viết phương trình mặt phằng (P) qua A ( $\sqrt{2}$, 0,0) , tiếp xúc với (S) : x2+y2+z2=1  và cắt oy , oz tại B,C ( yB >0. zC >0) sao cho S$\Delta $ABC = 2$\sqrt{2}$
$(S): x^2+y^2+z^2=1$ có tâm    $O(0;0;0) ;R=1$
Gọi $B(0;b;0)=(P)\cap Oy  C(0;0;c)=(P)\cap Oz   (b;c>0)$
$=>(P): \frac{x}{\sqrt2} + \frac{y}{b}  +\frac{z}{c}=1$hay $xbc+\sqrt2cy+\sqrt2bz-\sqrt2bc=0$
$(P)$ tiếp xúc với $(S)\Rightarrow |\sqrt2bc|=\sqrt{(bc)^2+2(b^2+c^2)}                (1)$
$\overrightarrow{AB}(-\sqrt2;b;0)                          $
$\overrightarrow{AC}(-\sqrt2;0;c)$
$\Rightarrow |[\overrightarrow{AB},\overrightarrow{AC}]|=|bc+\sqrt2c+\sqrt2b|=4\sqrt2                    (2)$
Từ $(1) (2)\begin{cases}b+c=2\sqrt2 \\ bc=2 \end{cases}\begin{cases}b=\sqrt2 \\ c=\sqrt2 \end{cases}$
pooh ơi bạn nhầm đâu rồi thì phải .mình giải thì đc kq là b=c= 2 –  nhutuyet12t7.1995 06-04-13 11:36 PM
b=c= 2 căn 2 à pooh –  nhutuyet12t7.1995 06-04-13 05:55 AM
pt đoạn chắn mà –  Pooh 05-04-13 08:33 PM
sao a= căn hai vậy hả bạn –  thekyrooney 05-04-13 03:02 PM

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