Vector Geometry

Vectors provide a convenient way for describing geometry in $\mathbf{R}^3$.

Lines

Theorem. Let $L$ be a line through a point $A: \vec{OA} = \mathbf{a}$ and parallel to a vector $\mathbf{t}$, then

\[L: (\mathbf{x} - \mathbf{a}) \times \mathbf{t} = \mathbf{0}\]

Proof.

Let $P: \vec{OP} = \mathbf{x}$ be a point on $L$. As $\vec{AP} = \vec{OP} - \vec{OA} = \mathbf{x} - \mathbf{a}$, for $\lambda \in \mathbb{R}$,

\[\mathbf{x} - \mathbf{a} = \lambda \mathbf{t}\]

We can eliminate $\lambda$ by taking cross product on both side with $\mathbf{t}$, i.e.

\[(\mathbf{x} - \mathbf{a}) \times \mathbf{t} = \mathbf{0}\]

Corollary. Let $L$ be a line through two points $A: \vec{OA} = \mathbf{a}$ and $B: \vec{OB} = \mathbf{b}$, then

\[L: (\mathbf{x} - \mathbf{a}) \times (\mathbf{b} - \mathbf{a}) = \mathbf{0}\]

or alternatively, for $\lambda \in \mathbb{R}$,

\[L: \mathbf{x} = (1 - \lambda)\mathbf{a} + \lambda\mathbf{b}\]

Proof.

$L$ is parallel to $\mathbf{t} = \vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}$ and then by the point-slope form.

Alternatively, we can write $\mathbf{x} = \mathbf{a} + \lambda (\mathbf{b} - \mathbf{a})$ to derive the second form.

Theorem. The equation $\mathbf{x} \times \mathbf{t} = \mathbf{u}$ has many solutions when $\mathbf{t} \cdot \mathbf{u} = 0$, in which is a line in direction $\mathbf{t}$ through $(\mathbf{t} \times \mathbf{u} / \vert \mathbf{t} \vert^2$, i.e.

\[\mathbf{x} = {\mathbf{t} \times \mathbf{u} \over \vert \mathbf{t} \vert^2} + \lambda \mathbf{t}\]

It has no solutions if $\mathbf{t} \cdot \mathbf{u} \not = 0$.

Proof.

By dotting with $\mathbf{t}$, we have

\[\mathbf{t} \cdot \mathbf{u} = \mathbf{t} \cdot (\mathbf{x} \times \mathbf{t}) = 0\]

which there are no solutions unless $\mathbf{t} \cdot \mathbf{u} = 0$. Geometrically, it is due to the fact that the vector product $\mathbf{x} \times \mathbf{t}$ has to be perpendicular to $\mathbf{t}$.

By crossing with $\mathbf{t}$, we have

\[\mathbf{t} \times \mathbf{u} = \mathbf{t} \times (\mathbf{x} \times \mathbf{t}) = (\mathbf{t} \cdot \mathbf{t})\mathbf{x} - (\mathbf{t} \cdot \mathbf{x})\mathbf{t}\]

Hence,

\[\mathbf{x} = {\mathbf{t} \times \mathbf{u} \over \vert \mathbf{t} \vert^2} + {(\mathbf{t} \cdot \mathbf{x}) \over \vert \mathbf{t} \vert^2 }\mathbf{t} = {\mathbf{t} \times \mathbf{u} \over \vert \mathbf{t} \vert^2} + \lambda \mathbf{t}\]

It provides a way to convert a line $L: \mathbf{x} \times \mathbf{t} = \mathbf{u}$ to other forms we know.

Planes

Theorem. Let $\Pi$ be a plane contains the point $A: \vec{OA} = \mathbf{a}$ and has normal in the direction of a vector $\mathbf{n}$, then

\[\Pi: (\mathbf{x} - \mathbf{a}) \cdot \mathbf{n} = 0\]

Proof.

Let $P: \vec{OP} = \mathbf{x}$ be a point on $\Pi$, then $\vec{PA} \perp \mathbf{n}$ and hence

\[(\mathbf{x} - \mathbf{a}) \cdot \mathbf{n} = 0\]

Corollary. Let $\Pi$ be a plane through three points $A: \vec{OA} = \mathbf{a}$, $B: \vec{OB} = \mathbf{b}$ and $C: \vec{OC} = \mathbf{c}$, then

\[\Pi: (\mathbf{x} - \mathbf{a}) \cdot [(\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a})] = [\mathbf{x} - \mathbf{a}, \mathbf{b} - \mathbf{a}, \mathbf{c} - \mathbf{a}] = 0\]

or

\[\Pi: [\mathbf{x}, \mathbf{b}, \mathbf{c}] + [\mathbf{a}, \mathbf{x}, \mathbf{c}] + [\mathbf{a}, \mathbf{b}, \mathbf{x}] = [\mathbf{a}, \mathbf{b}, \mathbf{c}]\]

or alternatively, for $\lambda, \mu \in \mathbb{R}$,

\[\Pi: \mathbf{x} = (1 - \lambda - \mu)\mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}\]

Theorem. The distance between the origin and the plane $\Pi: (\mathbf{x} - \mathbf{a}) \cdot \mathbf{\hat{n}} = 0$ is

\[d = \mathbf{a} \cdot \mathbf{\hat{n}}\]

Proof.

Let $\mathbf{x_O}$ be the position vector of the closest point on the plane $\Pi$ to the origin, we have

\[(\mathbf{x_O} - \mathbf{a}) \cdot \mathbf{\hat{n}} = 0\]

$\mathbf{x_O}$ must be orthogonal to the plane and therefore parallel to $\mathbf{\hat{n}}$, i.e.

\[\mathbf{x_O} = d\mathbf{\hat{n}}\]

where $\vert d \vert $ is the distance from the origin to the plane.

Hence,

\[d = \vert \mathbf{x_O} \cdot \mathbf{\hat{n}} \vert = \vert \mathbf{a} \cdot \mathbf{\hat{n}} \vert\]

Corollary. Let $\Pi$ be a plane which has unit normal $\mathbf{\hat{n}}$ and has distance $d$ from the origin, then

\[\Pi: \mathbf{x} \cdot \mathbf{\hat{n}} = d\]

It provides a way to convert a plane $\Pi: \mathbf{x} \cdot \mathbf{\hat{n}} = d$ to other forms we know and gives a geometric meaning to it.

Geometry Problems

Distance between Point and Line

Proposition. The distance between a point $Y: \vec{OY} = \mathbf{y}$ and a line $L: (\mathbf{x} - \mathbf{a}) \times \mathbf{t} = \mathbf{0}$ is

\[d = {|(\mathbf{y} - \mathbf{a}) \times \mathbf{t}| \over |\mathbf{t}|}\]

Proof.

Let $\theta$ be the non-reflex angle between $\vec{AY}$ and $\mathbf{t}$, we have

\[d = |\mathbf{y} - \mathbf{a}|\sin\theta\]

Since

\[\vert (\mathbf{y} - \mathbf{a}) \times \mathbf{t} \vert = \vert (\mathbf{y} - \mathbf{a}) \vert \vert \mathbf{t} \vert \sin\theta = d\vert t \vert\]

Hence,

\[d = {|(\mathbf{y} - \mathbf{a}) \times \mathbf{t}| \over |\mathbf{t}|}\]

Distance between Parallel Lines

Proposition. The distance between two parallel lines $L_1: (\mathbf{x} - \mathbf{a}) \times \mathbf{t} = \mathbf{0}$ and $L_2: (\mathbf{x} - \mathbf{b}) \times \mathbf{t} = \mathbf{0}$ is

\[d = {|(\mathbf{b} - \mathbf{a}) \times \mathbf{t}| \over |\mathbf{t}|}\]

Proof.

As $\mathbf{b}$ is a point on $L_2$, we can find the distance by substituting $\mathbf{y} = \mathbf{b}$ to the above.

Distance between Skew Lines

Definition. Two lines in $\mathbb{R}^3$ are said to be skew if they do not lie in any plane.

Proposition. The distance between two skew lines $L_1: (\mathbf{x} - \mathbf{a}) \times \mathbf{t} = \mathbf{0}$ and $L_2: (\mathbf{x} - \mathbf{b}) \times \mathbf{u} = \mathbf{0}$ is

\[d = {|(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{t} \times \mathbf{u})| \over |\mathbf{t} \times \mathbf{u}|}\]

Proof.

$L_1$ and $L_2$ are skew if they lie in a pair of parallel planes. Consider the planes

\[\Pi_1: (\mathbf{x} - \mathbf{a}) \cdot (\mathbf{t} \times \mathbf{u}) = 0 \quad \text{and} \quad \Pi_2: (\mathbf{x} - \mathbf{b}) \cdot (\mathbf{t} \times \mathbf{u}) = 0\]

$L_1$ is on $\Pi_1$ and $L_2$ is on $\Pi_2$. Also, $\Pi_1$ and $\Pi_2$ are parallel because they have common normal $\mathbf{t} \times \mathbf{u}$. Hence, the distance between $L_1$ and $L_2$ is equal to the distance between $\Pi_1$ and $\Pi_2$, which is

\[d = {|(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{t} \times \mathbf{u})| \over |\mathbf{t} \times \mathbf{u}|}\]

Intersection between Lines

Proposition. Two lines $L_1: (\mathbf{x} - \mathbf{a}) \times \mathbf{t} = \mathbf{0}$ and $L_2: (\mathbf{x} - \mathbf{b}) \times \mathbf{u} = \mathbf{0}$ intersects iff $\mathbf{t} \times \mathbf{u} \not = \mathbf{0}$ and the intersection is

\[\mathbf{x} = \mathbf{a} + {[\mathbf{t} \times \mathbf{u}, \mathbf{b} - \mathbf{a}, \mathbf{u}] \over |\mathbf{t} \times \mathbf{u}|^2}\mathbf{t}\]

Proof.

The two lines are parallel if $ \mathbf{t} \times \mathbf{u} = \mathbf{0}$. Hence, either they are the same line or there is no intersection.

If $ \mathbf{t} \times \mathbf{u} \not = \mathbf{0}$, consider a line $L_2’$ through $\mathbf{a}$ and parallel to $L_2$, $L_1$ and $L_2’$ forms a plane $\Pi$ with normal vector $\mathbf{t} \times \mathbf{u}$. Therefore, we have

\[\Pi: (\mathbf{x} - \mathbf{a}) \cdot (\mathbf{t} \times \mathbf{u}) = 0\]

As $L_2$ is parallel to $L_2’$, either $L_2$ intersects $\Pi$ nowhere (in which $L_1$ and $L_2$ has no intersection) or $L_2$ is on $\Pi$. Therefore, the condition for $L_1$ and $L_2$ to intersect is

\[(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{t} \times \mathbf{u}) = 0\]

Conversely, if the above is true, the plane $\Pi$ passes through the origin as the distance of the plane from origin is $0$, and we can express $\mathbf{b} - \mathbf{a}$ as

\[\mathbf{b} - \mathbf{a} = \mathbf{0} + \lambda \mathbf{t} + \mu \mathbf{u}\]

for some $\lambda, \mu \in \mathbb{R}$.

Consier

\[\mathbf{x} = \mathbf{a} + \lambda\mathbf{t} = \mathbf{b} - \mu\mathbf{u}\]

$\mathbf{x}$ is on both $L_1$ and $L_2$ and therefore it is the intersection of $L_1$ and $L_2$.

By some vector manipulations, we have

\[\begin{gather} \mathbf{a} + \lambda\mathbf{t} = \mathbf{b} - \mu\mathbf{u} \\ \lambda(\mathbf{t} \times \mathbf{u}) = (\mathbf{b} - \mathbf{a}) \times \mathbf{u} \\ \lambda |\mathbf{t} \times \mathbf{u}|^2 = [\mathbf{t} \times \mathbf{u}, \mathbf{b} - \mathbf{a}, \mathbf{u}] \\ \end{gather}\]

Hence,

\[\mathbf{x} = \mathbf{a} + {[\mathbf{t} \times \mathbf{u}, \mathbf{b} - \mathbf{a}, \mathbf{u}] \over |\mathbf{t} \times \mathbf{u}|^2}\mathbf{t}\]

Distance between Point and Plane

Proposition. The distance between a point $Y: \vec{OY} = \mathbf{y}$ and a plane $\Pi: (\mathbf{x} - \mathbf{a}) \cdot \mathbf{\hat{n}} = 0$ is

\[d = \vert \mathbf{a} \cdot \mathbf{n} - \mathbf{y} \cdot \mathbf{\hat{n}} \vert\]

Proof.

$P: \vec{OP} = \mathbf{y} + d\mathbf{\hat{n}}$ is a point on $\Pi$, where $\vert d \vert$ is the distance from $Y$ to $\Pi$, i.e.

\[(\mathbf{y} + d\mathbf{\hat{n}} - \mathbf{a}) \cdot \mathbf{\hat{n}} = 0\]

Hence,

\[d = \vert \mathbf{a} \cdot \mathbf{\hat{n}} - \mathbf{y} \cdot \mathbf{\hat{n}} \vert\]

Distance between Parallel Planes

Proposition. The distance between two parallel planes $\Pi_1: (\mathbf{x} - \mathbf{a}) \cdot \mathbf{\hat{n}} = 0$ and $\Pi_2: (\mathbf{x} - \mathbf{b}) \cdot \mathbf{\hat{n}} = 0$ is

\[d = \vert (\mathbf{b} - \mathbf{a}) \cdot \mathbf{\hat{n}} \vert\]

Proof.

$\mathbf{b}$ is a point on $\Pi_2$ hence

\[d = \vert \mathbf{b} \cdot \mathbf{\hat{n}} - \mathbf{a} \cdot \mathbf{\hat{n}} \vert = \vert (\mathbf{b} - \mathbf{a}) \cdot \mathbf{\hat{n}} \vert\]

Intersection between Line and Plane

Proposition. Given a line $L: (\mathbf{x} - \mathbf{a}) \times \mathbf{t} = \mathbf{0}$ and a plane $\Pi: (\mathbf{x} - \mathbf{b}) \cdot \mathbf{n} = 0$. If $\mathbf{n} \cdot \mathbf{t} = 0$, there is either no intersection or $L$ is on the plane $\Pi$. If $\mathbf{n} \cdot \mathbf{t} \not = 0$, then the intersection is

\[\mathbf{x} = \mathbf{a} + {\mathbf{n} \cdot \mathbf{b} - \mathbf{n} \cdot \mathbf{a} \over \mathbf{n} \cdot \mathbf{t}} \mathbf{t}\]

Proof.

If $\mathbf{n} \cdot \mathbf{t} = 0$, $L$ and $\Pi$ are parallel. If $\mathbf{a}$ is on $\Pi$, i.e. $(\mathbf{a} - \mathbf{b}) \cdot \mathbf{n} = 0$, then $L$ is on $\Pi$, otherwise, there is no intersection.

If $\mathbf{n} \cdot \mathbf{t} \not = 0$, by crossing by $\mathbf{n}$ on both side, we have

\[\begin{align*} \mathbf{n} \times (\mathbf{x} \times \mathbf{t}) &= \mathbf{n} \times (\mathbf{a} \times \mathbf{t}) \\ (\mathbf{n} \cdot \mathbf{t}) \mathbf{x} - (\mathbf{n} \cdot \mathbf{x}) \mathbf{t} &= (\mathbf{n} \cdot \mathbf{t}) \mathbf{a} - (\mathbf{n} \cdot \mathbf{a}) \mathbf{t} \\ (\mathbf{n} \cdot \mathbf{t}) \mathbf{x} &= (\mathbf{n} \cdot \mathbf{t}) \mathbf{a} + (\mathbf{n} \cdot \mathbf{b}) \mathbf{t} - (\mathbf{n} \cdot \mathbf{a}) \mathbf{t} \\ \end{align*}\]

Hence,

\[\mathbf{x} = \mathbf{a} + {\mathbf{n} \cdot \mathbf{b} - \mathbf{n} \cdot \mathbf{a} \over \mathbf{n} \cdot \mathbf{t}} \mathbf{t}\]

Intersection of Two Planes (Line)

Proposition. Given two planes $\Pi_1: \mathbf{x} \cdot \mathbf{n_1} = d_1$ and $\Pi_2: \mathbf{x} \cdot \mathbf{n_2} = d_2$. If $\mathbf{n_1} \times \mathbf{n_2} = \mathbf{0}$, then there are no intersections unless $d_1 = d_2$. If $\mathbf{n_1} \times \mathbf{n_2} \not = \mathbf{0}$, then the intersections are

\[\mathbf{x} \times (\mathbf{n_1} \times \mathbf{n_2}) = d_2\mathbf{n_1} - d_1\mathbf{n_2}\]

Proof.

Their line of intersection $L$ must be in the direction that is orthogonal to both $\mathbf{n_1}$ and $\mathbf{n_2}$, i.e. $\mathbf{n_1} \times \mathbf{n_2}$.

Therefore, the equation of $L$ is of the form

\[\mathbf{x} \times (\mathbf{n_1} \times \mathbf{n_2}) = \mathbf{c}\]

Expanding the L.H.S. we have

\[\begin{align*} \mathbf{c} &= \mathbf{x} \times (\mathbf{n_1} \times \mathbf{n_2}) \\ &= (\mathbf{x} \cdot \mathbf{n_2}) \mathbf{n_1} - (\mathbf{x} \cdot \mathbf{n_1}) \mathbf{n_n} \\ \end{align*}\]

Hence, as $\mathbf{x}$ is on $\Pi_1$ and $\Pi_2$, the intersections are

\[\mathbf{x} \times (\mathbf{n_1} \times \mathbf{n_2}) = d_2\mathbf{n_1} - d_1\mathbf{n_2}\]

Intersection of Three Planes (Point)

Proposition. Given three planes $\Pi_1: \mathbf{x} \cdot \mathbf{n_1} = d_1$, $\Pi_2: \mathbf{x} \cdot \mathbf{n_2} = d_2$, $\Pi_3: \mathbf{x} \cdot \mathbf{n_3} = d_3$. If $[\mathbf{n_1}, \mathbf{n_2}, \mathbf{n_3}] \not = 0$, their intersection is

\[\mathbf{x} = {d_1(\mathbf{n_2} \times \mathbf{n_3}) + d_2(\mathbf{n_3} \times \mathbf{n_1}) + d_3(\mathbf{n_1} \times \mathbf{n_2}) \over [\mathbf{n_1}, \mathbf{n_2}, \mathbf{n_3}]}\]

Proof.

For the three planes to intersect at one point, their normals can’t be co-planar, i.e. $[\mathbf{n_1}, \mathbf{n_2}, \mathbf{n_3}] \not = 0$.

As any point $\mathbf{p}$ in the 3D space can be expressed as

\[\mathbf{p} = {\mathbf{p} \cdot \mathbf{n_1} \over [\mathbf{n_1}, \mathbf{n_2}, \mathbf{n_3}]} (\mathbf{n_2} \times \mathbf{n_3}) + {\mathbf{p} \cdot \mathbf{n_2} \over [\mathbf{n_1}, \mathbf{n_2}, \mathbf{n_3}]} (\mathbf{n_3} \times \mathbf{n_1}) + {\mathbf{p} \cdot \mathbf{n_3} \over [\mathbf{n_1}, \mathbf{n_2}, \mathbf{n_3}]} (\mathbf{n_1} \times \mathbf{n_2})\]

Hence, their intersection is

\[\mathbf{x} = {d_1(\mathbf{n_2} \times \mathbf{n_3}) + d_2(\mathbf{n_3} \times \mathbf{n_1}) + d_3(\mathbf{n_1} \times \mathbf{n_2}) \over [\mathbf{n_1}, \mathbf{n_2}, \mathbf{n_3}]}\]

References