We will now learn about points, lines and planes in $\mathbb{R}^3$. The purpose of this section is to help you understand the geometrical objets both in terms of the equations that describe them as well as to visualize what they look like.

• $p=(p_x,p_y,p_z)$: a point in $\mathbb{R}^3$.
• $\vec{v}=(v_x,v_y,v_z)$: a vector in $\mathbb{R}^3$.
• $\hat{v}=\frac{ \vec{v} }{ |\vec{v}| }$: a unit vector in the direction of $\vec{v}$.
• $\ell: \{ p_o+t\:\vec{v}, t \in \mathbb{R} \}$:

the equation of a line with direction vector $\vec{v}$

  passing through the point $p_o$.
* $ \ell: \left\{ \frac{x - p_{0x}}{v_x} = \frac{y - p_{0y}}{v_y} = \frac{z - p_{0z}}{v_z} \right\}$:
  the symmetric equation of the line $\ell$.
* $P: \{ (x,y,z) \in \mathbb{R}^3 \ | \ (x,y,z)=p_o+s\:\vec{v} + t\:\vec{w}, \ s,t \in \mathbb{R} \}$:
  the //parametric// equation of a plane $P$.
* $P: \left\{ (x,y,z) \in \mathbb{R}^3 \ | \ \vec{n} \cdot [ (x,y,z) - p_o ] = 0 \right\}$. 
  the //geometric// equation of a plane
  which contains $p_o$ and has normal vector $\hat{n}$.
* $P: \left\{ Ax+By+Cz=D \right\}$: the //general// equation of a plane.
* $d(a,b)$: the shortest //distance// between two objects $a$ and $b$.

We can specify a point in $\mathbb{R}^3$ by its coordinates $p=(p_x,p_y,p_z)$, which is similar to how we specify vectors. In fact the two notions are equivalent: we can either talk about the destination point $p$ or the vector $\vec{p}$ that takes us from the origin to the point $p$. By this equivalence, it makes sense to add vectors and points.

We can also specify a point as the intersection of two lines. For example in $\mathbb{R}^2$ we can describe $p$ as the intersection of the lines $x + 2y = 5$ and $3x + 9y = 21$. To find the point, $p$ we would have to solve these equations in parallel. In other words, we are looking for a point which lies on both lines. The answer is the point $p=(1,2)$.

In three dimensions, a point can also be specified as the intersection of three planes. Indeed, this is precisely what is going on when we are solving equations of the form $A\vec{x}=\vec{b}$ with $A \in \mathbb{R}^{3 \times 3}$ and $\vec{b} \in \mathbb{R}^{3}$. We are looking for some $\vec{x}$ that is lies in all three planes.

A line $\ell$ is a one-dimensional space that is infinitely long. There are a number of ways to specify the equation of a line.

The parametric equation of a line is obtained as follows. Given a direction vector $\vec{v}$ and some point $p_o$ on the line, we can define the line as: \[ \ell: \ \{ (x,y,z) \in \mathbb{R}^3 \ | \ (x,y,z)=p_o+t\:\vec{v}, t \in \mathbb{R} \}. \] We say the line is parametrized by the variable $t$. The line consists of all the points $(x,y,z)$ which can be obtained starting from the point $p_o$ and adding any multiple of the direction vector $\vec{v}$.

The symmetric equation is an equivalent way for describing a line that does not require an explicit parametrization. Consider the equation that corresponds to each of the coordinates in the equation of the line: \[ x = p_{0x} + t\:v_x, \quad y = p_{0y} + t\:v_y, \quad z = p_{0z} + t\:v_z. \] When we solve for $t$ in each of these equations and equate the results, we obtain the symmetric equation for a line: \[ \ell: \ \left\{ \ \frac{x - p_{0x}}{v_x} = \frac{y - p_{0y}}{v_y} = \frac{z - p_{0z}}{v_z} \right\}, \] in which the parameter $t$ does not appear at all. The symmetric equation specifies the line as the relationship between the $x$,$y$ and $z$ coordinates that holds for all the points on the line.

You are probably most familiar with this type of equation in the special case $\mathbb{R}^2$ when there is no $z$ variable. For non-vertical lines, we can think of $y$ as being a function of $x$ and write the line the equivalent form: \[ \frac{x - p_{0x}}{v_x} = \frac{y - p_{0y}}{v_y}, \qquad \Leftrightarrow \qquad y(x) = mx + b, \] where $m=\frac{v_y}{v_x}$ and $b=p_{oy}-\frac{v_y}{v_x}p_{ox}$, assuming $v_x \neq 0$. This makes sense intuitively, since we always thought of the slope $m$ as the “rise over run”, i.e., how much of the line goes in the $y$ direction divided by how much the line goes in the $x$ direction.

Another way to describe a line is to specify two points that are part of the line. The equation of a line that contains the points $p$ and $q$ can be obtained as follows: \[ \ell: \ \{ \vec{x}=p+t \: (p-q), \ t \in \mathbb{R} \}, \] where $(p-q)$ plays the role of the direction vector $\vec{v}$ of the line. We said any vector could be used in the definition so long as it is in the same direction as the line: $\vec{v}=p-q$ certainly can play that role since $p$ and $q$ are two points on the line.

In three dimensions, the intersection of two planes forms a line. The equation of the line corresponds to the solutions of the equation $A\vec{x}=\vec{b}$ with $A \in \mathbb{R}^{2 \times 3}$ and $\vec{b} \in \mathbb{R}^{2}$.

A plane $P$ in $\mathbb{R}^3$ is a two-dimensional space with infinite extent. The orientation of the plane is specified by a normal vector $\vec{n}$, which is perpendicular to the plane.

A plane is specified as the subspace that contains all the vectors that are orthogonal to the plane's normal vector $\vec{n}$ and contain the point $p_o$. The formula in compact notation is \[ P: \ \ \vec{n} \cdot [ (x,y,z) - p_o ] = 0. \] Recall that the dot product of two vectors is zero if and only if these vectors are orthogonal. In the above equation, the expression $[(x,y,z) - p_o]$ forms an arbitrary vector with one endpoint at $p_o$. From all these vectors we select only those that are perpendicular to the $\vec{n}$, and thus we obtain all the points of the plane.

If we expand the above formula, we obtain the general equation of the plane: \[ P: \ \ Ax + By + Cz = D, \] where $A = n_x, B=n_y, C=n_z$ and $D = \vec{n} \cdot p_o = n_xp_{0x} + n_yp_{0y} + n_yp_{oz}$.

We can also give a parametric description of a plane $P$, provided we have some point $p_o$ in the plane and two linearly independent vectors $\vec{v}$ and $\vec{w}$ which lie inside the plane: \[ P: \ \{ (x,y,z) \in \mathbb{R}^3 \ | \ (x,y,z)=p_o+s\:\vec{v} + t\:\vec{w}, \ s,t \in \mathbb{R} \}. \] Note that since a plane is two-dimensional, we need two parameters $s$ and $t$ to describe it.

Suppose we're given three points $p$, $q$, and $r$ that lie in the plane. Can you find the equation for this plane in the form: $\vec{n} \cdot [ (x,y,z) - p_o ] = 0$? We can use the point $p$ as the point $p_o$, but how do we find the normal vector $\vec{n}$ for that plane. The trick is to use the cross product. First we build two vectors that lie in the plane $\vec{v} = q-p$ and $\vec{w} = r-p$ and then to find a vector that is perpendicular to them we compute: \[ \vec{n} = \vec{v} \times \vec{w} = (q - p) \times ( r - p ). \] We can then write down the equation of a plane $\vec{n} \cdot [ (x,y,z) - p ] = 0$ as usual. The key property we used was the fact that the cross product of two vectors results in a vector that is perpendicular to both vectors. The cross product is the perfect tool for finding the normal vector.

The distance between 2 points $p$ and $q$ is equal to the length of the vector that goes from $p$ to $q$: \[ d(p,q)=\| q - p \| = \sqrt{ (q_x-p_x)^2 + (q_y-p_y)^2 + (q_z-p_z)^2}. \]

The distance between the line $\ell: \{ (x,y,z) \in \mathbb{R}^3 \ | \ (x,y,z)=p_o+t\:\vec{v}, t \in \mathbb{R} \}$ and the origin $O=(0,0,0)$ is given by the formula: \[ d(\ell,O) = \left\| p_o - \frac{ p_o \cdot \vec{v} }{ \| \vec{v} \|^2 } \vec{v} \right\|. \]

The interpretation of this formula is as follows. The first step is to identify a vector that starts at the origin and goes to any point $p_o$. The projection of $p_o$ onto line $\ell$ is given by the formula $\frac{ p_o \cdot \vec{v} }{ \| \vec{v} \|^2 } \vec{v}$. This is the part of the vector $p_o$ which is entirely in the direction of $\vec{v}$. The distance $d(\ell,O)$ is equal to the orthogonal complement of this vector.

The distance between a plane $P: \ \vec{n} \cdot [ (x,y,z) - p_o ] = 0$ and the origin $O$ is given by: \[ d(P,O)= \frac{| \vec{n}\cdot p_o |}{ \| \vec{n} \| }. \]

The above distance formulas are somewhat complicated expressions which involve computing dot products and taking the length of vectors a lot. In order to understand what is going on, we need to learn a bit about projective geometry which will help us measure distances between arbitrary points, lines and planes. As you can see from the formulas above, there will be no new math: just vector $+$, $-$, $\|.\|$ and dot products. The new stuff is actually all in picture-proofs (formally called vector diagrams). Projections play a key role in all of this and this is why we will learn about them in great detail in the next section.

Find the plane which contains the line of intersection of the two planes $x+2y+z=1$ and $2x-y-z=2$ and is parallel to the line $x=1+2t$, $y=-2+t$, $z=-1-t$.

NOINDENT Sol: Find direction vector for the line of intersection $\vec{v}_1 = ( 1, 2,1 ) \times ( 2, -1, -1)$. We know that the plane is parallel to $\vec{v}_2=(2,1,-1)$. So the plane must be the $\textrm{span}\{\vec{v}_1, \vec{v}_2 \} + p_o$. To find a normal vector for the plane we do $\vec{n} = \vec{v}_1 \times \vec{v}_2$. Then choose a point that is on both of the above planes. Conveniently the point $(1,0,0)$ is in both of the above planes. So the anser is $\vec{n}\cdot[ (x,y,z) - (1,0,0) ]=0$.