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 >}

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download