-3s=−3 into equation 3: 1+6(−3)=−5+3(−3)1+6(-3)=-5+3(-3)1+6(−3)=−5+3(−3) → −17=−14-17=-14−17=−14

Since we get an equation with no solutions, it means that our lines are skewed and do not intersect.

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]

L2: [0,3,5]+s[2,−4,−6]L_2:\ \left[0,3,5\right]+s\left[2,−4,−6\right]L2​: [0,3,5]+s[2,−4,−6]

i. L1L_1L1​ and L2L_2L2​ are parallel
ii. L1L_1L1​ and L2L_2L2​ coincide
iii. L1L_1L1​ and L2L_2L2​ are skew
iv. L1L_1L1​ and L2L_2L2​ intersect at exactly one point

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.

a) 𝑙1: [1,3,4]+𝑡[−1,2,3]𝑙_1:\ \left[1,3,4\right]+𝑡\left[−1,2,3\right]l1​: [1,3,4]+t[−1,2,3]

    𝑙2: [−1,7,10]+𝑠[2,−4,−6]𝑙_2:\ \left[−1,7,10\right]+𝑠\left[2,−4,−6\right]l2​: [−1,7,10]+s[2,−4,−6]

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

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

The lines coincide and have infinitely many points of intersection

The lines are parallel and do not intersect

The lines are skewed and do not intersect

I don't know