Lines and curves in space

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Vector-valued functions are parameterized curves.

Vector-valued functions

A function $\vec {f}: \R \to \R ^3$ can be thought of as associating to each time $t$ a vector $\vector {x(t),y(t),z(t)}$ .

A vector-valued function maps real numbers to vectors in $\R ^n$ .

Vector-valued functions simply map numbers to lists of numbers, that we interpret as vectors:

Placing the tail of the vector at the origin, its head will sweep out a curve parameterized by $t$ . Below we see a plot of the vector-valued function: $\vec {f}(t) = \vector {t,sin(t),cos(t)}$

Use the slider to see how the vector-valued function is “drawn” by the tip of the vector

Consider the function $\vec {f}(t) = \vector {\cos (t),\sin (t),t}$ . The projection of the point $\vec {f}(t)$ into the $(x,y)$ -plane moves around the unit circle in the positive direction. The projection onto the $z$ axis moves at a constant rate in the positive direction. So we expect that $\vec {f}$ parameterizes

a straight line a circle around the $x$ -axis a circle around the $y$ -axis a circle around the $z$ -axis a spiral around the $x$ -axis a spiral around the $y$ -axis a spiral around the $z$ -axis

Here is the graph of $\vec {f}$ :

How are vector-valued functions useful?

To get your imagination going, here are a few examples of what a function $f: \R \to \R ^3$ could represent:

The $3$ -dimensional position of a rocket in space as a function of time.
The population of $3$ different species of bacteria found in a swimming pool as a function of the amount of chlorine in the water.
The performance of $3$ different stocks as a function of time.
The trunk width, height, and canopy radius of a tree as a function of time.
The average temperature, humidity, and air pressure at a given latitude as a function of that latitude.
The RGB color of a single pixel of a LCD screen varying over time.

Of the examples above, perhaps “position in space” is the best mental model to use to help you understand vector-valued functions.

Lines in space

It is easy to create a vector-valued function that passes through two points $\vec {p}$ and $\vec {q}$ : $\vecl (t) = \vec {p} + t(\vec {q}-\vec {p}).$

What is the value of $\vecl (0)$ ?

$\vecl (0)$ is unknowable $\vecl (0)=\vec {p}$ $\vecl (0)=\vec {q}$ $\vecl (0)=\vec {q}-\vec {p}$

What is the value of $\vecl (1)$ ?

$\vecl (1)$ is unknowable $\vecl (1)=\vec {p}$ $\vecl (1)=\vec {q}$ $\vecl (1)=\vec {q}-\vec {p}$

What value of $t$ gives the midpoint of the tips of vectors $\vec {p}$ and $\vec {q}$ ? $t = \answer {1/2}$

Here we see vectors $\vec {p}$ and $\vec {q}$ . In blue below we see the vector $t(\vec {p}-\vec {q})$ starting at point $\vec {p}$ . Convince yourself that $\vecl (t) = \vec {p} + t(\vec {p}-\vec {q})$ draws a line.

If we know that a line passes through two points (that we’ll notate with vectors) $\vec {p}$ and $\vec {q}$ , then we know that it points in the direction $\vec {v} = \vec {q} - \vec {p}$ , and passes through the tip of $\vec {p}$ . When we will know that the line passes through the tip of $\vec {p}$ , and points in a direction $\vec {v}$ . Then we write $\vecl (t) = \vec {p}+t\vec {v}.$ Play around with the interactive below to see if you get the idea:

Using the ideas above, find an expression in terms of $t$ parameterizing the line passing through $\vec {p} = \vector {0,2,4}$ when $t=0$ , and $\vec {q} = \vector {1,1,1}$ when $t=1$ . $\vecl (t) = \vector {\answer {t},\answer {2-t},\answer {4-3t}}$

The line passes through $\vec {p}$ and points in the direction $\vec {q} - \vec {p} = \vector {1,1,1} - \vector {0,2,4}$

Let $\vecl$ be a line that passes through the points $\vecl (0) = \vector {1,2,3}$ and $\vecl (1) = \vector {2,2,2}$ . What are the components of $\vecl (t)$ ? $\begin{align*} x(t) &= \answer{1+t}\\ y(t) &= \answer{2}\\ z(t) &= \answer{3-t} \end{align*}$

Let $\vecl (t) = \vec {p}+t\vec {v}$ , then $\vecl (0) = \vec {p}$ .

$\vec {v}= \vector {2,2,2}-\vector {1,2,3}$

There are an infinite number of ways to parameterize the same line. Try your hand at the following puzzlers:

Compare and contrast the curves $\vec {f}(t) = \vector {-3+t,5+2t,1+3t}$ and $\vec {g}(t)=\vector {-3+2t,5+4t,1+6t}$ .

They parameterize different lines. They parameterize the same line, but $\vec {f}(t)$ moves “twice as fast” as $\vec {g}(t)$ . They parameterize the same line, but $\vec {g}(t)$ moves “twice as fast” as $\vec {f}(t)$ . These are the same function!

Note, both lines start at the same point when $t=0$ .

We can rewrite $\vec {f}$ and $\vec {g}$ as: $\begin{align*} \vec{f}(t) &= \vector{-3,5,1}+t\vector{1,2,3}\\ \vec{g}(t) &= \vector{-3,5,1}+t\vector{2,4,6} \end{align*}$

We can further rewrite $\vec {g}$ as: $\vec {g}(t) = \vector {-3,5,1}+2t\vector {1,2,3}$

Compare and contrast the curves $\vec {f}(t) = \vector {-3+t,5+2t,1+3t}$ and $\vec {g}(t)=\vector {-3-t,5-2t,1-3t}$ .

They parameterize different lines. They parameterize the same line, but $\vec {f}(t)$ moves in the opposite direction compared with $\vec {g}(t)$ . They parameterize the same line, but $\vec {g}(t)$ moves “twice as fast” as $\vec {f}(t)$ . These are the same function!

Note both lines start at the same point when $t=0$ .

We can rewrite $\vec {f}$ and $\vec {g}$ as: $\begin{align*} \vec{f}(t) &= \vector{-3,5,1}+t\vector{1,2,3}\\ \vec{g}(t) &= \vector{-3,5,1}+t\vector{-1,-2,-3} \end{align*}$

We can further rewrite $\vec {g}$ as: $\vec {g}(t) = \vector {-3,5,1}-t\vector {1,2,3}$

We can use these ideas to parameterize any line in space. However, our parameterizations will not be unique as there are infinitely many different ways to parameterize the same line. Some parameterizations may “move faster” than others, or in the opposite direction, or even at uneven rates!

Distance between a point and a line

Given a point $\vec {p}$ , notated as the tip of a vector with its tail at the origin, and a line $\vecl (t) = \vec {q} + t\vec {v}$ we often want to know the distance between $\vec {p}$ and $\vecl$ .

This distance is the length of the shortest path from $\vec {p}$ to the line $\vecl$ . How do we find this distance? Well:

(a): Recalling that the magnitude of a vector $|\vec {w}| =\sqrt {\vec {w}\dotp \vec {w}}$ we could attempt to minimize the function $\mathrm {distance}(t)^2= \vecl (t) \dotp \vec {p}$ using the derivative. The square-root of the minimum value will be the distance.
(b): We could compute the distance between $\proj _{\vec {v}}(\vec {p}-\vec {q})$ and $\vec {p}-\vec {q}$ . This is: $|\proj _{\vec {v}}(\vec {p}-\vec {q})- (\vec {p}-\vec {q})|$ Checkout the diagram below:

However, both of these methods are somewhat involved. Perhaps the quickest method for determining the distance between a point and a line is by using the cross product. Since the cross product is only defined in $\R ^3$ , we need $3$ -dimensional vectors. If we consider the vector $\vec {p}-\vec {q}$ , we see by the definition of sine

that the distance we are looking for is given by $|\vec {p}-\vec {q}|\sin (\theta ).$ However, $|\vec {p}-\vec {q}||\vec {v}|\sin (\theta ) = |(\vec {p}-\vec {q})\cross \vec {v}|$ so we see that $\mathrm {distance} = |\vec {p}-\vec {q}|\sin (\theta ) = \frac {|(\vec {p}-\vec {q})\cross \vec {v}|}{|\vec {v}|}.$ Try your hand at it by answering the following questions:

What is the distance between the point $(1,2,3)$ and the line that passes through the origin and $(1,-2,2)$ ? $\mathrm {distance} = \answer {\sqrt {13}}$

Try to use a similar technique for points and lines in $\R ^2$ :

What is the distance between the point $(3,1)$ and the line that passes through the points $(1,1)$ and $(3,2)$ ?

To use the cross product, make these points $3$ -dimensional by adding a $z$ -component of $0$ to each point.

In this case $\vecl (t) = \vector {1,1,0} + t (\vector {3,2,0}-\vector {1,1,0})$ .

$\mathrm {distance} = \answer {2/\sqrt {5}}$

However, depending on the question, you might want to think before blindly applying formulas. Try your hand and this last question:

Consider the line: $\vecl (t) = \vector {-3,5,2} +t \vector {-1,-1,1}$ What point on $\vecl$ is nearest to the point $(1,-1,-3)$ ?

Here, we are not asking for the distance, we are asking for the nearest point.

It will be easiest to use an orthogonal projection to answer this question.

The point on $\vecl$ closest to $(1,-1,-3)$ is: $\left (\answer {-2},\answer {6},\answer {1}\right )$

Circles and ellipses

Given two orthogonal unit vectors, $\uvec {u}$ and $\uvec {v}$ , and any other vector $\vec {p}$ , the vector-valued function $\vec {f}(t) = \vec {p}+r\cdot \cos (t)\cdot \uvec {u} + r \cdot \sin (t)\cdot \uvec {v}$ gives a circle of radius $r$ , centered at the tip of $\vec {p}$ , lying in the plane containing $\uvec {u}$ and $\uvec {v}$ . Moreover, to produce an ellipse, we write: $\vec {g}(t) = \vec {p}+a\cdot \cos (t)\cdot \uvec {u} + b \cdot \sin (t)\cdot \uvec {v}$

Given an ellipse, the major axis of an ellipse is its longest diameter, and the minor axis is its smallest diameter. The semi-major axis is half of the major axis, and the semi-minor axis is half of the minor axis. Given an ellipse of the form $\vec {g}(t) = \vec {p}+a\cdot \cos (t)\cdot \uvec {u} + b \cdot \sin (t)\cdot \uvec {v}$ where $a>b$ , $a$ is the semi-major axis and $b$ is the semi-minor axis.

Let’s see an example.

Give a vector-valued formula for an ellipse that is drawn in the $(y,z)$ -plane centered at the point $(0,2,3)$ whose semi-major axis is $5$ on a line parallel to the $y$ -axis, and whose semi-minor axis is $1$ on a line parallel to the $z$ -axis.

There are actually infinitely many solutions to this problem, though we’ll just give one. Write: $\vec {g}(t) = \vector {\answer [given]{0},\answer [given]{2},\answer [given]{3}} + \answer [given]{5}\cdot \cos (t) \vecj + \answer [given]{1} \cdot \sin (t) \veck$

Can you find a vector-valued formula for a circle of radius $3$ in the plane $y=2$ centered at $(3,2,0)$ ? A circle we seek is: $\vec {f}(t) = \vector {\answer {3},\answer {2},\answer {0}} + \answer {3}\cos (t) \veci + \answer {3}\sin (t) \veck$

Lines and curves embedded in surfaces

Curves can lie on surfaces. Typically, the surface is defined implicitly, and the curve is a vector-valued function. To check if the curve lies on the surface, break the curve into components and substitute:

The $x$ -component of the curve for $x$ in the equation of the surface.
The $y$ -component of the curve for $y$ in the equation of the surface.
The $z$ -component of the curve for $z$ in the equation of the surface.

If the equation defining the surfaces holds after the substitution, the curve lies on the surface. Try your hand at these puzzles:

Consider the plane: $x+2y+3z = 6$ Which of the following lines are on this plane?

Separate each line into its component functions: $x$ , $y$ , and $z$ , and see if the equation defining the surface is valid for all $t$ .

$\vector {3,-3,3}+t\vector {-7,5,-1}$ $\vector {-1,-1,3}+t\vector {3,-2,3}$ $\vector {-4,2,2}+t\vector {5,-1,-1}$ $\vector {1,1,1}+t\vector {2,-4,2}$ $\vector {1,-1,-1}+t\vector {-7,5,1}$ $\vector {-2,-2,4}+t\vector {4,-2,2}$

Consider the planes: $\begin{align*} -2x+3y+4z &=6\\ 4x+y+6z &=12 \end{align*}$

Separate each line into its component functions: $x$ , $y$ , and $z$ , and see if the equation defining the surface is valid for all $t$ .

$\vector {22/7,38/7,-1}+t\vector {1,2,-1}$ $\vector {22/7,38/7,-1}+ t\vector {-2,-4,2}$ $\vector {8/7,10/7,1}+t\vector {2,4,-2}$ $\vector {15/7, 24/7, 0}+t\vector {7,-14,-7}$ $\vector {15/7, 24/7, 0}+t\vector {7,14,-7}$ $\vector {8/7,10/7,1}+t\vector {1,2,-1}$

Sometimes lines lie on surprising surfaces:

Consider the surface determined by all $x$ , $y$ and $z$ such that: $x^2+y^2=z^2+1$ This surface looks something like:

Which of the following lines lie on the surface $x^2+y^2=z^2+1$ ?

Separate each line into its component functions: $x$ , $y$ , and $z$ , and see if the equation defining the surface is valid for all $t$ .

$\vector {1,0,0} + t\vector {-5,0,5}$ $\vector {1,0,0} + t\vector {0,-3,3}$ $\vector {0,1,0} + t\vector {-7,0,-7}$ $\vector {0,1,0} + t\vector {-4,4,0}$ $\vector {1/2,\sqrt {3}/2,0} + t\vector {-\sqrt {3},1,2}$ $\vector {1/2,\sqrt {3}/2,0} + t\vector {-1,\sqrt {3},2}$

Though their formulation may be more complex, a vector-valued function that produces a curve is no different from that which produces a line (a line is a special type of curve!).

Consider the unit sphere: $x^2+y^2+z^2 = 1$ Which of the following curves lie on this sphere?

Separate each curve into its component functions: $x$ , $y$ , and $z$ , and see if the equation defining the surface is valid for all $t$ .

$\vector {\cos (t),\sin (t),0}$ $\vector {\sin (t),0,\cos (t)}$ $\vector {\cos (t),\sin (t),1}/\sqrt {2}$ $\vector {\sin (t),\sin (t),\cos (t)}/\sqrt {2}$

Consider the surface determined by all $x$ , $y$ and $z$ such that: $z^2=x^2+y^2$ Which of the following curves lie on the surface $z^2=x^2+y^2$ ?

Separate each curve into its component functions: $x$ , $y$ , and $z$ , and see if the equation defining the surface is valid for all $t$ .

$\vector {t\cos (t),\sin (t),t}$ $\vector {6\cos (t),6\sin (t),1}$ $\vector {t\cos (t),t\sin (t),t}$ $\vector {\cos (t),-2\sin (t),-2}$ $\vector {6\cos (t),6\sin (t),6}$ $\vector {-2\cos (t),-2\sin (t),-2}$