Površina paralelograma je
P = a b sin α , P=ab\sin\alpha, P = ab sin α , gdje je a = ∣ a ⃗ ∣ a=|\vec a| a = ∣ a ∣ b = ∣ b ⃗ ∣ b=|\vec b| b = ∣ b ∣ α = ∠ ( a ⃗ , b ⃗ ) \alpha=\angle(\vec a,\vec b) α = ∠ ( a , b )
Vrijedi da je
m ⃗ 2 = ∣ m ⃗ ∣ 2 = 1 , n ⃗ 2 = ∣ n ⃗ ∣ 2 = 1 \vec m^{\,2}=|\vec m|^2=1,\quad \vec n^{\,2}=|\vec n|^2=1 m 2 = ∣ m ∣ 2 = 1 , n 2 = ∣ n ∣ 2 = 1 i
m ⃗ ⋅ n ⃗ = ∣ m ⃗ ∣ ⋅ ∣ n ⃗ ∣ cos π 3 = 1 2 . \vec m\cdot\vec n=|\vec m|\cdot|\vec n|\cos\frac\pi3=\frac12. m ⋅ n = ∣ m ∣ ⋅ ∣ n ∣ cos 3 π = 2 1 . ∣ a ⃗ ∣ 2 = a ⃗ 2 = ( 2 m ⃗ + n ⃗ ) 2 = 4 m ⃗ 2 + 4 m ⃗ ⋅ n ⃗ + n ⃗ 2 = 4 + 2 + 1 = 7 , |\vec a|^2=\vec a^{\,2}=(2\vec m+\vec n)^2=4\vec m^{\,2}+4\vec m\cdot\vec n+\vec n^{\,2}=4+2+1=7, ∣ a ∣ 2 = a 2 = ( 2 m + n ) 2 = 4 m 2 + 4 m ⋅ n + n 2 = 4 + 2 + 1 = 7 , ∣ b ⃗ ∣ 2 = b ⃗ 2 = ( m ⃗ − n ⃗ ) 2 = m ⃗ 2 − 2 m ⃗ ⋅ n ⃗ + n ⃗ 2 = 1 − 1 + 1 = 1 , |\vec b|^2=\vec b^{\,2}=(\vec m-\vec n)^2=\vec m^{\,2}-2\vec m\cdot\vec n+\vec n^{\,2}=1-1+1=1, ∣ b ∣ 2 = b 2 = ( m − n ) 2 = m 2 − 2 m ⋅ n + n 2 = 1 − 1 + 1 = 1 , a ⃗ ⋅ b ⃗ = ( 2 m ⃗ + n ⃗ ) ( m ⃗ − n ⃗ ) = 2 m ⃗ 2 − m ⃗ ⋅ n ⃗ − n ⃗ 2 = 2 − 1 2 − 1 = 1 2 . \vec a\cdot\vec b=(2\vec m+\vec n)(\vec m-\vec n)=2\vec m^{\,2}-\vec m\cdot\vec n-\vec n^{\,2}=2-\frac12-1=\frac12. a ⋅ b = ( 2 m + n ) ( m − n ) = 2 m 2 − m ⋅ n − n 2 = 2 − 2 1 − 1 = 2 1 . cos α = a ⃗ ⋅ b ⃗ ∣ a ⃗ ∣ ⋅ ∣ b ⃗ ∣ = 1 2 7 ⋅ 1 = 1 2 7 . \cos\alpha=\frac{\vec a\cdot\vec b}{|\vec a|\cdot|\vec b|}=\frac{\frac12}{\sqrt7\cdot1}=\frac{1}{2\sqrt7}. cos α = ∣ a ∣ ⋅ ∣ b ∣ a ⋅ b = 7 ⋅ 1 2 1 = 2 7 1 . sin α = 1 − cos 2 α = 1 − 1 28 = 3 3 2 7 = 3 21 14 . \sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\frac1{28}}=\frac{3\sqrt3}{2\sqrt7}=\frac{3\sqrt{21}}{14}. sin α = 1 − cos 2 α = 1 − 28 1 = 2 7 3 3 = 14 3 21 . P = a b sin α = 7 ⋅ 1 ⋅ 3 3 2 7 P=ab\sin\alpha=\sqrt7\cdot1\cdot\frac{3\sqrt3}{2\sqrt7} P = ab sin α = 7 ⋅ 1 ⋅ 2 7 3 3 P = 3 3 2 . P=\frac{3\sqrt3}{2}. P = 2 3 3 . 