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Quiz #1
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#1(3 pt's) Find the parametric equations of the line that contains the point (1,2,3) and is parallel to
x  2 z+1

, y  4.
the line with symmetric equations
5
3
 x  1  5t
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M  (5,0,3) is parallel to the line. So the parametric equations are:  y  2
 z  3  3t
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#2(4 pt’s). Let x = (3, 4, –1) and y = (–5, 2, 1).
a.
Find x  32  42  (1)2  26 .
b.
8
 xy 
 12 16 4 
(3, 4, 1)  
,
, 
Find projx y  
x 
26
 xx 
 13 13 13 
 x  1  2sin t
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#3(4 pts) Describe the curve given parametrically by  y  3  sin t , 0  t  2 . (You need not
 z  2  3sin t
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graph the curve). Be as explicit as possible. Describe the shape of the curve and how it is
traced out as t goes from 0 to 2 .
This is a line segment. Since 1  sin t  1, this will be the line segment from
( x, y, z )  (1  2, 3  1, 2  3)  (1, 4,5) to the point (3, 2, 1) . It is traced out from (1, 3, 2) to
(3, 2,1) then back through (1, 3, 2) to (1, 4,5) and finally back to the point (1, 3, 2)
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#4(4 pt’s) Let u , v , w  3 . Determine whether each statement is true or false. If true, prove it. If
false, give an example that demonstrates it is false.
i.
If u  0 and u w  u v , then w  v
False:
Let u  (1,0,0), w  (3,1,1)and v  (3, 211,190). Then u w  3  u v , but w  v .
ii.
u v  u v
False: If u  iˆ and v  ˆj, then u  v  0 and u v  1. Note that this is true when the vectors are
parallel.
#5(5 pt's) Let x  (5, a,3a) and y  (10, 4, 6a)
a. Find all values of a such that x and y orthogonal.
We know that if x and y orthogonal, then
x  y  50  4a  18a 2  0  9a 2  2a  25  0  no such a exists.
b. Find all values of a such that x and y parallel.
If x and y parallel, then y is a scalar multiple of x. Looking at the first component, y  2x .
Then, 2a  4 and  2(3a)  6a. Therefore, a  2 .
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#6 (3 pt’s) Prove that the midpoints of a quadrilateral in
3
form a parallelogram.
Proof. Let ABCD be the quadrilateral as shown. (Note that is not necessary that the four
vertices lie in the same plane).
Let E , F , G and H be the midpoints of the sides ad pictured.
A
E
B
H
F
D
G
C
To show that quadrilateral EFGH is a parallelogram, it suffices to show that
EF  HG and FG  EH .
1
1
AB  BC =
AD  DC and it follows that
We note that: AB  BC = AD  DC . Then,
2
2
EB  BF = HD  DG . Observing the figure, this gives us immediately that EF  HG .
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
The proof that FG  EH is similar starting with BC  CD = BA  AD
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