Lines and planes
Intersection of lines - suggested problems - solutions
For each of the pairs of lines below
(a) Rewrite the given equations in parametric form, if necessary.
(b) Graph both lines on the same coordinate axes in MVT, and determine by inspection
whether the lines appear to be intersecting, parallel, skew, or the same line.
(c) Solve for point of intersection (or show algebraically that there is none). Remember to
use a different parameter for the second line when solving.
(d) Examine direction vectors and use along with your point of intersection results to determine parallel/skew/intersecting/same.
P1: L1 : < x, y, z >=< 4, 0, −1 > +t < 2, 3, 1 >
L2 : x = 2t + 2, y = 2t + 3, z = t + 1
Parametric:
x = 2t + 4
y = 3t
z = t−1
x = 2s + 2
y = 2s + 3
z = s+1
Graph attached at end; the lines appear to be skew.
Equate:
2t + 4 = 2s + 2
3t = 2s + 3
t−1 = s+1
→
−2s + 2t = −2
−2s + 3t = 3
−s + t = 2
Solve the first two:
−2s + 2t
−(−2s + 3t
−t
t
=
=
=
=
−2
3)
−5
5
−2s + 2(5) = −2
−2s = −12
s = 6
And check into the third: −s + t = 2, so −6 + 5 = 2. But −1 6= 2, so no solution, and
the lines do not intersect.
Direction vector for L1 is v1 =< 2, 3, 1 > and for L2 is v2 =< 2, 2, 1 >. Since v1 6= cv2
(the vectors are not scalar multiples of each other), the lines are not parallel.
Non-intersecting, non-parallel lines are skew.
P2: L1 :
L2 :
x = −3t + 6, y = 2t − 2, z = 4t + 5
x = y − 2 = z − 13
6
−4
−8
Parametric:
x = −3t + 6
y = 2t − 2
z = 4t + 5
x = 6s
y = −4s + 2
z = −8s + 13
Graph attached at end; the lines appear to be the same line.
Equate:
−3t + 6 = 6s
2t − 2 = −4s + 2
4t + 5 = −8s + 13
→
6s + 3t = 6
4s + 2t = 4
8s + 4t = 8
Solve the first two:
1 (6s + 3t = 6)
3
− 12 (4s + 2t = 4)
0 = 0
That indicates infinitely many solutions for the first two, so they are the same line, but
you still need to check against the third. The easiest way to see it is to simply reduce all
the equations:
1 (6s + 3t = 6)
3
1 (4s + 2t = 4)
2
1 (8s + 4t = 8)
4
→
2s + t = 2
2s + t = 2
2s + t = 2
These are all the same equation, so the system has infinitely many solutions for s and t,
implying that the original L1 and L2 are both parameterizations of the same line.
P3: L1 :
L2 :
x = y−2 =z+1
3
−1
x = 4t + 1, y = t − 2, z = −3t − 3
Parametric:
x = 3t
y = −t + 2
z = t−1
x = 4s + 1
y = s−2
z = −3s − 3
Graph attached at end; the lines appear to be skew.
Equate:
3t = 4s + 1
−t + 2 = s − 2
t − 1 = −3s − 3
→
−4s + 3t = 1
s+t = 4
3s + t = −2
Solve the first two:
−4s + 3t
+4(s + t
7t
t
=
=
=
=
1
4)
17
17
7
s+t = 4
s + 17
7 = 4
s = 11
7
11
50
And check into the third: 3s + t = −2, so 3 7 + 17
7 = −2. But 7 6= −2, so no
solution, and the lines do not intersect.
Direction vector for L1 is v1 =< 3, −1, 1 > and for L2 is v2 =< 4, 1, −3 >. Since v1 6= cv2
(the vectors are not scalar multiples of each other), the lines are not parallel.
Non-intersecting, non-parallel lines are skew.
P4: L1 :
L2 :
x = −3t + 6, y = 2t − 2, z = 4t + 5
x = 10 − 6t, y = 3 + 4t, z = 7 + 8t
Parametric:
x = −3t + 6
y = 2t − 2
z = 4t + 5
x = −6s + 10
y = 4s + 3
z = 8s + 7
Graph attached at end; the lines appear to be parallel.
Equate:
−3t + 6 = −6s + 10
2t − 2 = 4s + 3
4t + 5 = 8s + 7
Solve the first two:
→
6s − 3t = 4
−4s + 2t = 5
−8s + 4t = 2
1 (6s − 3t = 4)
3
1 (−4s + 2t = 5)
2
0 + 0 = = 23
6
We’ve already hit no solution, so no reason to check into the third. The lines do not
intersect.
Direction vector for L1 is v1 =< −3, 2, 4 > and for L2 is v2 =< −6, 4, 8 >. Since
v1 = 21 v2 (the vectors are scalar multiples of each other), the lines have the same direction. Since they do not intersect, they must be parallel.
P5: L1 : x = −1 + t, y = 4 + 2t, z = 5 − t
L2 :< x, y, z >=< −3, 7, 10 > +t < −2, 3, 5 >
Parametric:
x = t−1
y = 2t + 4
z = −t + 5
x = −2s − 3
y = 3s + 7
z = 5s + 10
Graph attached at end; the lines appear to intersect.
Equate:
t − 1 = −2s − 3
2t + 4 = 3s + 7
−t + 5 = 5s + 10
→
2s + t = −2
−3s + 2t = 3
−5s − t = 5
Solve the first two:
3(2s + t
2(−3s + 2t
7t
t
=
=
=
=
−2)
3)
0
0
2s + t = −2
2s + 0 = −2
s = −1
And check into the third: −5s − t = 5, so −5(−1) − (0) = 5. And 5 = 5, so this system
is consistent, and there is a solution.
Plugging s = −1 or t = 0 into either of the parameterizations produces the point of
intersection (−1, 4, 5).
Finally, I’ve shown the algebraic solve for each of the systems above. You may prefer to work in
matrix form and row reduce (using either your calculator, or SciLab or some similar program).
I’ll show the matrices for each of the preceding systems:
P1:
−2s + 2t = −2
−2s + 3t = 3
−s + t = 2
-->A = [-2 2 -2; -2 3 3; -1 1 2]
A =
- 2.
- 2.
- 1.
2.
3.
1.
- 2.
3.
2.
-->rref(A)
ans =
1.
0.
0.
0.
1.
0.
0.
0.
1.
Be careful when examining the row reduced form - you might be used to looking for the
identity matrix as indicating a solution, but watch out for where the equals sign lies:
1.
0.
0.
0.
1.
0.
|
|
|
0.
0.
1.
That last line is a 0 = 1, indicating no solution.
P2:
6s + 3t = 6
4s + 2t = 4
8s + 4t = 8
-->A = [6 3 6; 4 2 4; 8 4 8]
A =
6.
4.
8.
3.
2.
4.
6.
4.
8.
-->rref(A)
ans =
1.
0.
0.
0.5
0.
0.
|
|
|
1.
0.
0.
Infinitely many solutions, in the form s = s, t = 1 − .5s.
P3:
−4s + 3t = 1
s+t = 4
3s + t = −2
-->A = [-4 3 1; 1 1 4; 3 1 -2]
A =
- 4.
1.
3.
3.
1.
1.
1.
4.
- 2.
-->rref(A)
ans =
1.
0.
0.
0.
1.
0.
No solution.
|
|
|
0.
0.
1.
P4:
6s − 3t = 4
−4s + 2t = 5
−8s + 4t = 2
-->A = [6 -3 4; -4 2 5; -8 4 2]
A =
6.
- 4.
- 8.
- 3.
2.
4.
4.
5.
2.
-->rref(A)
ans =
1.
0.
0.
- 0.5
0.
0.
|
|
|
0.
1.
0.
No solution (the 0 = 1 in the second row).
P5: -->A = [2 1 -2; -3 2 3; -5 -1 5]
A =
2.
- 3.
- 5.
1.
2.
- 1.
- 2.
3.
5.
-->rref(A)
ans =
1.
0.
0.
0.
1.
0.
- 1.
0.
0.
Solution is s = −1, t = 0.
Plots for suggested problems: Intersection of lines