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September 2, 2009
NAME____________________________________________
Find an equation for the plane that has
x
intercept
X
(5
;
0
;
0)
,
y
intercept
Y
(0
;
3
;
0)
, and
z
intercept
Z
(0
;
0
;
8)
.
Solution 1 (using the cross product):
The vectors
XY
=
5
;
3
;
0
i
and
XZ
=
5
;
0
;
8
i
are both parallel to the plane in question. Thus a normal
vector to this plane is
n
=
XY
±
XZ
=
i
j
k
5
3 0
5
0
8
=
24
i
+ 40
j
15
k
.
An equation for the plane is thus
24 (
x
5) + 40 (
y
0)
15 (
z
0) = 0
which we can also write as
24 (
x
5)
40 (
y
0) + 15 (
z
0) = 0
or as
24
x
40
y
+ 15
z
= 120
.
Solution 2 (not using the cross product):
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Unformatted text preview: We know that the plane will have an equation of the form a ( x & 5) + b ( y & 0) + c ( z & 0) = 0 or ax + by + cz = 5 a . Since the point Y (0 ; & 3 ; 0) must be in this plane, we must have & 3 b = 5 a . Since the point Z (0 ; ; 8) must be in this plane, we must have 8 c = 5 a . Thus b = & 5 3 a and c = 5 8 a . If we choose a = 24 , we obtain b = & 40 and c = 15 : We thus obtain the same solution as given in Solution 1. 1...
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This note was uploaded on 02/05/2012 for the course MATH 2203 taught by Professor Ellermeyer during the Fall '10 term at FIU.
 Fall '10
 Ellermeyer
 Vectors

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