目录:
点-点
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\(\large A(x_1,y_1,z_1)\)
\(\large B(x_2,y_2,z_2)\)
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距离
\[\large d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2} \] -
所成直线(直线的两点式方程)
\[\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1},x_1\ne x_2,y_1 \ne y_2 \]
点-线
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\(\large P(x_{0},y_{0},z_{0})\)
\(\large M(x_1,y_1,z_1)\)
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\(\large L:\dfrac{x-x_{1}}{m}=\dfrac{y-y_{1}}{n}=\dfrac{z-z_{1}}{p}\)
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\(\large \vec{s}=(m,n,p)\)
\(\large \overrightarrow{PM}=(x_{0}-x_{1},y_{0}-y_{1},z_{0}-z_{1})\)
\(P \notin L\) 不在线上
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距离
\[d=\frac{\left|\overrightarrow{PM}\times \vec{s}\right|}{|\vec{s}|} \] -
所成平面 \(\pi\)
\[\pi:A(x-x_{0})+B(y-y_{0})+C(z-z_{0})=0 \]其中 \(\large \vec{n}_{\pi}=\overrightarrow{PM}\times \vec{s}=(A,B,C)\)
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垂线
\[L:\begin{cases}A(x-x_{0})+B(y-y_{0})+C(z-z_{0})=0\\m(x-x_{0})+n(y-y_{0})+p(z-z_{0})=0\end{cases} \] -
投影点(P 在 \(L\) 上的投影)
\[P'(mt+x_{1},nt+y_{1},pt+z_{1}) \]其中
\[t=-\frac{m\left(x_{1}-x_{0}\right)+n\left(y_{1}-y_{0}\right)+p\left(z_{1}-z_{0}\right)}{m^{2}+n^{2}+p^{2}} \]
\(P \in L\)
- 过点的垂面
点-面
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\(\large P(x_0,y_0,z_0)\)
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\(\large \vec{n}=(A,B,C)\)
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\(\large \pi:Ax+By+Cz+D=0\)
\(P\notin \pi\)点在面上
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距离
\[\large d=\frac{|Ax_{0}+By_{0}+Cz_{0}+D|}{\sqrt{A^{2}+B^{2}+C^{2}}} \\ \]对于平面点法式方程 \(\large \pi:A(x-x_1)+B(y-y_1)+C(z-z_1)=0\),其中点为 \(\large M(x_1,y_1,z_1)\)
\[\begin{aligned} d&=\frac{\left|\overrightarrow{PM}\cdot\vec{n}\right|}{\left|\vec{n}\right|} \end{aligned} \] -
投影点
\[\large H(At+X_{0},Bt+Y_{0},Ct+Z_{0}) \]其中
\[t=\frac{-(Ax_{0}+By_{0}+Cz_{0}+D)}{A^{2}+B^{2}+C^{2}} \] -
垂线
\[\large L:\frac{x-x_0}{A}=\frac{y-y_0}{B}=\frac{z-z_0}{C} \]
\(P \in \pi\) 略
线-线
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\(\large M_1(x_1,y_1,z_1),M_2(x_2,y_2,z_2)\)
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\(\large L_1:\dfrac{x-x_1}{m_1}=\dfrac{y-y_1}{n_1}=\dfrac{z-z_1}{p_1}\)
\(\large L_2:\dfrac{x-x_2}{m_2}=\dfrac{y-y_2}{n_2}=\dfrac{z-z_2}{p_2}\)
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\(\large \vec{s_1}=(m_1,n_1,p_1)\)
\(\large \vec{s_2}=(m_2,n_2,p_2)\)
\(\large \overrightarrow{M_{1}M_{2}}=(x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1})\)
位置关系
[注] 以下判定按照优先级顺序排序,具体表现如平行不包括重合的情况。
\(L_1=L_2\) (重合)
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判定条件
\[\large \begin{equation} \frac{m_1}{m_2}=\frac{n_1}{n_2}=\frac{p_1}{p_2},M_1\notin L_2(M2 \notin L_1) \end{equation} \]
\(L_1 // L_2\) (平行)
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判定条件
\[\large \frac{m_1}{m_2}=\frac{n_1}{n_2}=\frac{p_1}{p_2},M_1\in L_2(M2 \in L_1) \] -
距离
\[\large \begin{aligned} d = d_{M_1-L_2} = \dfrac{|\overrightarrow{M_1M_2}\times\vec{s_2}|}{|\vec{s_2}|} \end{aligned} \] -
所成平面
\[\large \pi:A(x-x_1)+B(y-y_1)+C(z-z_1)=0 \]其中 \(\large \vec{s_1}\times\overrightarrow{M_1M_2}=(A,B,C)\)
\(L_1 \cap L_2 = P\)(相交)
- 判定条件
\([\vec{a}\vec{b}\vec{c}]\)表示三个向量的混合积
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所成平面
\[\large \pi:A(x-x_1)+B(y-y_1)+C(z-z_1)=0 \]其中 \(\large \vec{s_1}\times\vec{s_2}=(A,B,C)\)
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交点
\[\large t_1=\frac{n_2(x_2-x_1)-m_2(y_2-y_1)}{m_1n_2-m_2n_1}\\ t_2=\frac{n_1(x_2-x_1)-m_1(y_2-y_1)}{m_1n_2-m_2n_1} \]交点坐标 \(\large P(m_1t_1+x_1,n_1t_1+y_1,p_1t_1+z_1)\) 或 \(\large P(m_2t_2+x_2,n_2t_2+y_2,p_2t_2+z_2)\)
PS:求点向式直线交点坐标的题目我没有找到,所以看看就行
\(L_1\) 与 \(L_2\) 异面
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判定条件
\[\large [\vec{s_{1}}\ \vec{s_{2}}\ \overrightarrow{M_{1}M_{2}}]\ne0 \] -
距离
\[\large d = \dfrac{|\overrightarrow{M_1M_2}\times\vec{s}|}{|\vec{s}|} \]其中 \(\large \vec{s}=\vec{s_1}\times \vec{s_2}\)
其他关系
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夹角
\[\large \cos \varphi = \frac{\left\vert m_1m_2+n_1n_2+p_1p_2\right\vert}{\sqrt{m_1^2+n_1^2+p_1^2}\sqrt{m_2^2+n_2^2+p_2^2}} \]*垂直:\(\large \vec{s_1} \bot \vec{s_2} \iff\vec{s_1}\cdot \vec{s_2} = m_1m_2+n_1n_2+p_1p_2=0\)
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一般式\(-\!-\!\!\!\!\!\!>\)标准式
对直线的一般式方程\(\large \begin{cases} A_1x+B_1y+C_1z+D_1=0\\ \\A_2x+B_2y+C_2z+D_2=0 \end{cases}\),带入一点 \(\large M(x_0,y_0,z_0)\)
\[\vec{s} = (A_1,B_1,C_1) \times (A_2,B_2,C_2) =(m,n,p)\\ L:\frac{x-x_0}{m} = \frac{y-y_0}{n} = \frac{z-z_0}{p} \] -
标准式\(-\!-\!\!\!\!\!\!>\)一般式
对于标准式方程 \(\large L:\dfrac{x-x_0}{m} = \dfrac{y-y_0}{n} = \dfrac{z-z_0}{p}\)
\[\begin{cases} nx-my+my_0-nx_0=0 \\ \\ py-nz+nz_0-py_0=0 \end{cases} \]
线-面
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\(\large M_1(x_0,y_0,z_0)\)
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\(\large L:\dfrac{x-x_0}{m}=\dfrac{y-y_0}{n}=\dfrac{z-z_0}{p}\)
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\(\large \pi:Ax+By+Cz+D=0\)
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\(\large \vec{s}=(m,n,p),\vec{n}=(A,B,C)\)
位置关系
\(L \subset \pi\)
- 判定条件\[\large \vec{s}\ \bot\ \vec{n} , M_0 \in \pi \iff \begin{cases} mA + nB + pC = 0 \\ \\ Ax_0+By_0+Cz_0+D=0 \end{cases} \]
\(L \ //\ \pi\)
- 判定条件\[\large \vec{s}\ \bot\ \vec{n} , M_0 \notin \pi \iff \begin{cases} mA + nB + pC = 0 \\ \\ Ax_0+By_0+Cz_0+D\ne 0 \end{cases} \]
\(L\ \cap\ \pi = P\)
- 判定条件\[\large \large \vec{s}\ \bot \!\!\!\!/ \ \vec{n} \iff mA+nB+pC \neq 0 \]
其他关系
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过 \(L\) 的垂面
\[\large \pi':A'(x-x_0)+B'(y-y_0)+C'(z-z_0)=0 \]其中 \(\large \vec{n'} = (A',B',C') = \vec{s} \times \vec{n}\)
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距离
\[\large d=d_{M-\pi} = \frac{|Ax_{0}+By_{0}+Cz_{0}+D|}{\sqrt{A^{2}+B^{2}+C^{2}}} \\ \] -
夹角 \(\varphi\)
\[\large \begin{aligned} \cos\left(\frac\pi2-\varphi\right)=\frac{\mid \vec{n}\cdot \vec{s}\mid}{\mid \vec{n}\mid\mid \vec{s}\mid} \end{aligned} \]【规定】\(0 \le \varphi \le \dfrac{\pi}{2}\)
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投影
\[\large L': \begin{cases} \pi:Ax+By+Cz+D=0 \\ \\ \pi':A'(x-x_0)+B'(y-y_0)+C'(z-z_0)=0 \end{cases} \]其中 \(\pi'\) 为 \(\pi\) 过 \(L\) 的垂面,上面已经讨论过。
面-面
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\(\large \pi_1:A_1x+B_1y+C_1z+D_1=0\)
\(\large \pi_2:A_2x+B_2y+C_2z+D_2=0\)
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\(\large\)\(\large\vec{n_1}=(A_1,B_1,C_1)\)
\(\large \vec{n_2}=(A_2,B_2,C_2)\)
位置关系
\(\pi_1=\pi_2\)
- 判定条件\[\large \frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2} = \frac{D_1}{D_2} \]
\(\pi_1\ //\ \pi_2\)
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判定条件
\[\large \frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}\ne \frac{D_1}{D_2} \] -
距离
\[\large d=d_{\pi_1-\pi_2} = \frac{|D_1-D_2|}{\sqrt{A^{2}+B^{2}+C^{2}}} \\ \]
\(\pi_1\ \cap\ \pi_2 = L\)
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判定条件
\[\large \vec{n_1} \ /\!/ \!\!\!\!\!\!\smallsetminus \vec{n_2} \]即,式
\[\large \frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2} \]不成立
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夹角 \(\varphi\)
\[\large \begin{aligned} \cos\varphi=\frac{\mid \vec{n_1}\cdot \vec{n_2}\mid}{\mid \vec{n_1}\mid\mid \vec{n_2}\mid} =\frac{\mid A_{1}A_{2}+B_{1}B_{2}+C_{1}C_{2}\mid}{\sqrt{A_{1}^{2}+B_{1}^{2}+C_{1}^{2}}\sqrt{A_{2}^{2}+B_{2}^{2}+C_{2}^{2}}} \end{aligned} \]【规定】\(0 \le \varphi \le \dfrac{\pi}{2}\)
其他关系
- 交线 (平面束方程)\[\large L: \begin{cases} \pi_1:A_1x+B_1y+C_1z+D_1=0 \\ \\ \pi_2:A_2x+B_2y+C_2z+D_2=0 \end{cases} \]其中 \(\pi_1\ \bot\ \pi_2\)
平面束
- \(\large L:\begin{cases}\pi_1:A_1x+B_1y+C_1z+D_1=0 \\ \pi_2:A_2x+B_2y+C_2z+D_2=0\end{cases}\)
- \(\large \pi:Ax+By+Cz+D= 0\)
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过L的平面与 \(\pi\) 的交线
\[L':\large \begin{cases}\pi':A_1x+B_1y+C_1z+D_1+\lambda(A_2x+B_2y+C_2z+D_2)=0 \\ \\\pi:Ax+By+Cz+D= 0\end{cases} \] -
过L且与 \(\pi\) 成 \(\theta\) 的平面
\[\cos \theta = \frac{\left|\vec{n_1}\cdot\vec{n_2}\right|}{\left|\vec{n_1}\right|\left|\vec{n_2}\right|} \]解上述方程,其中
\[\begin{aligned} \vec{n_1} &= (A_1 + \lambda A_2,B_1 + \lambda B_2,B_1 + \lambda B_2) \\ \vec{n_2} &= (A, B, C) \end{aligned} \]- 若 \(\lambda\) 有解,则 \(\pi'\) 为所求平面;
- 若 \(\lambda\) 无解,则 \(\pi_2\) 为所求平面。