Bài 17: Góc giữa hai đường thẳng

I.Lý thuyết
Cho \(\Delta _1\) có 1 VTCP \(\overrightarrow{u_1}=(a_1;b_1;c_1)\)
Cho \(\Delta _2\) có 1 VTCP \(\overrightarrow{u_2}=(a_2;b_2;c_2)\)​
\(cos(\Delta _1;\Delta _2)=\left | cos(\overrightarrow{u_1};\overrightarrow{u_2}) \right |=\frac{\left | \overrightarrow{u_1}\overrightarrow{u_2} \right |}{ \left | \overrightarrow{u_1} \right |.\left | \overrightarrow{u_2} \right |}\)
\(=\frac{\left | a_1a_2+b_1b_2+c_1c_2 \right |}{\sqrt{a^2_1+b^2_1+c^2_1} .\sqrt{a^2_2+b^2_2+c^2_2}}\)
Nhận xét:
1) \(0^0\leq (\Delta _1;\Delta _2)\leq 90^0\)
2) \(\Delta _1\perp \Delta _2\Leftrightarrow a_1a_2+b_1b_2+c_1c_2=0\)
II. Bài tập

VD1: Tính góc giữa \(\Delta :\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-2}{1}\) và trục Ox
Giải
\(\Delta\) có 1 VTCP \(\overrightarrow{u}=(2;1;1)\)
Ox có 1 VTCP \(\overrightarrow{i}=(1;0;0)\)
\(cos(\Delta ;Ox)=\left | cos(\overrightarrow{u};\overrightarrow{i}) \right | =\frac{\left | 2.1+1.0+1.0 \right |}{\sqrt{2^2+1^2+1^2}.\sqrt{1^2+0^2+0^2}}= \frac{2}{\sqrt{6}}\)
\((\Delta ;Ox)= arccos \frac{2}{\sqrt{6}}\)
VD2: Cho \(\Delta _1:\left\{\begin{matrix} x=1+mt\\ y=2+2t\\ z=3-2t \end{matrix}\right. \ \Delta _2: \frac{x-2}{2}=\frac{y}{2}=\frac{z+3}{1}\)
Tìm m để
\(a) \ \ \Delta _1\perp \Delta _2\)
\(b) \ \ (\widehat{\Delta _1; \Delta _2})=60^0\)
Giải
\(\Delta _1\) có 1 VTCP \(\overrightarrow{u_1}=(m;2;-2)\)
\(\Delta _2\) có 1 VTCP \(\overrightarrow{u_2}=(2;2;1)\)
a)
\(\Delta _1\perp \Delta _2\Leftrightarrow \overrightarrow{u_1}.\overrightarrow{u_2} =0\Leftrightarrow 2m+4-2=0\Leftrightarrow m=-1\)
b)
\((\widehat{\Delta _1; \Delta _2})=60^0\Leftrightarrow cos(\Delta _1; \Delta _2)=\frac{1}{2}\)
\(\Leftrightarrow \frac{\left | 2m+4-2 \right |}{\sqrt{m^2+2^2(-2)^2}.\sqrt{2^2+2^2+1^2} }=\frac{1}{2}\)
\(\Leftrightarrow \frac{\left | 2m+2 \right |}{\sqrt{m^2+8}.3}=\frac{1}{2}\)
\(\Leftrightarrow 4\left | m+1 \right |=3.\sqrt{m^2+8}\)
\(\Leftrightarrow 16(m^2+2m+1)=9(m^2+8)\)
\(\Leftrightarrow 7m^2+32m-56=0\)
\(\Delta '=16^2+7.56=256+392=648\)
\(\Bigg \lbrack\begin{matrix} m=\frac{-16-\sqrt{648}}{7}\\ \\ m=\frac{-16+\sqrt{648}}{7} \end{matrix}\)
VD3: Viết phương trình \(\Delta\) đi qua M(1;-5;3) và tạo với Ox, Oy các góc bằng 60⁰
Giải
Gọi 1 VTCP của \(\Delta\) là \(\overrightarrow{u}=(a;b;c) \ \ a^2+b^2+c^2\neq 0\)
Ox có 1 VTCP \(\overrightarrow{i}=(1;0;0)\)
Oy có 1 VTCP \(\overrightarrow{j}=(0;1;0)\)
\(\left\{\begin{matrix} (\widehat{\Delta , Ox})=60^0\\ (\widehat{\Delta , Oy})=60^0 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} cos(\Delta , Ox)=\frac{1}{2}\\ \\ cos(\Delta , Oy)=\frac{1}{2} \end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \frac{\left | \overrightarrow{u}.\overrightarrow{i} \right |}{\left | \overrightarrow{u} \right |.\left | \overrightarrow{i} \right |}=\frac{1}{2}\\ \\ \\ \frac{\left | \overrightarrow{u}.\overrightarrow{j} \right |}{\left | \overrightarrow{u} \right |.\left | \overrightarrow{j} \right |}=\frac{1}{2} \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} \frac{\left | a \right |}{\sqrt{a^2+b^2+c^2}}=\frac{1}{2}\\ \\ \frac{\left | b \right |}{\sqrt{a^2+b^2+c^2}}=\frac{1}{2} \end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \left | a \right | = \left | b \right |\\ 4a^2=a^2+b^2+c^2 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \left | a \right |=\left | b \right |\\ c^2=2a^2 \end{matrix}\right.\)
+ Chọn a = 1 \(\Rightarrow \bigg \lbrack \begin{matrix} c=\sqrt{2}\\ c=-\sqrt{2} \end{matrix}\Rightarrow \bigg \lbrack \begin{matrix} b=1\\ b=-1 \end{matrix}\)
TH1:
\(a=1,c=\sqrt{2},b=1, \overrightarrow{u}=(1;1;\sqrt{2})\)
\(pt \ \Delta : \frac{x-1}{1}=\frac{y+5}{1}=\frac{z-3}{\sqrt{2}}\)
TH2:
\(a=1,b=1,c=-\sqrt{2} \ \ \ \overrightarrow{u}=(1;1;-\sqrt{2})\)
\(pt \ \Delta : \frac{x-1}{1}=\frac{y+5}{1}=\frac{z-3}{-\sqrt{2}}\)
TH3:
\(a=1,b=-1,c=\sqrt{2}\)
\(pt \ \Delta : \frac{x-1}{1}=\frac{y+5}{-1}=\frac{z-3}{\sqrt{2}}\)
TH4:
\(a=1,b=-1,c=-\sqrt{2}\)
\(pt \ \Delta : \frac{x-1}{1}=\frac{y+5}{-1}=\frac{z-3}{-\sqrt{2}}\)