Cho hình chóp $S.ABC$ có ba cạnh $SA, SB, SC$ vuông góc với nhau từng đôi một và $SA=a, SB=b, SC=c$
a) Tính thể tích khối chóp $S.ABC$. Chứng minh rằng hình chiếu vuông góc của đỉnh $S$ trên mặt phẳng $(ABC)$ là trực tâm tam giác $ABC$
b) Tìm tâm và bán kính mặt cầu ngoại tiếp hình chóp $S.ABC$

a) Ta có: $V=\frac{1}{3}S_{SAB}.SC=\frac{1}{3}.\frac{1}{2}SA.SB.SC=\frac{1}{6}abc    $
Vẽ $SH\bot (ABC)\Rightarrow  SH\bot AC$
$SB\bot (SAC)\Rightarrow  SB\bot AC$
$\Rightarrow  AC\bot (SBH)\Rightarrow  AC\bot BH (1)$
Tương tự: $SH\bot (ABC)\Rightarrow  SH\bot AB; SC\bot (SAB)\Rightarrow  SC\bot AB$
$\Rightarrow  AB\bot (SCH)\Rightarrow  AH\bot CH  (2)$
$Từ (1), (2)\Rightarrow  H$ là trực tâm $\Delta ABC$

b) Chọn hệ trục tọa độ $Oxyz$ như sau:
- Gốc $O\equiv S$
- Trục $Ox$ đi qua $SA$
- Trục $Oy$ đi qua $SB$
- Trục $Oz$ đi qua $SC$

Khi đó: $S(0;0;0), A(a;0;0), B(0;b;0), C(0;0;c)$ mặt cầu ngoại tiếp $S.ABC$ có dạng: $(S):x^2+y^2+z^2-2a'x-2b'y-2c'z=0$
Ta có hệ:
$\begin{array}{l}
\left\{ \begin{array}{l}
A \in (S)\\
B \in (S)\\
C \in (S)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{a^2} - 2a'a\,\,\,\,\,\,\,\,\,\,(1)\\
{b^2} - 2b'b = 0\,\,\,(2)\\
{c^2} - 2c'c = 0\,\,\,(3)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a' = \frac{a}{2}\\
b' = \frac{b}{2}\\
c' = \frac{c}{2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
I\left( {\frac{a}{2};\frac{b}{2};\frac{c}{2}} \right)\\
R = \sqrt {\frac{{{a^2}}}{4} + \frac{{{b^2}}}{4} + \frac{{{c^2}}}{4}}  = \frac{{\sqrt {{a^2} + {b^2} + {c^2}} }}{2}
\end{array} \right.
\end{array}$

Thẻ

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