Mathe HA 29 Juni (PDF)




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Mathe Hausaufgabe
Tilmann Hart
29. Juni 2011

1 Buch Seite 265 Nr. 9
a)





1
~a =  3 
1

cos(ϕ) = √

12


5
~b =  0 
3

1·5+3·0+1·3
5+0+3
√ ≈ 0.4136701

=√
2
2
2
2
2
+3 +1 · 5 +0 +3
11 · 34
=⇒ ϕ = 65, 56◦

b)



1
~a =  3 
5

cos(ϕ) = √

12




5
~b =  3 
1

5+9+5
19
1·5+3·3+5·1

√ =
=√
≈ 0, 5428571
2
2
2
2
2
35
+3 +5 · 5 +3 +1
35 · 35
=⇒ ϕ = 57, 12◦

1

c)



−11
~a =  4 
1




1
~b =  2 
3

0
(−11) · 1 + 4 · 2 + 1 · 3


=p
=0
cos(ϕ) = p
(−11)2 + 42 + 12 · 12 + 22 + 32
(−11)2 + 42 + 12 · 12 + 22 + 32
=⇒ ϕ = 90◦

2 Seite 265 Nr. 11
b)



2
−→  
0
AB =
4
cos(α) = √

22




0
−→ 
−4 
AC =
4

2 · 0 + 0 · −4 + 4 · 4
16
√ ≈ 0.6324555
p
=√
2
2
2
2
2
20 · 32
+ 0 + 4 · 0 + (−4) + 4
=⇒ α = 50, 77◦



−2
−→ 
0 
BA =
−4



−2
−−→ 
−4 
BC =
0

1
(−2) · (−2) + 0 · −4 + (−4) · 0
4
4
p
√ =
cos(β) = p
=√
= = 0, 2
20
5
20 · 20
(−2)2 + 02 + (−4)2 · (−2)2 + (−4)2 + 02
=⇒ β = 78, 46◦




2
−→  
0
CA =
4




0
−→ 
−4 
AC =
4

2

γ = 180◦ − α − β = 180◦ − 50, 77◦ − 78, 46◦ = 50, 77◦
=⇒ Dreieck ABC ist gleichschenklig.
Länge:



22 + 02 + 42 = 20 ≈ 4, 472 LE
p

AC = 02 + (−4)2 + 42 = 32 ≈ 5, 857 LE
p

BC = AC = 02 + (−4)2 + 42 = 32 ≈ 5, 857 LE
AB =

3 Seite 265 Nr. 13
b)
∆DCB :



−1
−7
−−→ 
−−→ 
−7, 5 
−4 
DC =
DB =
1
−3


(−1) · (−7) + (−7, 5) · (−4) + 1 · (−3)
34
p
=p
cos(β) = p
√ = 0, 53865
(−1)2 + (−7, 5)2 + 12 · (−7)2 + (−4)2 + (−3)2
2 + 7, 52 · 74
=⇒ β = 57, 41◦

4 Seite 266 Nr. 15
∆AM B

M(0|0|0)

A(4| − 2, 5| − 1, 5)



4
−−→ 
−2, 5 
MA =
−1, 5

B(4|2, 5| − 1, 5)



4
−−→ 
2, 5 
MB =
−1, 5

4 · 4 + (−2, 5) · 2, 5 + (−1, 5) · (−1, 5)
7, 5
p

cos(ϕ) = p
=√
= 0, 306122
2
2
2
2
2
2
24, 5 · 24, 5
4 + (−2, 5) + (−1, 5) · 4 + 2, 5 + (−1, 5)
=⇒ β = 72, 17◦

3






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