Normal
0
false
false
false
MicrosoftInternetExplorer4
Câu 1
$d_1 ;d_2 $ lầ\.vntime""="">n lượt có VTCP $\overrightarrow{u_1}$(0;-3;1)
$;\overrightarrow{u_2}(1;-1;0)$
Gọ\.vntime""="">i $\overrightarrow{n}(a;b;c)$ là VTPT của
$(P)$
ycbt$\Leftrightarrow \begin{cases}\overrightarrow{n}.\overrightarrow{u_2}=0 \\
\frac{|\overrightarrow{n}.\overrightarrow{u_1}|}{|\overrightarrow{n}|.|\overrightarrow{u_1}|}=\sqrt{\frac{13}{15}}
\end{cases}$
$\Leftrightarrow \begin{cases}a-b=0 \\
\frac{|3b-c|}{\sqrt{a^2+b^2+c^2}.\sqrt{10}}=\sqrt{\frac{13}{15}} \end{cases}$
thế\.vntime""=""> $a=b$ vào pt $(2)$ và bình phương ta có
$15(9b^2-6bc+c^2)=130(2b^2+c^2)\Leftrightarrow $
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mso-tstyle-rowband-size:0;
mso-tstyle-colband-size:0;
mso-style-noshow:yes;
mso-style-parent:"";
mso-padding-alt:0in 5.4pt 0in 5.4pt;
mso-para-margin:0in;
mso-para-margin-bottom:.0001pt;
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mso-ansi-language:#0400;
mso-fareast-language:#0400;
mso-bidi-language:#0400;}
Normal
0
false
false
false
MicrosoftInternetExplorer4
Câu 1
$d_1 ;d_2 $ lần lượt có VTCP $\overrightarrow{u_1}(0;-3;1)
;\overrightarrow{u_2}(1;-1;0)$
Gọi $\overrightarrow{n}(a;b;c)$ là VTPT của
$(P)$
ycbt$\Leftrightarrow \begin{cases}\overrightarrow{n}.\overrightarrow{u_2}=0 \\
\frac{|\overrightarrow{n}.\overrightarrow{u_1}|}{|\overrightarrow{n}|.|\overrightarrow{u_1}|}=\sqrt{\frac{13}{15}}
\end{cases}$
$\Leftrightarrow \begin{cases}a-b=0 \\
\frac{|3b-c|}{\sqrt{a^2+b^2+c^2}.\sqrt{10}}=\sqrt{\frac{13}{15}} \end{cases}$
thế $a=b$ vào pt $(2)$ và bình phương ta có
$15(9b^2-6bc+c^2)=130(2b^2+c^2)\Leftrightarrow $
/* Style Definitions */
table.MsoNormalTable
{mso-style-name:"Table Normal";
mso-tstyle-rowband-size:0;
mso-tstyle-colband-size:0;
mso-style-noshow:yes;
mso-style-parent:"";
mso-padding-alt:0in 5.4pt 0in 5.4pt;
mso-para-margin:0in;
mso-para-margin-bottom:.0001pt;
mso-pagination:widow-orphan;
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font-family:"Times New Roman";
mso-ansi-language:#0400;
mso-fareast-language:#0400;
mso-bidi-language:#0400;}
Normal
0
false
false
false
MicrosoftInternetExplorer4
Câu 1
$d_1 ;d_2 $ lầ
\.vntime""="">n lượt có VTCP $\overrightarrow{u_1}$(0;-3;1)
$;\overrightarrow{u_2}(1;-1;0)$
Gọ\.vntime""="">i $\overrightarrow{n}(a;b;c)$ là VTPT của
$(P)$
ycbt$\Leftrightarrow \begin{cases}\overrightarrow{n}.\overrightarrow{u_2}=0 \\
\frac{|\overrightarrow{n}.\overrightarrow{u_1}|}{|\overrightarrow{n}|.|\overrightarrow{u_1}|}=\sqrt{\frac{13}{15}}
\end{cases}$
$\Leftrightarrow \begin{cases}a-b=0 \\
\frac{|3b-c|}{\sqrt{a^2+b^2+c^2}.\sqrt{10}}=\sqrt{\frac{13}{15}} \end{cases}$
thế\.vntime""=""> $a=b$ vào pt $(2)$ và bình phương ta có
$15(9b^2-6bc+c^2)=130(2b^2+c^2)\Leftrightarrow $
/* Style Definitions */
table.MsoNormalTable
{mso-style-name:"Table Normal";
mso-tstyle-rowband-size:0;
mso-tstyle-colband-size:0;
mso-style-noshow:yes;
mso-style-parent:"";
mso-padding-alt:0in 5.4pt 0in 5.4pt;
mso-para-margin:0in;
mso-para-margin-bottom:.0001pt;
mso-pagination:widow-orphan;
font-size:10.0pt;
font-family:"Times New Roman";
mso-ansi-language:#0400;
mso-fareast-language:#0400;
mso-bidi-language:#0400;}