zad.1
Dane są punkty A(-2,2)B(1,1)C((4,2). Oblicz \(\displaystyle{ \vec{AB} o \vec{AC} , \vec{AB} o \vec{BC} , \vec{BC} o \vec{AC}}\)
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Morgus
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\(\displaystyle{ |AB|=\sqrt{(1+2)^{2}+(1-2)^{2}}=\sqrt{9+1}=\sqrt{10}}\)
\(\displaystyle{ |AC|=\sqrt{(4+2)^{2}+(2-2)^{2}}=\sqrt{6^{2}}=6}\)
\(\displaystyle{ cos\sphericalangle ( \vec{AB}, \vec{AC})= \frac{1+2}{\sqrt{3^{2}+1^{2}}}= \frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{10}}\)
\(\displaystyle{ \vec{AB}o \vec{AC}=|AB| |AC| cos ( \vec{AB}, \vec{AC})=\sqrt{10} 6 \frac{3\sqrt{10}}{10}=18}\)
\(\displaystyle{ |AB|=\sqrt{10}}\)
\(\displaystyle{ |BC|=\sqrt{(4-1)^{2}+(2-1)^{2}}=\sqrt{3^{2}+1}=\sqrt{10}}\)
\(\displaystyle{ cos\sphericalangle ( \vec{AB}, \vec{BC})= \frac{\sqrt{2}}{|AB|}= \frac{\sqrt{2}}{\sqrt{10}}= \frac{\sqrt{20}}{10}= \frac{2\sqrt{5}}{10}= \frac{\sqrt{5}}{5}}\)
\(\displaystyle{ \vec{AB}o \vec{BC}=|AB| |BC| ( \vec{AB}, \vec{BC})=\sqrt{10} \sqrt{10} \frac{\sqrt{5}}{5}=2\sqrt{5}}\)
\(\displaystyle{ |BC|=\sqrt{10}}\)
\(\displaystyle{ |AC|=6}\)
\(\displaystyle{ cos\sphericalangle ( \vec{BC}, \vec{AC})= cos\sphericalangle ( \vec{AB}, \vec{AC})=\frac{3\sqrt{10}}{10}}\) (trójkąt ABC jest równoramienny, \(\displaystyle{ AB=BC=\sqrt{10}}\))
\(\displaystyle{ \vec{BC}o \vec{AC}=|BC| |AC| cos ( \vec{BC}, \vec{AC})=\sqrt{10} 6 \frac{3\sqrt{10}}{10}=18}\)
\(\displaystyle{ |AC|=\sqrt{(4+2)^{2}+(2-2)^{2}}=\sqrt{6^{2}}=6}\)
\(\displaystyle{ cos\sphericalangle ( \vec{AB}, \vec{AC})= \frac{1+2}{\sqrt{3^{2}+1^{2}}}= \frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{10}}\)
\(\displaystyle{ \vec{AB}o \vec{AC}=|AB| |AC| cos ( \vec{AB}, \vec{AC})=\sqrt{10} 6 \frac{3\sqrt{10}}{10}=18}\)
\(\displaystyle{ |AB|=\sqrt{10}}\)
\(\displaystyle{ |BC|=\sqrt{(4-1)^{2}+(2-1)^{2}}=\sqrt{3^{2}+1}=\sqrt{10}}\)
\(\displaystyle{ cos\sphericalangle ( \vec{AB}, \vec{BC})= \frac{\sqrt{2}}{|AB|}= \frac{\sqrt{2}}{\sqrt{10}}= \frac{\sqrt{20}}{10}= \frac{2\sqrt{5}}{10}= \frac{\sqrt{5}}{5}}\)
\(\displaystyle{ \vec{AB}o \vec{BC}=|AB| |BC| ( \vec{AB}, \vec{BC})=\sqrt{10} \sqrt{10} \frac{\sqrt{5}}{5}=2\sqrt{5}}\)
\(\displaystyle{ |BC|=\sqrt{10}}\)
\(\displaystyle{ |AC|=6}\)
\(\displaystyle{ cos\sphericalangle ( \vec{BC}, \vec{AC})= cos\sphericalangle ( \vec{AB}, \vec{AC})=\frac{3\sqrt{10}}{10}}\) (trójkąt ABC jest równoramienny, \(\displaystyle{ AB=BC=\sqrt{10}}\))
\(\displaystyle{ \vec{BC}o \vec{AC}=|BC| |AC| cos ( \vec{BC}, \vec{AC})=\sqrt{10} 6 \frac{3\sqrt{10}}{10}=18}\)
