Đáp án:
$V_{S.ABC} = \dfrac{a^3\sqrt {30} } {12}$
Lời giải:
$$\eqalign{ & A\left( {0;0;0} \right),\,\,B\left( {\sqrt 2 ;0;0} \right);\,\,C\left( {0;\sqrt 5 ;0} \right) \cr & \Rightarrow H\left( {{{\sqrt 2 } \over 2};{{\sqrt 5 } \over 2};0} \right) \Rightarrow S\left( {{{\sqrt 2 } \over 2};{{\sqrt 5 } \over 2};x} \right)\,\,\left( {x > 0} \right) \cr & \text{Ta có:} \cr & \left\{ \matrix{ \overrightarrow {AS} = \left( {{{\sqrt 2 } \over 2};{{\sqrt 5 } \over 2};x} \right) \hfill \cr \overrightarrow {AB} = \left( {\sqrt 2 ;0;0} \right) \hfill \cr} \right. \Rightarrow \left[ {\overrightarrow {AS} ;\overrightarrow {AB} } \right] = \left( {0;\sqrt 2 x; - {{\sqrt {10} } \over 2}} \right) = {\overrightarrow n _{\left( {SAB} \right)}} \cr & \left\{ \matrix{ \overrightarrow {AS} = \left( {{{\sqrt 2 } \over 2};{{\sqrt 5 } \over 2};x} \right) \hfill \cr \overrightarrow {AC} = \left( {0;\sqrt 5 ;0} \right) \hfill \cr} \right. \Rightarrow \left[ {\overrightarrow {AS} ;\overrightarrow {AC} } \right] = \left( {\sqrt 5 x;0; - {{\sqrt {10} } \over 2}} \right) = {\overrightarrow n _{\left( {SAC} \right)}} \cr & \widehat {\left( {\left( {SAB} \right);\left( {SAC} \right)} \right)} = {60^0} \cr & \Rightarrow {{\left| {{{10} \over 4}} \right|} \over {\sqrt {2{x^2} + {5 \over 2}} \sqrt {5{x^2} + {5 \over 2}} }} = {1 \over 2} \Leftrightarrow \left( {2{x^2} + {5 \over 2}} \right)\left( {5{x^2} + {5 \over 2}} \right) = 25 \cr & \Leftrightarrow 10{x^4} + 5{x^2} + {{25} \over 2}{x^2} + {{25} \over 4} = 25 \cr & \Leftrightarrow 10{x^4} + {{35} \over 2}{x^2} - {{75} \over 4} = 0 \cr & \Leftrightarrow {x^2} = {3 \over 4} \Leftrightarrow x = {{\sqrt 3 } \over 2} \cr & \Rightarrow SH = {{a\sqrt 3 } \over 2} \cr & {S_{ABC}} = {1 \over 2}AB.AC = {1 \over 2}a\sqrt 2 .a\sqrt 5 = {{{a^2}\sqrt {10} } \over 2} \cr & \Rightarrow {V_{S.ABC}} = {1 \over 3}{{a\sqrt 3 } \over 2}.{{{a^2}\sqrt {10} } \over 2} = {{{a^3}\sqrt {30} } \over {12}} \cr} $$
