Đáp án:

$A.\ \dfrac16$

Giải thích các bước giải:

Chọn hệ trục toạ độ sao cho:

$A(0;0;0);\ B(1;0;0);\ C(0;1;0);\ S(0;0;x)\quad (x > 0)$

$\Rightarrow \begin{cases}M\left(\dfrac12;0;0\right)\\\overrightarrow{SB}= (1;0;-x)\\\overrightarrow{BC}= (-1;1;0)\\\overrightarrow{SM}=\left(\dfrac12;0;-x\right)\\\overrightarrow{MC}=\left(-\dfrac12;1;0\right)\end{cases}$

$\Rightarrow\begin{cases}\overrightarrow{n_{(SBC)}}= (x;x;1)\\\overrightarrow{n_{(SMC)}}= (2x;x;1)\end{cases}$

Ta được:

$\quad \cos\varphi = \dfrac{\left|\overrightarrow{n_{(SBC)}}\cdot \overrightarrow{n_{(SMC)}}\right|}{\left|\overrightarrow{n_{(SBC)}}\right|\cdot \left|\overrightarrow{n_{(SMC)}}\right|}$

$\Rightarrow \cos^2\varphi = \dfrac{\left|\overrightarrow{n_{(SBC)}}\cdot \overrightarrow{n_{(SMC)}}\right|^2}{\left|\overrightarrow{n_{(SBC)}}\right|^2\cdot \left|\overrightarrow{n_{(SMC)}}\right|^2}$

$\Leftrightarrow 1 - \sin^2\varphi = \dfrac{\left(2x^2 + x^2 +1\right)^2}{\left(x^2 + x^2 + 1\right)\left(4x^2 + x^2 + 1\right)}$

$\Leftrightarrow 1 - \dfrac19 = \dfrac{(3x^2+1)^2}{(2x^2 +1)(5x^2+1)}$

$\Leftrightarrow \dfrac{(x^2-1)^2}{9(2x^2+1)(5x^2+1)}= 0$

$\Leftrightarrow x^2 - 1 = 0$

$\Leftrightarrow \left[\begin{array}{l}x = -1\quad (loại)\\x = 1\quad\ (nhận)\end{array}\right.$

$\Rightarrow S(0;0;1)$

$\Rightarrow SA = 1$

Khi đó:

$V_{S.ABC}=\dfrac16SA.AB.AC = \dfrac16\cdot 1\cdot 1\cdot 1 =\dfrac16$