This preview shows page 1. Sign up to view the full content.
Section 8.8
1. The given line has equations
x
=
3 +
t
(13

3) = 3 + 10
t,
y
=

2 +
t
(3 + 2) =

2 + 5
t,
z
=
7 +
t
(

8

7) = 7

15
t.
The line meets the plane
y
= 0 in the point (
x,
0
, z
), where 0 =

2+5
t
, or
t
= 2
/
5. The corresponding values for
x
and
z
are 7 and 1, respectively.
2.
E
=
1
2
(
B
+
C
),
F
= (1

t
)
A
+
t
E
, where
t
=
AF
AE
=
AF
AF
+
FE
=
AF/FE
(
AF/FE
) + 1
=
2
3
.
Hence
F
=
1
3
A
+
2
3
µ
1
2
(
B
+
C
)
¶
=
1
3
A
+
1
3
(
B
+
C
)
=
1
3
(
A
+
B
+
C
)
.
3. Let
A
= (2
,
1
,
4)
, B
= (1
,

1
,
2)
, C
= (3
,
3
,
6). Then we prove

AC
=
t

AB
for some real
t
. We have

AC
=
1
2
2
,

AB
=

1

2

2
.
Hence

AC
= (

1)

AB
and consequently
C
is on the line
This is the end of the preview. Sign up
to
access the rest of the document.
Unformatted text preview: AB . In fact A is between C and B , with AC = AB . 4. The points P on the line AB which satisfy AP = 2 5 PB are given by P = A + tAB , where  t/ (1t )  = 2 / 5. Hence t/ (1t ) = ± 2 / 5. The equation t/ (1t ) = 2 / 5 gives t = 2 / 7 and hence P = 2 31 + 2 7 1 4 5 = 16 / 7 29 / 7 3 / 7 . 87...
View
Full
Document
This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.
 Fall '10
 Dreibelbis

Click to edit the document details