Directed line segments
A segment tells you how far apart two points are. How do you also say which way you went?
Where you are 4 in Vectors
Why this exists
The segment \overline{AB} and the segment \overline{BA} are the same set of points. A segment has a length and nothing more: it does not know which end it starts from.
A move does care. An air hockey puck sliding from one end of the table to the other has not made the same trip as a puck sliding back. To describe a move, a force or a velocity you need two facts at once: how far and which way.
So pick one end as the start, the other as the finish, and put an arrow on the segment. That arrow is a directed line segment, the first way to see a vector.
Definition
A directed line segment is a segment with one endpoint chosen as its start. The directed segment from P to Q is written \overrightarrow{PQ}. Its initial point (or tail) is P and its terminal point (or head) is Q. It is drawn as the segment \overline{PQ} with an arrowhead at Q.
A directed segment carries two things.
- Its length is the length PQ of the segment \overline{PQ}.
- Its direction is the way the arrow points. Two directed segments of positive length have the same direction when they lie on parallel lines (or on one line) and point the same way. They have opposite directions when they lie on parallel lines (or on one line) and point opposite ways.
Swapping the ends gives \overrightarrow{QP}, the opposite of \overrightarrow{PQ}: same length, opposite direction.
When the start and the finish are the same point, \overrightarrow{PP} is the zero directed segment. Its length is 0, and it has no direction, since there is no arrow to point anywhere.
Two directed segments are equivalent when they have the same length and the same direction. Any two zero directed segments count as equivalent to each other. Equivalent directed segments may sit in different places: equivalence ignores where the trip starts and keeps only how far and which way it goes. A vector is exactly that shared information, and each directed segment that carries it is one drawing of the vector.
In geometry the same symbol \overrightarrow{AB} names a ray, which starts at A and runs forever through B. With vectors it names the directed segment, which stops at B.
Intuition
Mark where Kealoha’s puck starts and stops, and draw an arrow between them. The arrow ignores the speed and any bounce off a rail. It records only the net trip: so far, that way.
Rustam’s puck starts elsewhere and makes the same trip. His arrow is Kealoha’s slid to a new place without turning, a translation, and the two arrows are equivalent. Figure 2 shows five such copies.
On a grid, count how far each arrow goes across and up, from tail to head. Copies have the same two counts: any arrow going 4 across and 3 up is equivalent to \overrightarrow{PQ} in Figure 1. Writing a vector by those counts is its component form, the next lesson; here the counts are only a test.
Results
Across and up. For P(x_1, y_1) and Q(x_2, y_2), the directed segment \overrightarrow{PQ} goes x_2 - x_1 across and y_2 - y_1 up. A negative count means left or down. Its length comes from the distance formula:
The equivalence test. Two directed segments in the coordinate plane are equivalent exactly when they have the same across count and the same up count.
The opposite. \overrightarrow{QP} goes x_1 - x_2 across and y_1 - y_2 up: both counts of \overrightarrow{PQ} with their signs changed.
Parallelograms. For four points A, B, C, D not all on one line, ABCD is a parallelogram exactly when \overrightarrow{AB} and \overrightarrow{DC} are equivalent.
Why it is true
Call the counts of one directed segment a across and b up, and of another c across and d up.
Same counts give equivalent segments. Suppose c = a and d = b. Both lengths are \displaystyle \sqrt{a^2 + b^2}, so they match. If a = b = 0, both are zero segments and are equivalent by definition. Otherwise, if a \neq 0, both segments have slope \displaystyle \frac{b}{a}, so they are parallel or on one line; if a = 0, both are vertical. Either way the lines match. They point the same way because both go right (or both left, or neither) and both go up (or both down, or neither) by the same amounts.
Equivalent segments have the same counts. Suppose the two are equivalent and not zero. Being parallel or on one line, they have equal slopes or are both vertical, so c = ka and d = kb for one number k (take \displaystyle k = \frac{c}{a} if a \neq 0, else \displaystyle k = \frac{d}{b}). Pointing the same way makes the signs agree, so k > 0. The lengths are \displaystyle \sqrt{a^2 + b^2} and \displaystyle \sqrt{k^2a^2 + k^2b^2} = k\sqrt{a^2 + b^2}, and these are equal only when k = 1. So c = a and d = b.
Parallelograms. If \overrightarrow{AB} and \overrightarrow{DC} are equivalent, the sides \overline{AB} and \overline{DC} are equal and parallel, and a quadrilateral with one pair of opposite sides equal and parallel is a parallelogram. Conversely, in a parallelogram ABCD those sides are equal and parallel, and B and C lie on the same side of line AD, so the arrows from A to B and from D to C point the same way.
Worked examples
Reading a directed segment
On a grid drawn on an air hockey table, in feet, Kealoha’s puck slides from P(-2.5, 1) to Q(0.5, -1). Give the initial and terminal points of its trip, the counts across and up, and the length. Then describe the trip back from Q to P.
SolutionThe trip is \overrightarrow{PQ}, so the initial point is P(-2.5, 1) and the terminal point is Q(0.5, -1). Subtract the start from the finish, coordinate by coordinate:
The puck goes 3 feet across (to the right) and -2 feet up, which is 2 feet down. Its length is
The trip back is \overrightarrow{QP}, with counts -3 across and 2 up. It has the same length, \displaystyle \sqrt{13} feet, and the opposite direction.
Which of these are equivalent?
Figure 3 shows \overrightarrow{AB} from A(1, -3) to B(4, 1), \overrightarrow{CD} from C(-5, 0) to D(-2, 4), \overrightarrow{EF} from E(0, 3) to F(-3, -1), and \overrightarrow{GH} from G(3, -4) to H(7, -1). Which are equivalent to \overrightarrow{AB}?
Count each one, finish minus start:
| Directed segment | Across | Up | Length |
|---|---|---|---|
| \overrightarrow{AB} | 4 - 1 = 3 | 1 - (-3) = 4 | 5 |
| \overrightarrow{CD} | -2 - (-5) = 3 | 4 - 0 = 4 | 5 |
| \overrightarrow{EF} | -3 - 0 = -3 | -1 - 3 = -4 | 5 |
| \overrightarrow{GH} | 7 - 3 = 4 | -1 - (-4) = 3 | 5 |
Each length is \displaystyle \sqrt{3^2 + 4^2} = 5, so length alone cannot tell them apart.
\overrightarrow{CD} has the same counts as \overrightarrow{AB}, so it is equivalent.
\overrightarrow{EF} has both counts with their signs changed: a parallel line, pointing the other way. It has the opposite direction, so it is not equivalent.
\overrightarrow{GH} has its counts swapped. Its slope is \displaystyle \frac{3}{4}, not \displaystyle \frac{4}{3}, so it is not even parallel to \overrightarrow{AB}. Same length, different direction: not equivalent.
Only \overrightarrow{CD} is equivalent to \overrightarrow{AB}.
The same slide, a different stone
A curling sheet has a grid in feet, x across the sheet and y along it. Rustam’s stone starts at R(-2, 30) and stops at S(1, 54). Munkh’s stone starts at U(3, 28) and makes a trip equivalent to Rustam’s. Where does it stop? Kealoha’s stone makes an equivalent trip and stops at W(0, 50). Where did it start?
SolutionA curling stone’s path bends, but the directed segment keeps only the start and the stop.
Rustam’s trip \overrightarrow{RS} goes 1 - (-2) = 3 across and 54 - 30 = 24 up. An equivalent trip has the same counts, so Munkh’s stone stops at
For Kealoha’s stone the finish is known, so undo the counts: subtract them from the finish.
Check. \overrightarrow{VW} goes 0 - (-3) = 3 across and 50 - 26 = 24 up, the same as \overrightarrow{RS}.
Four points on one line
The points A(-6, -3), B(-2, -1), C(2, 1) and D(6, 3) all lie on the line \displaystyle y = \frac{1}{2}x. Compare \overrightarrow{CD}, \overrightarrow{BD} and \overrightarrow{DB} with \overrightarrow{AB}.
Solution\overrightarrow{AB} goes 4 across and 2 up.
\overrightarrow{CD} goes 6 - 2 = 4 across and 3 - 1 = 2 up. Same counts, so it is equivalent to \overrightarrow{AB}. Equivalent segments may share a line.
\overrightarrow{BD} goes 8 across and 4 up. It is on the same line and points the same way, so it has the same direction, but its length is \displaystyle \sqrt{8^2 + 4^2} = 4\sqrt{5} against \displaystyle \sqrt{4^2 + 2^2} = 2\sqrt{5}. Twice as long, so not equivalent.
\overrightarrow{DB} goes -8 across and -4 up. Same line, but it points the other way: the opposite direction, and twice the length. Not equivalent.
Finishing a parallelogram
Three vertices of parallelogram ABCD are A(-2, -1), B(2, 0) and C(4, 4). Find D.
ABCD is a parallelogram exactly when \overrightarrow{AB} and \overrightarrow{DC} are equivalent. \overrightarrow{AB} goes 4 across and 1 up. The trip from D to C must have those counts, and its finish is C, so start from C and undo them:
Check. The other pair of sides should match too. \overrightarrow{AD} goes 0 - (-2) = 2 across and 3 - (-1) = 4 up, and \overrightarrow{BC} goes 4 - 2 = 2 across and 4 - 0 = 4 up. They are equivalent, as opposite sides of a parallelogram must be.
Both counts must agree
Let A(k, 2), B(5, k + 1), C(1, -1) and D(4, 0). Find every k that makes \overrightarrow{AB} equivalent to \overrightarrow{CD}, and every k that makes it opposite to \overrightarrow{CD}.
Solution\overrightarrow{AB} goes 5 - k across and (k + 1) - 2 = k - 1 up. \overrightarrow{CD} goes 3 across and 1 up.
Equivalent. Both counts must match, so both equations must hold at once:
The first gives k = 2, and the second gives k = 2 as well. So k = 2, which makes A(2, 2) and B(5, 3): 3 across and 1 up, as required.
Opposite. Now the counts must be -3 and -1:
The first needs k = 8 and the second k = 0, so no value of k makes \overrightarrow{AB} opposite to \overrightarrow{CD}. Matching one count is never enough.
Boundaries
Direction needs length. Only a directed segment of positive length has a direction. The zero segment \overrightarrow{PP} has none, so it has neither the same direction as another segment nor the opposite direction; it is equivalent only to other zero segments.
Equal length is not equivalence, and neither is parallel. In Example 2, \overrightarrow{EF} and \overrightarrow{GH} have the length of \overrightarrow{AB}, and \overrightarrow{EF} is parallel to it, yet neither is equivalent. Same length and same direction, including which way it points, are both needed.
Equivalent is not equal. Equivalent directed segments are different sets of points. What they share is the vector they represent.
A directed segment is a net trip. It records a start and a finish, not the path.
The counts test needs coordinates. Without a grid, go back to the definition: compare lengths, and check that the segments are parallel and point the same way.
Where this goes next
A directed segment is a vector drawn in one place. Keeping only its two counts is component form, and the same two counts give its length as the magnitude of the vector, along with an angle for its direction. Stretching an arrow while keeping its line gives scalar multiples, which is what \overrightarrow{BD} was to \overrightarrow{AB} in Example 4. Placing one arrow’s tail at another’s head is vector addition, the way two trips in a row make one.
Exercises
Reading a directed segment
Use M(4, -1) and N(-2, 3) for Exercises 1 to 3.
- Name the initial and terminal points of \overrightarrow{MN}, and give its counts across and up.
- Give the counts across and up of \overrightarrow{NM}. How is it related to \overrightarrow{MN}?
- Find the length of \overrightarrow{MN} in simplest radical form.
- Find the counts across and up and the length of the directed segment from O(0, 0) to (-5, -12).
Equivalent, opposite, or neither
Decide whether the two directed segments are equivalent, opposite (same length, opposite direction), or neither. Explain.
- \overrightarrow{AB} with A(2, -3), B(6, -1), and \overrightarrow{CD} with C(-4, 5), D(0, 7).
- \overrightarrow{AB} with A(1, 1), B(-2, 5), and \overrightarrow{CD} with C(3, 0), D(6, -4).
- \overrightarrow{AB} with A(0, 2), B(5, 2), and \overrightarrow{CD} with C(-1, -3), D(-1, 2).
- \overrightarrow{AB} with A(-6, -1), B(-2, 1), and \overrightarrow{CD} with C(1, 3), D(9, 7).
- \overrightarrow{PP} with P(3, 3), and \overrightarrow{RR} with R(-1, 4).
- \overrightarrow{AB} with A(0.5, 2), B(2, -1.5), and \overrightarrow{CD} with C(-3, 0), D(-1.5, -3.5).
Finding missing points
Use G(-1, 5) and H(3, -1) for Exercises 11 to 13.
- A directed segment equivalent to \overrightarrow{GH} starts at (5, 5). Where does it end?
- A directed segment equivalent to \overrightarrow{GH} ends at (0, 0). Where does it start?
- A directed segment starts at (2, -3), has the same length as \overrightarrow{GH} and the opposite direction. Where does it end?
- Three vertices of parallelogram JKLM are J(-2, 0), K(1, -2) and L(4, 1). Find M.
- The points A(0, 0), B(4, 1) and C(1, 3) are three vertices of a parallelogram. Find all three possible positions of the fourth vertex.
In context
- Rustam’s air hockey puck slides from (-1.5, -1) to (2, 0.5), in feet. Munkh’s puck starts at (-3, 0) and makes an equivalent trip. Where does it stop, and how far does each puck travel from start to stop, to the nearest hundredth of a foot?
- On a curling sheet with a grid in feet, Kealoha’s stone starts at (-1, 40) and stops at (3, 100). A second stone makes an equivalent trip and stops at (-2, 105). Where did it start?
Unknowns in the coordinates
- Find a and b so that \overrightarrow{AB} with A(a, 1), B(4, b) is equivalent to \overrightarrow{CD} with C(-2, 3), D(1, -1).
- Let P(t, 1), Q(2, t), R(0, 0) and S(2, -3). Find every t that makes \overrightarrow{PQ} equivalent to \overrightarrow{RS}, and every t that makes it opposite to \overrightarrow{RS}.
Reasoning
- Explain why \overline{AB} and \overline{BA} name the same segment, while \overrightarrow{AB} and \overrightarrow{BA} are not equivalent unless A and B are the same point.
- Show that if \overrightarrow{AB} is equivalent to \overrightarrow{CD}, and \overrightarrow{CD} is equivalent to \overrightarrow{EF}, then \overrightarrow{AB} is equivalent to \overrightarrow{EF}.
- Show that if \overrightarrow{AB} is equivalent to \overrightarrow{CD}, then \overrightarrow{AC} is equivalent to \overrightarrow{BD}.
- Show that the only directed segments equivalent to their own opposites are the zero segments.
- Munkh says: “\overrightarrow{AB} and \overrightarrow{CD} have the same length and lie on parallel lines, so they are equivalent.” Give coordinates that show the claim is false, and say what is missing from it.
- Show that \overrightarrow{AM} is equivalent to \overrightarrow{MB} exactly when M is the midpoint of \overline{AB}.
Answers
- Initial point M, terminal point N; -6 across and 4 up. Across: -2 - 4 = -6; up: 3 - (-1) = 4.
- 6 across and -4 up. Both counts change sign: \overrightarrow{NM} is the opposite of \overrightarrow{MN}, with the same length and the opposite direction.
- \displaystyle 2\sqrt{13}. \displaystyle \sqrt{(-6)^2 + 4^2} = \sqrt{52} = 2\sqrt{13}.
- -5 across, -12 up, length 13. \displaystyle \sqrt{25 + 144} = \sqrt{169} = 13.
- Equivalent. Both go 4 across and 2 up.
- Opposite. \overrightarrow{AB} goes -3 across and 4 up; \overrightarrow{CD} goes 3 across and -4 up, the same counts with signs changed.
- Neither. \overrightarrow{AB} goes 5 across and 0 up, \overrightarrow{CD} goes 0 across and 5 up. Both have length 5, but one is horizontal and the other vertical.
- Neither. \overrightarrow{AB} goes 4 across and 2 up, \overrightarrow{CD} goes 8 across and 4 up. Same direction, but \overrightarrow{CD} is twice as long.
- Equivalent. Both are zero directed segments, and any two zero segments are equivalent.
- Equivalent. \overrightarrow{AB} goes 2 - 0.5 = 1.5 across and -1.5 - 2 = -3.5 up; \overrightarrow{CD} goes -1.5 - (-3) = 1.5 across and -3.5 - 0 = -3.5 up.
- (9, -1). \overrightarrow{GH} goes 4 across and -6 up, and (5 + 4, 5 - 6) = (9, -1).
- (-4, 6). Undo the counts from the finish: (0 - 4, 0 + 6).
- (-2, 3). The opposite trip goes -4 across and 6 up: (2 - 4, -3 + 6).
- M(1, 3). \overrightarrow{JK} goes 3 across and -2 up, and \overrightarrow{ML} must match, so M = (4 - 3, 1 + 2). Check: \overrightarrow{JM} and \overrightarrow{KL} both go 3 across and 3 up.
- (-3, 2), (3, -2) and (5, 4). Any of the three given points can be the vertex opposite the new one. For ABCD, \overrightarrow{DC} matches \overrightarrow{AB} (4 across, 1 up): D = (1 - 4, 3 - 1) = (-3, 2). For ABDC, \overrightarrow{CD} matches \overrightarrow{AB}: D = (1 + 4, 3 + 1) = (5, 4). For ADBC, \overrightarrow{AD} matches \overrightarrow{CB} (3 across, -2 up): D = (3, -2).
- (0.5, 1.5); each travels \displaystyle \sqrt{14.5} \approx 3.81 feet. Rustam’s trip goes 3.5 across and 1.5 up, so Munkh’s ends at (-3 + 3.5, 0 + 1.5). Length: \displaystyle \sqrt{3.5^2 + 1.5^2} = \sqrt{12.25 + 2.25}, the same for both since the trips are equivalent.
- (-6, 45). Kealoha’s trip goes 4 across and 60 up; undo it from the finish: (-2 - 4, 105 - 60).
- a = 1 and b = -3. \overrightarrow{CD} goes 3 across and -4 up, so 4 - a = 3 and b - 1 = -4.
- No t makes them equivalent; t = 4 makes them opposite. \overrightarrow{PQ} goes 2 - t across and t - 1 up. Equivalent needs 2 - t = 2 and t - 1 = -3, that is t = 0 and t = -2 at once, impossible. Opposite needs 2 - t = -2 and t - 1 = 3, and both give t = 4: then P(4, 1), Q(2, 4), which is -2 across and 3 up.
- \overline{AB} and \overline{BA} are the same set of points. A directed segment also has a start, A for \overrightarrow{AB} and B for \overrightarrow{BA}. When A \neq B the two point opposite ways along one line. When A = B both are the zero segment.
- On a grid, equivalent means equal counts. \overrightarrow{AB} has the same counts as \overrightarrow{CD}, which has the same counts as \overrightarrow{EF}, so \overrightarrow{AB} and \overrightarrow{EF} have the same counts and are equivalent.
- Give the points coordinates and say \overrightarrow{AB} and \overrightarrow{CD} both go a across and b up. Then B = (x_A + a, y_A + b) and D = (x_C + a, y_C + b). \overrightarrow{BD} goes (x_C + a) - (x_A + a) = x_C - x_A across and, the same way, y_C - y_A up, which are exactly the counts of \overrightarrow{AC}.
- If a directed segment goes a across and b up, its opposite goes -a and -b. They are equivalent when a = -a and b = -b, which forces a = 0 and b = 0: the start and finish are the same point. A zero segment is its own opposite, so it is equivalent to it.
- For example A(0, 0), B(2, 0), C(2, 1), D(0, 1): both have length 2 and lie on horizontal lines, but \overrightarrow{AB} goes 2 across and \overrightarrow{CD} goes -2. The claim leaves out that the two must point the same way.
- Say A(x_1, y_1), B(x_2, y_2) and M(m, n). Equivalence means equal counts: m - x_1 = x_2 - m and n - y_1 = y_2 - n. These solve to \displaystyle m = \frac{x_1 + x_2}{2} and \displaystyle n = \frac{y_1 + y_2}{2}, which is the midpoint of \overline{AB}. Every step reverses, so the midpoint makes the two equivalent.