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A point and a direction: the vector, parametric and symmetric equations of a line, the distance from a point to it, and the third possibility the plane never offered — two lines that are neither parallel nor meeting.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to write a line in space in vector, parametric and symmetric form, decide whether a given point lies on it, find the distance from a point to a line, and classify two lines as parallel, intersecting or skew. You will also be able to say why a line has no single correct equation and how to check two descriptions against each other.
A point, a direction and the arithmetic of adding a vector to a point. A line in the plane was described by a slope; in space there is no such thing as one slope, and the description that replaces it is the one this lesson is about.
A direction vector is any non-zero vector along the line; any multiple of it will do. The parameter $t$ says how far along, in units of the direction vector. Parametric equations give the three coordinates as functions of $t$; the symmetric equations eliminate $t$ between them. Two lines are skew when they are neither parallel nor intersecting — a situation with no counterpart in the plane.
A line in space is fixed by one point on it and one direction along it. If $\mathbf{r}_0$ is the position vector of a point on the line and $\mathbf{v}$ is a direction vector, the line is
$$\mathbf{r}(t) = \mathbf{r}_0 + t\,\mathbf{v}, \qquad t \in \mathbb{R}.$$
Parametric form. Writing $\mathbf{r}_0 = \langle x_0, y_0, z_0 \rangle$ and $\mathbf{v} = \langle v_1, v_2, v_3 \rangle$ and reading off the coordinates:
$$x = x_0 + v_1 t, \qquad y = y_0 + v_2 t, \qquad z = z_0 + v_3 t.$$
Each coordinate moves at its own steady rate. The three numbers $v_1, v_2, v_3$ are the direction numbers, and they are what replaces a slope.
Symmetric form. When no $v_i$ is zero, solving each equation for $t$ and setting the three results equal gives
$$\frac{x - x_0}{v_1} = \frac{y - y_0}{v_2} = \frac{z - z_0}{v_3}.$$
This is the closest thing to a slope form a line in space has. A zero direction number breaks it, and the honest repair is to write that coordinate as a constant beside the other two fractions rather than to divide by zero.
Two lines. Given $L_1$ and $L_2$ with directions $\mathbf{v}_1$ and $\mathbf{v}_2$:
Distance from a point to a line. For a point $Q$ and a line through $P$ with direction $\mathbf{v}$,
$$d = \frac{|\overrightarrow{PQ} \times \mathbf{v}|}{|\mathbf{v}|}.$$
The cross product deletes the part of $\overrightarrow{PQ}$ along the line and keeps only the part across it, which is what a perpendicular distance is.
Another way: picture
Stand at $\mathbf{r}_0$ facing along $\mathbf{v}$ and walk. At $t = 1$ you have travelled exactly one $\mathbf{v}$; at $t = -2$ you are two $\mathbf{v}$s behind where you started. The whole line is the set of places you can reach, and the parameter is nothing more than a bookkeeping label saying how far along you are.
Another way: steps
To write the line through two points $P$ and $Q$:
In the plane, two lines are parallel or they meet; there is no third option, because two directions that are not multiples of each other must eventually cross. In space that argument fails. The directions can be independent and the lines can still miss each other entirely, passing at different heights — one along a corridor floor, the other along a corridor ceiling at right angles to it.
This is worth dwelling on because it changes what an answer has to contain. Solving for an intersection in the plane means solving two equations in two unknowns and the system almost always has a solution. In space, setting the two parametric points equal gives three equations in two parameters: one equation more than unknowns. Two of them fix the parameters and the third is left over, and whether it happens to hold is exactly the question of whether the lines meet.
So the leftover equation is not an inconvenience in the algebra; it is the content. A system with more equations than unknowns usually has no solution, which is the algebraic form of the statement that two lines in space usually miss.
Using the same parameter for both lines. $L_1$ and $L_2$ need their own parameters, $t$ and $s$. Sharing one asks whether the lines meet at the same time, which is a different and much stronger question than whether they meet at all.
Dividing by a zero direction number. A line with $v_2 = 0$ has no symmetric equation in the usual shape; write $y = y_0$ alongside the other fractions.
Thinking a line has one equation. Any point on it and any multiple of the direction give a correct answer, so two correct descriptions can look entirely different. Compare them by testing a point, never by comparing the coefficients.
Reading a parametric description as a journey with a start and an end. $t$ runs over every real number, so the line is infinite in both directions; restricting $t$ to an interval gives a segment, which is a different object. A third coordinate is not a third term bolted onto a two-dimensional formula. The cross product exists only in space, a plane is fixed by a direction rather than by a slope, and a curve in space can bend without ever leaving a surface. Checking a new formula against the plane is worth doing; assuming the plane's formula still governs is how most of the errors in this course begin.
Every point on the line can serve as $\mathbf{r}_0$, and every non-zero multiple of $\mathbf{v}$ can serve as the direction, so one line has infinitely many correct descriptions. Two students can hand in answers with no number in common and both be right.
The consequence for checking work is concrete. You cannot compare two line equations by comparing their coefficients. What you can do is take a point from one, substitute into the other, and see whether some parameter value produces it; then check that the two directions are multiples. Those two tests together settle it, and nothing shorter does.
The parameter itself is a label, not geometry. Doubling $\mathbf{v}$ halves every parameter value and changes no point of the line at all.
Through $P = (1, 2, 3)$ and $Q = (4, 6, 3)$: the direction is $\overrightarrow{PQ} = \langle 3, 4, 0 \rangle$.
A direction is the displacement between the two points.
Vector form $\mathbf{r}(t) = \langle 1, 2, 3 \rangle + t\langle 3, 4, 0 \rangle$; parametric form $x = 1 + 3t$, $y = 2 + 4t$, $z = 3$.
Reading off the coordinates one at a time.
The symmetric form is $\dfrac{x-1}{3} = \dfrac{y-2}{4}$ with $z = 3$. The third direction number is zero, so $z$ never moves and is written as a constant rather than as a fraction with zero underneath.
A zero direction number becomes a constant coordinate.
$L_1: \langle 1 + t, 2t, 3 - t \rangle$ and $L_2: \langle 2 + s, 2 + s, 1 \rangle$. The directions $\langle 1, 2, -1 \rangle$ and $\langle 1, 1, 0 \rangle$ are not multiples, so not parallel.
Directions first; different parameters for the two lines.
The third coordinates give $3 - t = 1$, so $t = 2$. The first then give $1 + 2 = 2 + s$, so $s = 1$.
Two equations fix the two parameters.
The second coordinates are the leftover test: $2t = 4$ and $2 + s = 3$. They disagree, so the lines are skew after all — which is why the third equation is kept back rather than used in the solving.
The leftover equation is the test, not an afterthought.
The first coordinate needs $1 + 2t = 7$, so $t = 3$.
One coordinate is enough to propose a parameter value.
Check the second: $3 - 3 = 0$, but the point has second coordinate $1$.
So the point is not on the line, and the third coordinate need not even be looked at. One disagreement settles it — and notice that had the first coordinate been checked alone, the point would have been accepted wrongly.
The line is $\mathbf{r}(t) = \langle 4, -6, 0 \rangle + t\langle 1, 1, 2 \rangle$. Mark the value of $t$ at which it passes through $(6, -4, 4)$.
-5 |——————————| 5
Mark the position with a cross, then write the value:
Put the $2$-second test for how two lines in space are related into the order it must be carried out.
Number the steps in order (write the number in the box):
How far is the point $(-6, 6, 8)$ from the $x$ axis?
Answer:
Line $L_1$ is $\langle t, 0, 0 \rangle$ and line $L_2$ is $\langle 0, s, -5 \rangle$. How are they related?
The line is $\mathbf{r}(t) = \langle 1, 1, 1 \rangle + t\langle 4, 4, 2 \rangle$. Mark the value of $t$ at which it passes through $(9, 9, 5)$.
-5 |——————————| 5
Mark the position with a cross, then write the value:
One robot follows $\mathbf{r}_1(t) = \mathbf{a} + t\mathbf{v}$ and another follows $\mathbf{r}_2(t) = \mathbf{a} + 3t\mathbf{v}$, both for every real $t$. What is different?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the $6$-second test for how two lines in space are related into the order it must be carried out.
Number the steps in order (write the number in the box):
You can describe a line by a point and a direction, test a point against it, and tell parallel from intersecting from skew. Next: the object a single direction rules out rather than picks, which is a plane.
10. Your turn: does the point $(7, 1, -3)$ lie on $\mathbf{r}(t) = \langle 1, 3, 1 \rangle + t\langle 2, -1, -2 \rangle$?, step 3