Vector Geometry of Lines and Planes

Vector Geometry of Lines

Definition Parallel (Collinear) Vectors

Two vectors $\Vect{u}$ and $\Vect{v}$ are parallel (or collinear) if one is a scalar multiple of the other. That is, if there exists a real number $k$ such that:$$\Vect{u} = k \Vect{v}.$$

Proposition Alignment

Three points $A, B$, and $C$ are aligned if and only if the vectors $\Vect{AB}$ and $\Vect{AC}$ are collinear.

Definition Characterizing a Line

A line $d$ is defined by a point $A$ and a direction vector $\Vect{u}$ ($\Vect{u} \neq \Vect{0}$). We denote it $d(A, \Vect{u})$.
A point $M$ belongs to $d$ if there exists a real $k$ such that $\Vect{AM} = k\Vect{u}$.

Proposition Relative Position of Two Lines

Let $d(A, \Vect{u})$ and $d'(B, \Vect{v})$ be two lines in 3D space.

If $\Vect{u}$ and $\Vect{v}$ are collinear, the lines are parallel (either strictly parallel or coincident).
If $\Vect{u}$ and $\Vect{v}$ are not collinear:
- They are secant if they have a common point (they are then coplanar).
- They are skew (non-coplanar) if they have no common point.

Vector Geometry of Planes

Definition Coplanar Vectors

Three vectors $\Vect{u}, \Vect{v}$, and $\Vect{w}$ are coplanar if one is a linear combination of the others. That is, if there exist real numbers $x$ and $y$ such that:$$\Vect{w} = x \Vect{u} + y \Vect{v}.$$

Proposition Coplanarity of Four Points

Four points $A, B, C$, and $D$ belong to the same plane if and only if the vectors $\Vect{AB}, \Vect{AC}$, and $\Vect{AD}$ are coplanar.

Definition Characterizing a Plane

A plane $\mathscr{P}$ is defined by a point $A$ and two non-collinear direction vectors $\Vect{u}$ and $\Vect{v}$. We denote it $\mathscr{P}(A, \Vect{u}, \Vect{v})$. A point $M$ belongs to $\mathscr{P}$ if there exist reals $x$ and $y$ such that $\Vect{AM} = x\Vect{u} + y\Vect{v}$.

Proposition Relative Position of a Line and a Plane

Let $d(A, \Vect{u})$ be a line and $\mathscr{P}(B, \Vect{v}, \Vect{w})$ be a plane.

The line $(d)$ is parallel to the plane $\mathscr{P}$ if $\Vect{u}, \Vect{v}$, and $\Vect{w}$ are coplanar.
Otherwise, the line and the plane are secant at exactly one point.

Proposition Relative Position of Two Planes

Two planes are parallel if and only if the direction vectors of one are linear combinations of the direction vectors of the other.
Otherwise, they are secant, and their intersection is always a line.

Normal Vectors to a Plane

Definition Normal Vector

Given a plane $\mathscr{P}$ defined by a point $A$ and two direction vectors $\Vect{u}$ and $\Vect{v}$, a non-zero vector $\Vect{n}$ is said to be normal to the plane if it is orthogonal to both $\Vect{u}$ and $\Vect{v}$.

Equivalently, $\Vect{n}$ is normal to $\mathscr{P}$ if for every point $M$ in the plane, the vectors $\Vect{AM}$ and $\Vect{n}$ are orthogonal.

Proposition Perpendicular Planes

Two planes $\mathscr{P}_1$ and $\mathscr{P}_2$ are perpendicular if and only if their respective normal vectors $\Vect{n_1}$ and $\Vect{n_2}$ are orthogonal.$$\mathscr{P}_1 \perp \mathscr{P}_2 \iff \Vect{n_1} \cdot \Vect{n_2} = 0$$

Example

In a cube $ABCDEFGH$, consider the planes $\mathscr{P}(A, \Vect{AB}, \Vect{AC})$ and $\mathscr{P}(A, \Vect{AD}, \Vect{AE})$.

A normal vector to plane $\mathscr{P}(A, \Vect{AB}, \Vect{AC})$ is $\Vect{AE}$ (since it is perpendicular to $\Vect{AB}$ and $\Vect{AC}$).
A normal vector to plane $\mathscr{P}(A, \Vect{AD}, \Vect{AE})$ is $\Vect{AB}$ (since it is perpendicular to edges $\Vect{AD}$ and $\Vect{AE}$).
Since $ABCDEFGH$ is a cube, $\Vect{AE} \cdot \Vect{AB} = 0$.

Therefore, the planes $\mathscr{P}(A, \Vect{AB}, \Vect{AC})$ and $\mathscr{P}(A, \Vect{AD}, \Vect{AE})$ are perpendicular.

Cartesian Equation of a Plane

Proposition Cartesian Equation

In an orthonormal coordinate system, the plane passing through point $A(x_A, y_A, z_A)$ with a normal vector $\Vect{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ has a Cartesian equation of the form:$$ax + by + cz + d = 0$$where $(a, b, c) \neq (0, 0, 0)$ and $d$ is a real constant.

Proof

A point $M(x, y, z)$ belongs to the plane if and only if $\Vect{AM} \cdot \Vect{n} = 0$.$$ \begin{aligned}\Vect{AM} \cdot \Vect{n} = 0 &\iff a(x - x_A) + b(y - y_A) + c(z - z_A) = 0 \\ &\iff ax + by + cz - (ax_A + by_A + cz_A) = 0\end{aligned} $$By setting $d = -(ax_A + by_A + cz_A)$, we obtain the form $ax + by + cz + d = 0$.

Example

Find the Cartesian equation of the plane passing through $A(-1, 2, 3)$ with normal vector $\Vect{n}=\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$.

Answer

Method 1: The equation is $2x - 3y + z + d = 0$. Since $A \in \mathscr{P}$: $$\begin{aligned} 2(-1) - 3(2) + 3 + d &= 0\\ -2 - 6 + 3 + d &= 0\\ d &= 5\end{aligned}$$Equation: $2x - 3y + z + 5 = 0$.
Method 2: $$\begin{aligned}\Vect{AM} \cdot \Vect{n} = 0 &\iff 2(x - (-1)) - 3(y - 2) + 1(z - 3) = 0\\ &\iff 2x - 3y + z + 5 = 0\end{aligned}$$

Distance from a Point to a Plane

Definition Orthogonal Projection

The orthogonal projection of a point $M$ onto a plane $\mathscr{P}$ is the point $H$, where $H$ is the intersection of the plane and the line passing through $M$ perpendicular to $\mathscr{P}$.

Definition Distance from a Point to a Set

The distance between a point $M$ and a set of points $\mathcal{S}$ is the minimum length $MA$ among all possible points $A$ belonging to $\mathcal{S}$.

Proposition Shortest Distance

The distance from a point $M$ to a plane $\mathscr{P}$ is the length $MH$, where $H$ is the orthogonal projection of $M$ onto $\mathscr{P}$.

Proof

Let $H$ be the orthogonal projection of point $M$ onto the plane $\mathscr{P}$, and let $A$ be any point belonging to the plane. We can decompose the vector $\Vect{MA}$ using Chasles' relation:$$\Vect{MA} = \Vect{MH} + \Vect{HA}$$By calculating the square of the norm (the scalar square), we obtain:$$ \begin{aligned}\|\Vect{MA}\|^2 &= (\Vect{MH} + \Vect{HA}) \cdot (\Vect{MH} + \Vect{HA}) \\ &= \|\Vect{MH}\|^2 + 2(\Vect{MH} \cdot \Vect{HA}) + \|\Vect{HA}\|^2\end{aligned} $$Since the line $(MH)$ is perpendicular to the plane $\mathscr{P}$ and the vector $\Vect{HA}$ lies within the plane, the vectors $\Vect{MH}$ and $\Vect{HA}$ are orthogonal. Their scalar product is therefore zero: $\Vect{MH} \cdot \Vect{HA} = 0$. The expression simplifies to:$$\|\Vect{MA}\|^2 = \|\Vect{MH}\|^2 + \|\Vect{HA}\|^2$$Since $\|\Vect{HA}\|^2 \geq 0$, it follows that $\|\Vect{MA}\|^2 \geq \|\Vect{MH}\|^2$, and thus $MA \geq MH$. This confirms that $MH$ is the minimum distance between $M$ and any point in the plane.