Mathe HA 29 Juni

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