« Line in three dimensions

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| {} \right| = 0

Example 2:

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

Solution:

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| {} \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| {} \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.