Chapter 12 Section 5 Lines and Planes in Space - Department of Mathematics
Chapter 12 Section 5 Lines and Planes in Space
Example 1
Show that the line through the points (0, 1, 1) and (1, -1, 6) is perpendicular to the line through the points (-4, 2, 1) and (-1, 6, 2) .
Vector equation for the first line: r1(t) =. < 0, 1, 1 > + t (< 1, -1, 6 > - < 0, 1, 1 >) =< 0, 1, 1 > + t < 1, -2, 5 >
Vector equation for the second line: r2(s) =. < -4, 2, 1 > + s (< -1, 6, 2 > - < -4, 2, 1 >)
=< -4, 2, 1 > + s < 3, 4, 1 >
< 1, -2, 5 > ? < 3, 4, 1 > cos =
| < 1, -2, 5 > | | < 3, 4, 1 > | (1)(3) + (-2)(4) + (5)(1)
= 12 + (-2)2 + 52 32 + 42 + 12 0
= = 0 30 26
Remark: These two lines are skew.
Example 2
(a) Find parametric equations for the line through (5, 1, 0) that is perpendicular to the plane 2x - y + z = 1
A normal vector to the plane is:
n =< 2, -1, 1 > r(t) =< 5, 1, 0 > + t < 2, -1, 1 >
(b) In what points does this line intersect the
coordinate planes?
xy-plane: 0 =. 0 + t 1
t = 0 r(0) =< 5, 1, 0 > yz-plane: 0 =. 5 + t 2
-5
-5
7 -5
t = r( ) =< 0, , >
zx-plane:
0
=.
2 1+
t
2 (-1)
22
t = 1 r(1) =< 7, 0, 1 >
Example 3
Parallelism, intersection for:
L1 : r1(t)
L2 : r2(s)
x-1 y z-1
==
2
1
4
=< 1, 0, 1 > + t < 2, 1, 4 >
=< 1 + 2t, t, 1 + 4t >
x y+2 z+2
=
=
1
2
3
=< 0, -2, -2 > + s < 1, 2, 3 >
=< s, -2 + 2s, -2 + 3s >
< 2, 1, 4 > ? < 1, 2, 3 > = < -5, -2, 3 > = 0 L1 L2
r1(t) =. r2(s)
< 1 + 2t, t, 1 + 4t > =. < s, -2 + 2s, -2 + 3s >
1 + 2t = s t = -2 + 2s
1 + 4t = -2 + 3s Solving the first two equations:
t = 0, s = 1 Checking the third equation:
1 + 4(0) = -2 + 3(1) (satisfied) Consequently: L1 L2 = {r1(0)} = {r2(1)} = {< 1, 0, 1 >}
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