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Precalculus · N.VM.A.2

Component form of a vector

Every arrow that makes the same trip is the same vector. What short name can every one of those arrows share?

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Why this exists

A directed line segment is one drawing of a vector, and the same vector has endless drawings, one starting at every point of the plane. Talking about “the vector drawn from (2, 7) to (5, 9)” picks one drawing out of all of them and makes you carry two points around when the vector itself is only a trip.

On a grid, every drawing of the same vector goes the same distance across and the same distance up. Those two numbers are what all the drawings share, so they can serve as the vector’s name. That name is the vector’s component form, and it turns arrows into numbers you can compute with.

Definition

Let P(x_1, y_1) be the initial point (tail) and Q(x_2, y_2) the terminal point (head) of a directed segment. The component form of the vector it represents is

\overrightarrow{PQ} = \langle x_2 - x_1,\ y_2 - y_1 \rangle

terminal minus initial, one coordinate at a time. A vector \mathbf{v} = \langle a, b \rangle has horizontal component a (the change in x) and vertical component b (the change in y). A negative component means the arrow goes left or down.

Angle brackets mark a vector; parentheses mark a point. The point (5, 3) is a location. The vector \langle 5, 3 \rangle is a trip, 5 right and 3 up, which can start anywhere.

The drawing of \langle a, b \rangle whose tail is the origin O(0, 0) is its position vector drawing. Its head is (0 + a, 0 + b) = (a, b), so the position vector of the point (a, b) is \langle a, b \rangle.

The zero vector is \mathbf{0} = \langle 0, 0 \rangle, the vector of a trip that ends where it starts.

Two vectors are equal when they are the same vector: every drawing of one is a drawing of the other. With directed segments the word was equivalent; once a drawing is read as the vector it stands for, the vectors are equal.

Intuition

x y P Q 5 3 O (5, 3)
Figure 1The arrow from P to Q goes 5 across and 3 up, so it is the vector \langle 5, 3 \rangle. Its copy with tail at the origin, the position vector, ends at the point (5, 3).

In Figure 1 the arrow from P(-4, 2) to Q(1, 5) is one drawing of a vector. Walk from tail to head along the grid: 5 to the right, then 3 up. Those two legs are the components, so the vector is \langle 5, 3 \rangle.

Slide that arrow, without turning it, until its tail sits on the origin. Now its head is at (5, 3), and the components can be read straight off the head. That is the whole job of the position vector: it is the one drawing whose head spells out the vector’s name.

Component form forgets where an arrow was drawn. That is the point, not a loss: a translation moves every point of a figure by the same vector, and it can only do that because the vector does not belong to any one starting point.

Results

Component form. The vector from P(x_1, y_1) to Q(x_2, y_2) is

\overrightarrow{PQ} = \langle x_2 - x_1,\ y_2 - y_1 \rangle

Position vectors. The point (a, b) and the vector \langle a, b \rangle match one to one: the drawing of \langle a, b \rangle with tail at the origin ends at (a, b). So the components of \overrightarrow{PQ} are the components of the position vector of Q minus those of the position vector of P.

Equal vectors. \langle a, b \rangle = \langle c, d \rangle exactly when a = c and b = d.

Head and tail. If \mathbf{v} = \langle a, b \rangle is drawn with tail (x_1, y_1), its head is (x_1 + a,\ y_1 + b). If it is drawn with head (x_2, y_2), its tail is (x_2 - a,\ y_2 - b).

Reversing. If \overrightarrow{PQ} = \langle a, b \rangle, then \overrightarrow{QP} = \langle -a, -b \rangle.

Why it is true

The name is well defined. A vector is what a family of equivalent directed segments has in common. The equivalence test says two directed segments on a grid are equivalent exactly when they have the same count across and the same count up. So every drawing of one vector gives the same pair \langle x_2 - x_1, y_2 - y_1 \rangle: the pair does not depend on which drawing you measure.

Different vectors get different names. By the same test, drawings with the same pair are equivalent, so they draw the same vector. Together the two facts say that vectors and pairs match one to one, which is the rule for equal vectors: equal pairs, and only equal pairs, name the same vector.

Head and tail. For a drawing with tail (x_1, y_1) and head (x_2, y_2), the components say x_2 - x_1 = a and y_2 - y_1 = b. Solve each for the unknown end: x_2 = x_1 + a and y_2 = y_1 + b, or x_1 = x_2 - a and y_1 = y_2 - b. With the tail at the origin, the head is (0 + a, 0 + b) = (a, b), which is the position vector result. Reading the components of \overrightarrow{PQ} as x_2 - x_1 and y_2 - y_1 is then exactly “position of Q minus position of P”, one coordinate at a time.

Reversing. Swapping head and tail swaps the two terms in each subtraction: x_1 - x_2 = -(x_2 - x_1) and y_1 - y_2 = -(y_2 - y_1).

Worked examples

Example 1

From two points to component form

Write the vector from A(-3, 4) to B(2, -1) in component form. Then write \overrightarrow{BA}, and give the head of the position vector drawing of \overrightarrow{AB}.

Solution

The tail is A and the head is B, so subtract A’s coordinates from B’s:

\overrightarrow{AB} = \langle 2 - (-3),\ -1 - 4 \rangle = \langle 5, -5 \rangle

The trip goes 5 right and 5 down. Reversing the arrow changes both signs: \overrightarrow{BA} = \langle -5, 5 \rangle. Check directly: \langle -3 - 2,\ 4 - (-1) \rangle gives the same.

The position vector drawing starts at the origin and goes 5 right and 5 down, so it ends at the point (5, -5). Notice the brackets: (5, -5) is where that one drawing ends, and \langle 5, -5 \rangle is the vector itself.

Example 2

Fractions and decimals

Find the component form of the vector from A\left(-\tfrac{3}{2}, \tfrac{2}{3}\right) to B\left(\tfrac{1}{2}, -\tfrac{7}{3}\right). Is it equal to the vector from C(0.25, 1.5) to D(2.25, -1.5)?

Solution

Terminal minus initial, coordinate by coordinate:

\begin{aligned} \overrightarrow{AB} &= \left\langle \tfrac{1}{2} - \left(-\tfrac{3}{2}\right),\ -\tfrac{7}{3} - \tfrac{2}{3} \right\rangle \\ &= \langle 2, -3 \rangle \end{aligned}

The fractions cancel neatly because the denominators match in each coordinate. For the second vector,

\overrightarrow{CD} = \langle 2.25 - 0.25,\ -1.5 - 1.5 \rangle = \langle 2, -3 \rangle

The components agree, 2 = 2 and -3 = -3, so \overrightarrow{AB} = \overrightarrow{CD}. The two arrows sit in different places and share no point, yet they are the same vector.

Example 3

A pass on a lacrosse field

A lacrosse field carries a grid in yards, x along the length of the field and y across it. Sanvi stands at S(14, 25) and passes. The ball’s trip from her stick to the catch is \langle 18, -9 \rangle. Where is Vedika, who catches it? Later Daisuke catches a pass at D(20, 40), and that pass was the vector \langle -7, 12 \rangle. Where was it thrown from?

Solution

The pass is a drawing of \langle 18, -9 \rangle with its tail at Sanvi, so the head is the tail plus the components:

(14 + 18,\ 25 + (-9)) = (32, 16)

Vedika is at (32, 16).

For Daisuke’s catch the head is known and the tail is not, so subtract the components from the head:

(20 - (-7),\ 40 - 12) = (27, 28)

Check. The vector from (27, 28) to (20, 40) is \langle 20 - 27,\ 40 - 28 \rangle = \langle -7, 12 \rangle, as given.

Example 4

Equal vectors give two equations

Find x and y so that

\langle 2x - 1,\ y + 3 \rangle = \langle x + 4,\ 2y - 5 \rangle

and name the vector. Then find every t with \langle t^2,\ 3t \rangle = \langle 4, -6 \rangle.

Solution

Equal vectors have equal horizontal components and equal vertical components, so one vector equation is two number equations:

2x - 1 = x + 4 \qquad\text{and}\qquad y + 3 = 2y - 5

The first gives x = 5 and the second y = 8. Then 2x - 1 = 9 and y + 3 = 11, so both sides are \langle 9, 11 \rangle. Check the right side: x + 4 = 9 and 2y - 5 = 11.

For the second equation, t^2 = 4 allows t = 2 or t = -2, but 3t = -6 allows only t = -2. Both must hold, so t = -2 alone. With t = 2 the vector would be \langle 4, 6 \rangle, which matches in one component and is a different vector.

Example 5

From position vectors on a warehouse floor

A forklift’s charging station is the origin of a grid on a warehouse floor, in meters. A pallet sits at the point with position vector \langle 8, -3 \rangle, and the shelf it must go to has position vector \langle -2, 9 \rangle. What vector carries the pallet to the shelf, and what vector brings the empty forklift back?

x y O P S
Figure 2The dashed arrows are the position vectors of P and S. The solid arrow, from P to S, is \langle -10, 12 \rangle.
Solution

A position vector ends at its point, so the pallet is at P(8, -3) and the shelf at S(-2, 9). The trip runs from P to S, so the head is S:

\overrightarrow{PS} = \langle -2 - 8,\ 9 - (-3) \rangle = \langle -10, 12 \rangle

That is 10 meters in the negative x direction and 12 in the positive y direction: the components of the shelf’s position vector minus those of the pallet’s. The trip back is the reverse, \overrightarrow{SP} = \langle 10, -12 \rangle.

Example 6

Building a square from one side

The square ABCD has its vertices in counterclockwise order, with A(-1, 1) and B(4, 2). Find C and D.

x y A B C D
Figure 3\overrightarrow{AB} = \langle 5, 1 \rangle, and a quarter turn counterclockwise gives \overrightarrow{AD} = \langle -1, 5 \rangle.
Solution

The side from A is \overrightarrow{AB} = \langle 5, 1 \rangle. The side \overrightarrow{AD} has the same length and is turned a quarter turn counterclockwise from it. Turning \langle a, b \rangle that way gives \langle -b, a \rangle, here \langle -1, 5 \rangle. Check it does what is needed. The slopes are \displaystyle \frac{1}{5} and \displaystyle \frac{5}{-1} = -5, whose product is -1, so the sides are perpendicular. The lengths, by the distance formula, are \displaystyle \sqrt{5^2 + 1^2} and \displaystyle \sqrt{(-1)^2 + 5^2}, both \displaystyle \sqrt{26}. And \langle -1, 5 \rangle points up and to the left, the counterclockwise side.

So D is the head of \langle -1, 5 \rangle drawn from A:

D = (-1 + (-1),\ 1 + 5) = (-2, 6)

In a square, \overrightarrow{BC} and \overrightarrow{AD} are opposite sides pointing the same way, so they are equal vectors, and C is the head of \langle -1, 5 \rangle drawn from B:

C = (4 + (-1),\ 2 + 5) = (3, 7)

Check. \overrightarrow{DC} = \langle 3 - (-2),\ 7 - 6 \rangle = \langle 5, 1 \rangle, equal to \overrightarrow{AB}, as the fourth side must be. Turning the other way, with \langle 1, -5 \rangle, would build the clockwise square on the other side of \overline{AB}, with D(0, -4) and C(5, -3).

Boundaries

Order matters. Component form is terminal minus initial. Initial minus terminal gives the reverse vector, \langle -a, -b \rangle.

A vector is not a point. \langle a, b \rangle has no location. To place a drawing you need one point, a tail or a head, as well as the components. Only the position vector drawing ties \langle a, b \rangle to the point (a, b).

Components depend on the grid. The numbers a and b are measured along the chosen axes in the chosen unit. Change the unit from yards to feet, or turn the axes, and the same arrow gets different components.

The zero vector has no direction. \langle 0, 0 \rangle has components like any vector, but its drawing is a single point with no arrow to point anywhere.

Two components belong to the plane. In space a vector needs three components; see vectors in space.

Where this goes next

With a vector named by two numbers, its length is \displaystyle \sqrt{a^2 + b^2} and its direction an angle, the magnitude and direction of the vector. Writing \langle a, b \rangle as a steps of a horizontal unit vector plus b of a vertical one is i and j notation. Stretching a vector multiplies both components by one number, a scalar multiple, and two trips in a row add component by component, vector addition. Components also make the dot product a two-term calculation.

Exercises

Component form from two points

  1. Write \overrightarrow{PQ} in component form for P(2, -5) and Q(-4, 1).
  2. Using the points of Exercise 1, write \overrightarrow{QP} in component form.
  3. Write the vector from M(0, 7) to N(-3, 0) in component form.
  4. Write the vector from E(-2.5, 1.25) to F(1, -0.75) in component form.
  5. Write the vector from G\left(\tfrac{1}{3}, -\tfrac{1}{2}\right) to H\left(-\tfrac{2}{3}, \tfrac{3}{2}\right) in component form.
  6. Write the vector from (h, k) to (h + 2,\ 2k) in component form.

Position vectors

  1. A vector is drawn from (-1, 6) to (4, 2). Write it in component form, and give the head of its position vector drawing.
  2. The vector \langle -2, 7 \rangle is drawn with its tail at (3, -4). Where is its head?
  3. On the warehouse floor of Example 5, Daisuke’s forklift picks up a box at the point with position vector \langle -5, 4 \rangle and drops it at the point with position vector \langle 7, 9 \rangle. What vector carries the box?

Equal vectors

  1. Is \overrightarrow{AB} equal to \overrightarrow{CD}, for A(3, -1), B(-2, 4), C(0, 6) and D(-5, 11)?
  2. Is \overrightarrow{AB} equal to \overrightarrow{CD}, for A(-3, 1), B(1, 5), C(2, -3) and D(6, 0)?
  3. Find x and y so that the vectors \langle 3x + 2,\ 5 - y \rangle and \langle x - 6,\ 2y + 11 \rangle are equal, and name the vector.
  4. Find every t with \langle t^2 - 1,\ 2t \rangle = \langle 8, -6 \rangle.
  5. Find every t with \langle t^2, t \rangle = \langle 4, 2 \rangle. Then find every t with \langle t^2, t \rangle = \langle 4, 3 \rangle.

Missing tail or head

  1. The vector \langle -7, 3 \rangle is drawn with its tail at (4, -2). Where is its head?
  2. The vector \langle 2.5, -6 \rangle is drawn with its head at (0, -1). Where is its tail?
  3. On the lacrosse field of Example 3, Vedika at (40, 18) passes to Daisuke, and the pass is the vector \langle -13, 9 \rangle. Where is Daisuke?
  4. Sanvi catches a pass at (35, 12). The pass was the vector \langle 16, -10 \rangle. Where was it thrown from?
  5. The square ABCD has its vertices in counterclockwise order, with A(0, -2) and B(2, 4). Find C and D.

Unknowns in the points

  1. The vector from A(a, 4) to B(-1, b) is \langle 3, -6 \rangle. Find a and b.
  2. Find the point P whose position vector equals \overrightarrow{PQ}, where Q(6, -4).

Reasoning

  1. Explain how the point (3, -1) and the vector \langle 3, -1 \rangle differ, and how they are linked.
  2. Show that if \overrightarrow{PQ} = \langle a, b \rangle, then \overrightarrow{QP} = \langle -a, -b \rangle.
  3. Vedika says the vector from (1, 2) to (4, 6) is \langle 4, 6 \rangle, since the arrow ends at (4, 6). Say what she has found instead, and give the correct component form.
  4. Show that a vector \langle a, b \rangle has exactly one drawing with its tail at a given point (h, k).
Answers
  1. \langle -6, 6 \rangle. \langle -4 - 2,\ 1 - (-5) \rangle.
  2. \langle 6, -6 \rangle. Reversing changes both signs: \langle 2 - (-4),\ -5 - 1 \rangle.
  3. \langle -3, -7 \rangle. \langle -3 - 0,\ 0 - 7 \rangle.
  4. \langle 3.5, -2 \rangle. \langle 1 - (-2.5),\ -0.75 - 1.25 \rangle.
  5. \langle -1, 2 \rangle. -\tfrac{2}{3} - \tfrac{1}{3} = -1 and \tfrac{3}{2} - \left(-\tfrac{1}{2}\right) = 2.
  6. \langle 2, k \rangle. (h + 2) - h = 2 and 2k - k = k.
  7. \langle 5, -4 \rangle; the position vector ends at (5, -4). \langle 4 - (-1),\ 2 - 6 \rangle.
  8. (1, 3). Tail plus components: (3 + (-2),\ -4 + 7).
  9. \langle 12, 5 \rangle. The box goes from (-5, 4) to (7, 9): \langle 7 - (-5),\ 9 - 4 \rangle.
  10. Equal. Both are \langle -5, 5 \rangle: \langle -2 - 3,\ 4 - (-1) \rangle and \langle -5 - 0,\ 11 - 6 \rangle.
  11. Not equal. \overrightarrow{AB} = \langle 4, 4 \rangle but \overrightarrow{CD} = \langle 4, 3 \rangle; the vertical components differ.
  12. x = -4, y = -2, vector \langle -10, 7 \rangle. From 3x + 2 = x - 6, 2x = -8; from 5 - y = 2y + 11, -6 = 3y. Then 3(-4) + 2 = -10 and 5 - (-2) = 7, and the right side agrees: -4 - 6 = -10, 2(-2) + 11 = 7.
  13. t = -3. t^2 - 1 = 8 gives t = 3 or t = -3, and 2t = -6 gives only t = -3.
  14. t = 2; then no t at all. For \langle 4, 2 \rangle, t^2 = 4 and t = 2 agree at t = 2. For \langle 4, 3 \rangle, t = 3 would make t^2 = 9, not 4.
  15. (-3, 1). (4 + (-7),\ -2 + 3).
  16. (-2.5, 5). Head minus components: (0 - 2.5,\ -1 - (-6)).
  17. (27, 27). (40 + (-13),\ 18 + 9).
  18. (19, 22). (35 - 16,\ 12 - (-10)). Check: \langle 35 - 19,\ 12 - 22 \rangle = \langle 16, -10 \rangle.
  19. C(-4, 6) and D(-6, 0). \overrightarrow{AB} = \langle 2, 6 \rangle; a quarter turn counterclockwise gives \langle -6, 2 \rangle, which is \overrightarrow{AD} and also \overrightarrow{BC}. Then D = (0 - 6,\ -2 + 2) and C = (2 - 6,\ 4 + 2). Check: \overrightarrow{DC} = \langle -4 - (-6),\ 6 - 0 \rangle = \langle 2, 6 \rangle, equal to \overrightarrow{AB}.
  20. a = -4 and b = -2. -1 - a = 3 and b - 4 = -6.
  21. P(3, -2). Write P(p, q). Its position vector is \langle p, q \rangle and \overrightarrow{PQ} = \langle 6 - p,\ -4 - q \rangle. Equal components: p = 6 - p and q = -4 - q, so p = 3 and q = -2. P is the midpoint of O and Q.
  22. (3, -1) is one location in the plane. \langle 3, -1 \rangle is a trip, 3 right and 1 down, drawable from any starting point. They are linked by the position vector: the drawing of \langle 3, -1 \rangle that starts at the origin ends at (3, -1).
  23. Say P(x_1, y_1) and Q(x_2, y_2), so a = x_2 - x_1 and b = y_2 - y_1. Then \overrightarrow{QP} = \langle x_1 - x_2,\ y_1 - y_2 \rangle, and each component is the negative of one of \overrightarrow{PQ}: x_1 - x_2 = -(x_2 - x_1) = -a and y_1 - y_2 = -(y_2 - y_1) = -b. So \overrightarrow{QP} = \langle -a, -b \rangle.
  24. She has written the head’s coordinates, which are the components only when the tail is the origin. Here the tail is (1, 2), so the vector is \langle 4 - 1,\ 6 - 2 \rangle = \langle 3, 4 \rangle.
  25. A drawing with tail (h, k) and head (x, y) is \langle x - h,\ y - k \rangle. It equals \langle a, b \rangle exactly when x - h = a and y - k = b, that is when the head is (h + a,\ k + b). That head works, and no other does, so there is exactly one such drawing.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0