Quay lại đáp án

$(S): x^2+y^2+z^2=1$ 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 $(1) (2)\begin{cases}b+c= \\ bc= \end{cases}$

$(S): x^2+y^2+z^2=1$ 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 $(1) (2)\begin{cases}b+c=2\sqrt2 \\ bc=2 \end{cases}\begin{cases}b=\sqrt2 \\ c=\sqrt2 \end{cases}$