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=
1
6

(

A
1
A
2
×

A
1
A
3
)
·

A
1
A
4

.
19. We have

CB
= [1
,
4
,

1]
t
,

CD
= [

3
,
3
,
0]
t
,

AD
= [3
,
0
,
3]
t
. Hence

CB
×

CD
= 3
i
+ 3
j
+ 15
k
,
so the vector
i
+
j
+ 5
k
is perpendicular to the plane
BCD
.
Now the plane
BCD
has equation
x
+
y
+5
z
= 9, as
B
= (2
,
2
,
1) is on
the plane.
Also the line through
A
normal to plane
BCD
has equation
x
y
z
=
1
1
5
+
t
1
1
5
= (1 +
t
)
1
1
5
.
Hence
x
= 1 +
t, y
= 1 +
t, z
= 5(1 +
t
).
[We remark that this line meets plane
BCD
in a point
E
which is given
by a value of
t
found by solving
(1 +
t
) + (1 +
t
) + 5(5 + 5
t
) = 9
.
So
t
=

2
/
3 and
E
= (1
/
3
,
1
/
3
,
5
/
3).]
The distance from
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Unformatted text preview: A to plane BCD is  1 × 1 + 1 × 1 + 5 × 59  1 2 + 1 2 + 5 2 = 18 √ 27 = 2 √ 3 . To Fnd the distance between lines AD and BC , we Frst note that (a) The equation of AD is P = 1 1 5 + t 3 3 = 1 + 3 t 1 5 + 3 t ; (b) The equation of BC is Q = 2 2 1 + s 1 41 = 2 + s 2 + 4 s 1s . 97...
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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

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