Question

đáp án bài này là (P):x+y-z-4=0  hoặc (P):7x-17y+5z-4=0

Trong không gian với hệ trục Oxyz, cho mặt cầu (S):

\[ x^2 + y^2 + z^2 + 2x - 2y + 2z - 1 = 0 \]

và đường thẳng \( d \):

\[
\begin{cases}
x - y - 2 = 0 \\
2x - z - 6 = 0
\end{cases}
\]

Viết phương trình mặt phẳng (P) chứa \( d \) và cắt mặt cầu (S) theo một đường tròn có bán kính \( r = 1 \).

Answer 1

\(\left(S\right):x^2+y^2+z^2+2x-2y+2z-1=0\)

\(\Rightarrow\left(S\right):\left(x+1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=4\)

\(\Rightarrow\) Tâm \(I\left(-1;1;-1\right);R=2\)

\(\left(d\right):\left\{{}\begin{matrix}x-y-2=0\\2x-z-6=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=t\\y=t-2\\z=2t-6\end{matrix}\right.\) \(\Rightarrow A\left(0;-2;-6\right)\in\left(d\right);\overrightarrow{u_d}=\left(1;1;2\right)\)

\(\left(d\right)\subset\left(P\right)\Leftrightarrow\left(P\right):a\left(x-y-2\right)+b\left(2x-z-6\right)=0\)

\(\Rightarrow\left(P\right):\left(a+2b\right)x-ay-bz-2a-6b=0\)

\(d\left(I;\left(P\right)\right)=\sqrt{R^2-r^2}=\sqrt{4-1}=\sqrt{3}\)

\(\Rightarrow\dfrac{\left|\left(a+2b\right).\left(-1\right)-a.1-b.\left(-1\right)-2a-6\right|}{\sqrt{\left(a+2b\right)^2+\left(-a\right)^2+\left(-b\right)^2}}=\sqrt{3}\)

\(\Leftrightarrow\dfrac{\left|-4a-7b\right|}{\sqrt{2a^2+4ab+5b^2}}=\sqrt{3}\)

\(\Leftrightarrow\left(4a+7b\right)^2=3\left(2a^2+4ab+5b^2\right)\)

\(\Leftrightarrow10a^2+44ab+34b=0\)

\(\Leftrightarrow5a^2+22ab+17b=0\)

\(\Leftrightarrow\left(5a+17b\right)\left(a+b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\â=-\dfrac{17}{5}\end{matrix}\right.\)

\(TH_1:a=-b\)

\(\left(P\right):\left(-b+2b\right)x+by-bz+2b-6b=0\)

\(\Rightarrow bx+by-bz-4b=0\) hay \(\left(P\right):x+y-z-4=0\)

\(TH_2:a=-\dfrac{17}{5}\)

\(\Rightarrow\left(P\right):\left(-\dfrac{17}{5}b+2b\right)x+\dfrac{17}{5}by-bz+\dfrac{34}{5}b-6b=0\)

\(\Rightarrow-7x+17y-5z+4=0\) hay \(\left(P\right):7x-17y+5z-4=0\)