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« Line in three dimensions
Analytic geometry of three dimensions: (lesson 2 of 2)

Planes

Plane through $(x_0, y_0, z_0)$ and perpendicular to the direction $(a, b, c)$:

$a (x - x_0) + b (y - y_0) + c (z - z_0) = 0$

Example 1

Find the equation for the plane through the point $(0, 1, 2)$ perpendicular to the vector $(2, 1, -3)$.

Solution:

$(x_0, y_0, z_0) = (0, 1, 2)$

$(a, b, c) = (2, 1, -3)$

The plane: $2 (x - 0) + 1 (y - 1) - 3 (z - 2) = 0$

$2x + y -3z = -5$.

Plane through $(x_0, y_0, z_0)$, $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:

$$ \left| {\begin{array}{*{20}{c}} {x - {x_0}}&{y - {y_0}}&{z - {z_0}}\\ {{x_1} - {x_0}}&{{y_1} - {y_0}}&{{z_1} - {z_0}}\\ {{x_2} - {x_0}}&{{y_2} - {y_0}}&{{z_2} - {z_0}} \end{array}} \right| = 0 $$

Example 2:

Find the equation for the plane through the points $(0, 1, 2)$, $(2, 1, 3)$ and $(3, 1, 0)$

Solution:

$$ \begin{aligned} &\left| {\begin{array}{*{20}{c}} {x - 0}&{y - 1}&{z - 2}\\ {2 - 0}&{1 - 1}&{3 - 2}\\ {3 - 0}&{1 - 1}&{0 - 2} \end{array}} \right| = 0 \\ &\left| {\begin{array}{*{20}{c}} x&{y - 1}&{z - 2}\\ 2&0&1\\ 3&0&{ - 2} \end{array}} \right| = 0 \\ &3(y - 1) + 4(y - 1) = 0 \\ &y = 1 \\ &0x + 1y + 0z = 0 \end{aligned} $$

Plane through $(x_0, y_0, z_0)$ and parallel to the vectors $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$:

$$ \left| {\begin{array}{*{20}{c}} {x - {x_0}}&{y - {y_0}}&{z - {z_0}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 0 $$

Plane through $(x_0, y_0, z_0)$ and $(x_1, y_1, z_1)$ and parallel to the direction $(a, b, c)$:

$$ \left| {\begin{array}{*{20}{c}} {x - {x_0}}&{y - {y_0}}&{z - {z_0}}\\ {{x_1} - {x_0}}&{{y_1} - {y_0}}&{{z_1} - {z_0}}\\ a&b&c \end{array}} \right| = 0 $$

The distance from the point $(x_0, y_0, z_0)$ to the plane $ax + by + cz + d = 0$ is

$$ d = \frac{{\left| {a{x_0} + b{x_0} + c{z_0}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }} $$

The angle between two planes $a_0 x + b_0 y +c_0 z + d_0 = 0$ and $a_1 x + b_1 y +c_1 z + d_1 = 0$ is

$$ \varphi = \arccos \frac{{a0a1 + b0b1 + c0c1}}{{\sqrt {{a_0}^2 + {b_0}^2 + {c_0}^2} \sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} }} $$

Two planes are parallel if their normal vectors are parallel (constant multiples of one another). It is easy to recognize parallel planes written in the form $ax + by + cz = d$, since a quick comparison of the normal vectors $n=< a, b, c >$ can be made.