12.5 - Lines and planes in Space

Line Defined by a Reference Point and Direction Vector

Vector form: $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v} = \langle x_0 + at, y_0 + bt, z_0 + ct \rangle$. Parametric form: $x(t) = x_0 + at$, $y(t) = y_0 + bt$, $z(t) = z_0 + ct$. Symmetric form (for $a, b, c \neq 0$): $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$, where $P_0(x_0, y_0, z_0)$ is a reference point and $\mathbf{v} = \langle a, b, c \rangle$ is the direction vector.

Line Passing Through Two Points

Given points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$: 1. Direction vector: $\mathbf{v} = \overrightarrow{PQ} = \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle$. 2. Parametric equations: $x(t) = x_1 + (x_2 - x_1)t$, $y(t) = y_1 + (y_2 - y_1)t$, $z(t) = z_1 + (z_2 - z_1)t$. 3. Symmetric equations: $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$.

Parameterization of a Finite Line Segment

For a line segment joining initial point $P$ to terminal point $Q$: $\mathbf{r}(t) = (1 - t)\mathbf{r}_P + t\mathbf{r}_Q = \mathbf{r}_P + t\overrightarrow{PQ}$, with the parameter restricted to the interval $0 \le t \le 1$.

Condition for Coplanar Lines

Two lines with direction vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ through base points $P_1$ and $P_2$ are coplanar if and only if their scalar triple product is zero: $(\mathbf{v}_1 \times \mathbf{v}_2) \cdot \overrightarrow{P_1P_2} = 0$.

Plane Defined by a Point and a Normal Vector

Vector equation: $\mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0$. Standard component equation: $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$. General linear equation: $Ax + By + Cz + D = 0$, where $D = -(Ax_0 + By_0 + Cz_0)$, for point $P_0(x_0, y_0, z_0)$ and normal vector $\mathbf{n} = \langle A, B, C \rangle$.

Plane Containing Three Non-Collinear Points

Given non-collinear points $A$, $B$, and $C$: 1. Determine two coplanar vectors: $\mathbf{u} = \overrightarrow{AB}$ and $\mathbf{w} = \overrightarrow{AC}$. 2. Compute the normal vector: $\mathbf{n} = \mathbf{u} \times \mathbf{w} = \langle A, B, C \rangle$. 3. Apply standard component form with point $A(x_0, y_0, z_0)$: $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$.

Distance from a Point to a Line

The shortest distance $d$ from point $S$ to a line passing through point $P$ with direction vector $\mathbf{v}$ is given by: $d = \frac{\|\overrightarrow{PS} \times \mathbf{v}\|}{\|\mathbf{v}\|}$.

Distance from a Point to a Plane

The shortest distance $d$ from point $S(x_0, y_0, z_0)$ to plane $Ax + By + Cz + D = 0$ is given by: $d = \frac{\|Ax_0 + By_0 + Cz_0 + D\|}{\sqrt{A^2 + B^2 + C^2}}$.

Angle Between Two Planes

The acute dihedral angle $\theta$ between two planes with normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$ satisfies: $\cos(\theta) = \frac{\|\mathbf{n}_1 \cdot \mathbf{n}_2\|}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|}$, giving $\theta = \arccos\left(\frac{\|\mathbf{n}_1 \cdot \mathbf{n}_2\|}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|}\right)$.

Intersection of a Line and a Plane

1. Substitute the line parametric expressions $x(t)$, $y(t)$, $z(t)$ into the plane equation $Ax + By + Cz + D = 0$. 2. Solve the resulting equation for the scalar parameter $t$. 3. Substitute $t$ back into the line equations to obtain the coordinates of the intersection point.

Kinematic Modeling of Uniform 3D Linear Motion

For linear motion at constant speed $v$ from initial position $\mathbf{r}_0$ along vector $\mathbf{d}$: 1. Unit direction vector: $\mathbf{u} = \frac{\mathbf{d}}{\|\mathbf{d}\|}$. 2. Position vector at time $t$: $\mathbf{r}(t) = \mathbf{r}_0 + (v \times t)\mathbf{u}$.

Intersection Line of Two Planes

For two intersecting planes with normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$: 1. Direction vector of the line: $\mathbf{v} = \mathbf{n}_1 \times \mathbf{n}_2$. 2. Find a common base point $P_0$ by setting one coordinate to zero and solving the system for the remaining two coordinates. 3. Assemble the parametric equations of the line using $P_0$ and $\mathbf{v}$.