Bài toán. Cho hình chóp $S.ABC$ có\[SA = a;{\mkern 1mu} \,SB = b;{\mkern 1mu} \,SC = c;\,\widehat {BSC} = \alpha ;\,\widehat {CSA} = \beta ;\,\widehat {ASB} = \gamma \]Tính thể tích của hình chóp $S.ABC$?
Lời giải. Đặt $\overrightarrow {SA} = \overrightarrow a ;\,\,\overrightarrow {SB} = \overrightarrow b ;\,\,\overrightarrow {SA} = \overrightarrow c$, có\[\left| {\overrightarrow a } \right| = a;\,\left| {\overrightarrow b } \right| = b;\,\left| {\overrightarrow c } \right| = c;\,\,\overrightarrow a .\overrightarrow b = ab\cos \gamma ;\,\overrightarrow b .\overrightarrow c = bc\cos \alpha ;\,\overrightarrow c .\overrightarrow a = ca\cos \beta \]Ta lại có\[{V_{S.ABC}} = \frac{{\left| {\overrightarrow {SA} .\left( {\overrightarrow {SB} \wedge \overrightarrow {SC} } \right)} \right|}}{6} = \frac{{\left| {\overrightarrow a .\left( {\overrightarrow b \wedge \overrightarrow c } \right)} \right|}}{6}\]Do bộ ba $\overrightarrow {SA} = \overrightarrow a ;\,\,\overrightarrow {SB} = \overrightarrow b ;\,\,\overrightarrow {SA} = \overrightarrow c$ là các vector không đồng phẳng nên giả sử \[\overrightarrow b \wedge \overrightarrow c = k\overrightarrow a + l\overrightarrow b + m\overrightarrow {c\,} ;\,\,k;{\mkern 1mu} l;{\mkern 1mu} m \in \mathbb{R}\]Vì $\overrightarrow b .\left( {\overrightarrow b \wedge \overrightarrow c } \right) = \overrightarrow c .\left( {\overrightarrow b \wedge \overrightarrow c } \right) = 0$, cho nên dẫn đến\[kab\cos \gamma + l{b^2} + mbc\cos \alpha = kca\cos \beta + lbc\cos \alpha + m{c^2} = 0.\]Từ đó giải hệ ẩn $k;\,l$ tham số $m$ sau\[\begin{cases}ka\cos \gamma + lb + mc\cos \alpha &= 0\\ka\cos \beta + lb\cos \alpha + mc &= 0\end{cases}\]Ta sẽ có\[k = \frac{{mc}}{a}.\frac{{{{\sin }^2}\alpha }}{{\cos \beta – \cos \gamma \cos \alpha }},\qquad l = \frac{{mc}}{b}.\frac{{\cos \gamma – \cos \alpha \cos \beta }}{{\cos \beta – \cos \gamma \cos \alpha }}.\]Kết hợp với $\left| {\overrightarrow b \wedge \overrightarrow c } \right| = bc\sin \alpha $ ta có\[\begin{array}{l}{b^2}{c^2}\sin \alpha & = {k^2}{a^2} + {l^2}{b^2} + {m^2}{c^2} + 2\left( {klab\cos \gamma + lmbc\cos \alpha + mkca\cos \beta } \right)\\ &= {m^2}{c^2}.T\end{array}\]Ở đây\[\begin{array}{l}T& = {\left( {\frac{{{{\sin }^2}\alpha }}{{\cos \beta – \cos \gamma \cos \alpha }}} \right)^2} + {\left( {\frac{{\cos \gamma – \cos \alpha \cos \beta }}{{\cos \beta – \cos \gamma \cos \alpha }}} \right)^2} + 1\\ &+ 2\cos \gamma \left( {\frac{{{{\sin }^2}\alpha }}{{\cos \beta – \cos \gamma \cos \alpha }}} \right)\left( {\frac{{\cos \gamma – \cos \alpha \cos \beta }}{{\cos \beta – \cos \gamma \cos \alpha }}} \right) + 2\cos \alpha + 2\cos \beta \end{array}\]Từ đó có $|m| = b\sqrt {\frac{{\sin \alpha }}{T}} $, vậy nên thể tích cần tính là\[\begin{array}{l}V &= \frac{1}{6}\left| {k{a^2} + lab\cos \gamma + mca\cos \alpha } \right|\\ &= \frac{1}{6}\left| {mca} \right|\left| {\frac{{{{\sin }^2}\alpha }}{{\cos \beta – \cos \gamma \cos \alpha }} + \frac{{\cos \gamma – \cos \alpha \cos \beta }}{{\cos \beta – \cos \gamma \cos \alpha }} + \cos \alpha } \right|\\ &= \frac{{abc}}{6}\left| {\frac{{{{\sin }^2}\alpha }}{{\cos \beta – \cos \gamma \cos \alpha }} + \frac{{\cos \gamma – \cos \alpha \cos \beta }}{{\cos \beta – \cos \gamma \cos \alpha }} + \cos \alpha } \right|\sqrt {\frac{{\sin \alpha }}{T}} \end{array}\]Với \[\begin{array}{l}T &= {\left( {\frac{{{{\sin }^2}\alpha }}{{\cos \beta – \cos \gamma \cos \alpha }}} \right)^2} + {\left( {\frac{{\cos \gamma – \cos \alpha \cos \beta }}{{\cos \beta – \cos \gamma \cos \alpha }}} \right)^2} + 1\\& + 2\cos \gamma \left( {\frac{{{{\sin }^2}\alpha }}{{\cos \beta – \cos \gamma \cos \alpha }}} \right)\left( {\frac{{\cos \gamma – \cos \alpha \cos \beta }}{{\cos \beta – \cos \gamma \cos \alpha }}} \right) + 2\cos \alpha + 2\cos \beta \end{array}\]
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