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Point direction form:
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$$ a(x-x_1) + b(y-y_1) + c(z-z_1) = 0 $$ |
where $P(x_1, y_1, z_1)$ lies in the plane, and the direction $(a,b,c)$ is normal to the plane.
General form:
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$$ Ax + By + Cz + D = 0 $$ |
where direction $(A,B,C)$ is normal to the plane.
Intercept form:
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$$ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 $$ |
this plane passes through the points $(a,0,0), (0,b,0)$ and $(0,0,c)$.
Three point form:
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$$ \begin{vmatrix} x-x_3 & y-y_3 & z-z_3 \\ x_1-x_3 & y_1-y_3 & z_1-z_3 \\ x_2-x_3 & y_2-y_3 & z_2-z_3 \end{vmatrix} = 0 $$ |
Normal form:
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$$ x\,\cos \alpha + y\,\cos\beta + z\,\cos\gamma = p $$ |
Parametric form:
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$$ \begin{aligned} x &= x_1 + a_1\,s + a_2\,t \\ y &= y_1 + b_1\,s + b_2\,t \\ z &= z_1 + c_1\,s + c_2\,t \end{aligned} $$ |
where the directions $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are parallel to the plane.
The angle between planes $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$ is:
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$$ \alpha = \arccos \frac{A_1A_2 + B_1B_2 + C_1C_2} {\sqrt{A_1^2 + B_1^2 + C_1^2} \cdot \sqrt{A_2^2 + B_2^2 + C_2^2}} $$ |
The planes are parallel if and only if
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$$ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} $$ |
The equation of a plane through $P_1(x_1, y_1, z_1)$ and parallel to directions $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ has an equation:
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$$ \begin{vmatrix} x-x_1 & y - y_1 & z - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0 $$ |
The equation of a plane through $P_1(x_1, y_1, z_1)$ and$P_1(x_2, y_2, z_2)$), and parallel to direction $(a,b,c)$, has equation
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$$ \begin{vmatrix} x-x_1 & y - y_1 & z - z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a & b & c \end{vmatrix} = 0 $$ |
The equation of a plane through $P_1(x_1, y_1, z_1)$ , $P_2(x_2, y_2, z_2)$ and $P_3(x_3, y_3, z_3)$ , has equation
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$$ \begin{vmatrix} x-x_1 & y - y_1 & z - z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0 $$ |
The distance of $P_1(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is
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$$ d = \frac{Ax_1 + By_1 + Cz_1}{\sqrt{A^2 + B^2 + C^2}} $$ |
The intersection of planes $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0 $ is the line:
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$$ \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} $$ |
where
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$$ \begin{aligned} a &= \begin{vmatrix} B_1 & C_1 \\ B_2 & C_2 \end{vmatrix}~~ b = \begin{vmatrix} C_1 & A_1 \\ C_2 & A_2 \end{vmatrix}~~ c = \begin{vmatrix} A_1 & B_1 \\ A_2 & B_2 \end{vmatrix} \\ x_1&= \frac{b\begin{vmatrix}D_1& C_1 \\ D_2 & C_2 \end{vmatrix} - c\begin{vmatrix}D_1& B_1 \\ D_2 & B_2 \end{vmatrix} }{a^2 + b^2 + c^2} \\ y_1&= \frac{c\begin{vmatrix}D_1& A_1 \\ D_2 & A_2 \end{vmatrix} - a\begin{vmatrix}D_1& C_1 \\ D_2 & C_2 \end{vmatrix} }{a^2 + b^2 + c^2} \\ z_1&= \frac{a\begin{vmatrix}D_1& B_1 \\ D_2 & B_2 \end{vmatrix} - b\begin{vmatrix}D_1& A_1 \\ D_2 & A_2 \end{vmatrix} }{a^2 + b^2 + c^2} \end{aligned} $$ |
If $ a = b = c = 0 $, then the planes are parallel.
Please tell me how can I make this better.