forked from logzhan/NotesUESTC
33 lines
864 B
Markdown
33 lines
864 B
Markdown
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数学基础
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1、从点到面的距离推导
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求三维空间内点$p(x_p,y_p,z_p)$到平面$Ax+By+Cz+D$的距离。
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很容易知道，平面方程$Ax+By+Cz+D=0$的法向量是$\vec n = (A,B,C)$。对于过空间点p且方向向量为$\vec n$的直线为：
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$$
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\frac{x-x_P}{A}=\frac{y-y_p}{B}=\frac{z-z_p}{C}
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$$
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联立方程组：
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$$
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\frac{x-x_P}{A}=\frac{y-y_p}{B}=\frac{z-z_p}{C} \\
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Ax+By+Cz+D=0
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$$
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假设三维直线和三维平面的交点为$(x_0,y_0,z_0)$
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$$
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\frac{x_0-x_P}{A}=\frac{y_0-y_p}{B} \\
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\frac{y_0-y_p}{B}=\frac{z_0-z_p}{C} \\
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Ax_0+By_0+Cz_0+D_0=0
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$$
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那么根据距离公式：
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$$
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d=\sqrt{(x_0-x_p)^2+(y_0-y_p)^2+(z_0-z_p)^2}\\
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d=\sqrt{(\frac{A}{B})^2\cdot(y_0-y_p)^2+(y_0-y_p)^2+(\frac{C}{B})^2\cdot(y_0-y_p)^2}\\
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d^2=\frac{A^2\cdot(y_0-y_p)^2}{B^2}+(y_0-y_p)^2+\frac{C^2}{B^2}\cdot(y_0-y_p)^2
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=\frac{(A^2+B^2+C^2)}{B^2}\cdot(y_0-y_p)^2
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$$
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