Oto kolejne zadanie z którym mam niemały problem.
Oblicz pole trójkąta o wierzchołkach
\(\displaystyle{ A(2,-1,1)}\)
\(\displaystyle{ B(4,6,2)}\)
\(\displaystyle{ C(1,0,5)}\)
z góry dzieki za pomoc
Oblicz pole trójkąta
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Oblicz pole trójkąta
\(\displaystyle{ A(2,-1,1)}\)
\(\displaystyle{ B(4,6,2)}\)
\(\displaystyle{ C(1,0,5)}\)
\(\displaystyle{ \vec{AB}=[2,7,1]}\)
\(\displaystyle{ \vec{AC}=[-1,1,4]}\)
\(\displaystyle{ \frac{1}{2}|\vec{AB}\times \vec{AC}| =\frac{1}{2} | ft[\begin{array}{ccc}i&j&k \\ 2&7&1 \\ -1&1&4 \end{array}\right] | =}\)
\(\displaystyle{ =\frac{1}{2} ft| i 7 4 + 2 1 k + (-1)\cdot j 1 - (-1) 7 k - i 1 1 - 2 j 4\right| =}\)
\(\displaystyle{ = \frac{1}{2} | 27i - 9j + 9k|=}\)
\(\displaystyle{ = \frac{1}{2} \sqrt{27^2 + (-9)^2 + 9^2} = \frac{9\sqrt{11}}{2}}\)
[ Dodano: 27 Stycznia 2008, 13:15 ]
sprawdzenie rozwiązania (II sposób)
\(\displaystyle{ \vec{AB}=[2,7,1]}\)
\(\displaystyle{ \vec{AC}=[-1,1,4]}\)
\(\displaystyle{ \vec{BC}=[-3,-6,3]}\)
\(\displaystyle{ |\vec{AB}|=\sqrt{2^2+7^2+1^2} = \sqrt{4+49+1} = \sqrt{54} = 3\sqrt{6}}\)
\(\displaystyle{ |\vec{AC}|=\sqrt{(-1)^2+1^2+4^2} = \sqrt{18} = 3\sqrt{2}}\)
\(\displaystyle{ |\vec{BC}|=\sqrt{(-3)^2+(-6)^2+3^2} = \sqrt{9+36+9} = \sqrt{54} = 3\sqrt{6}}\)
wzór Herona na pole trójkąta:
\(\displaystyle{ P_{\Delta} = \sqrt{p(p-a)(p-b)(p-c)}}\)
\(\displaystyle{ p=\frac{a+b+c}{2}}\)
\(\displaystyle{ p=\frac{3\sqrt{6}+3\sqrt{2}+3\sqrt{6}}{2} = \frac{6\sqrt{6} + 3\sqrt{2}}{2} = 3\sqrt{6} +\frac{3\sqrt{2}}{2}}\)
\(\displaystyle{ P_{\Delta} = \sqrt{(3\sqrt{6} +\frac{3\sqrt{2}}{2})(3\sqrt{6} +\frac{3\sqrt{2}}{2}-3\sqrt{6})(3\sqrt{6} +\frac{3\sqrt{2}}{2}-3\sqrt{2})(3\sqrt{6} +\frac{3\sqrt{2}}{2}-3\sqrt{6})}}\)
\(\displaystyle{ = \sqrt{(3\sqrt{6} +\frac{3\sqrt{2}}{2}) \frac{3\sqrt{2}}{2} (3\sqrt{6} - \frac{3\sqrt{2}}{2}) \frac{3\sqrt{2}}{2}} = \sqrt{(54 - \frac{9}{2}) \frac{9}{2} }= \sqrt{\frac{99}{2} \frac{9}{2}}}\)
\(\displaystyle{ =\frac{9\sqrt{11}}{2}}\)
\(\displaystyle{ B(4,6,2)}\)
\(\displaystyle{ C(1,0,5)}\)
\(\displaystyle{ \vec{AB}=[2,7,1]}\)
\(\displaystyle{ \vec{AC}=[-1,1,4]}\)
\(\displaystyle{ \frac{1}{2}|\vec{AB}\times \vec{AC}| =\frac{1}{2} | ft[\begin{array}{ccc}i&j&k \\ 2&7&1 \\ -1&1&4 \end{array}\right] | =}\)
\(\displaystyle{ =\frac{1}{2} ft| i 7 4 + 2 1 k + (-1)\cdot j 1 - (-1) 7 k - i 1 1 - 2 j 4\right| =}\)
\(\displaystyle{ = \frac{1}{2} | 27i - 9j + 9k|=}\)
\(\displaystyle{ = \frac{1}{2} \sqrt{27^2 + (-9)^2 + 9^2} = \frac{9\sqrt{11}}{2}}\)
[ Dodano: 27 Stycznia 2008, 13:15 ]
sprawdzenie rozwiązania (II sposób)
\(\displaystyle{ \vec{AB}=[2,7,1]}\)
\(\displaystyle{ \vec{AC}=[-1,1,4]}\)
\(\displaystyle{ \vec{BC}=[-3,-6,3]}\)
\(\displaystyle{ |\vec{AB}|=\sqrt{2^2+7^2+1^2} = \sqrt{4+49+1} = \sqrt{54} = 3\sqrt{6}}\)
\(\displaystyle{ |\vec{AC}|=\sqrt{(-1)^2+1^2+4^2} = \sqrt{18} = 3\sqrt{2}}\)
\(\displaystyle{ |\vec{BC}|=\sqrt{(-3)^2+(-6)^2+3^2} = \sqrt{9+36+9} = \sqrt{54} = 3\sqrt{6}}\)
wzór Herona na pole trójkąta:
\(\displaystyle{ P_{\Delta} = \sqrt{p(p-a)(p-b)(p-c)}}\)
\(\displaystyle{ p=\frac{a+b+c}{2}}\)
\(\displaystyle{ p=\frac{3\sqrt{6}+3\sqrt{2}+3\sqrt{6}}{2} = \frac{6\sqrt{6} + 3\sqrt{2}}{2} = 3\sqrt{6} +\frac{3\sqrt{2}}{2}}\)
\(\displaystyle{ P_{\Delta} = \sqrt{(3\sqrt{6} +\frac{3\sqrt{2}}{2})(3\sqrt{6} +\frac{3\sqrt{2}}{2}-3\sqrt{6})(3\sqrt{6} +\frac{3\sqrt{2}}{2}-3\sqrt{2})(3\sqrt{6} +\frac{3\sqrt{2}}{2}-3\sqrt{6})}}\)
\(\displaystyle{ = \sqrt{(3\sqrt{6} +\frac{3\sqrt{2}}{2}) \frac{3\sqrt{2}}{2} (3\sqrt{6} - \frac{3\sqrt{2}}{2}) \frac{3\sqrt{2}}{2}} = \sqrt{(54 - \frac{9}{2}) \frac{9}{2} }= \sqrt{\frac{99}{2} \frac{9}{2}}}\)
\(\displaystyle{ =\frac{9\sqrt{11}}{2}}\)