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Precalculus · N.VM.A.1 · N.VM.A.3

Magnitude and direction of a vector

A vector is a trip of so much across and so much up. How long is the trip, which way does it point, and how do you get the two components back from a length and a direction?

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When you reach for it

Emi, Juho and Lotta drive hovercraft for a club on a large frozen lake. The club’s map puts the launch ramp at the origin, with x in meters east and y in meters north, and a logger on each craft records a run in component form: so many meters east, so many north. A driver, though, asks different questions. How far did I go? Which way was I pointed? Those are the vector’s magnitude and direction.

The traffic also runs the other way. A course card says “drive 150 meters at 64^\circ”, and to put that on the map, or to add it to another run later, you need its components.

Component form is the name to compute with; magnitude and direction are the name to describe and to steer by. This skill turns either name into the other.

The procedure

Magnitude. The magnitude of \mathbf{v} = \langle a, b \rangle, written \|\mathbf{v}\|, is the length of any of its drawings. All drawings of one vector have the same length, so measure the easiest, the position drawing from O(0, 0) to (a, b). The distance formula gives \displaystyle \sqrt{(a - 0)^2 + (b - 0)^2}:

\|\mathbf{v}\| = \sqrt{a^2 + b^2}

In Figure 1 this is the Pythagorean theorem on a right triangle with legs |a| and |b|. Squaring removes the signs, so the formula is the same in every quadrant. For the vector from P to Q it is simply the distance PQ.

x y
\theta
a
b
\|\mathbf{v}\|
(a, b)
 O
Figure 1The position drawing of \mathbf{v} = \langle a, b \rangle is the long side of a right triangle with legs a and b. Its length is \|\mathbf{v}\| and its direction is \theta.

Direction angle. The direction angle \theta of a nonzero vector is the angle in standard position whose terminal side runs along the position drawing: measured counterclockwise from the positive x-axis, with 0^\circ \le \theta < 360^\circ. The point (a, b) sits on that terminal side at distance \|\mathbf{v}\| from O, so by the definition of the trig functions of any angle

\begin{aligned} \cos\theta &= \frac{a}{\|\mathbf{v}\|}, \qquad \sin\theta = \frac{b}{\|\mathbf{v}\|}, \\ \tan\theta &= \frac{b}{a} \quad (a \ne 0) \end{aligned}

The quickest route to \theta is the tangent, but the inverse tangent only ever returns an angle strictly between -90^\circ and 90^\circ, that is, a ray pointing to the right. Tangent repeats every 180^\circ: the rays at \theta and \theta + 180^\circ make up one line through O, and both have slope \displaystyle \frac{b}{a}. So \displaystyle \arctan\frac{b}{a} always finds the right line, and the sign of a decides which of its two rays the vector is on. Writing \displaystyle \alpha = \arctan\frac{b}{a}:

ComponentsWhere \mathbf{v} pointsDirection angle
a > 0,\ b \ge 0quadrant I or along the positive x-axis\theta = \alpha
a < 0quadrant II or III, or along the negative x-axis\theta = \alpha + 180^\circ
a > 0,\ b < 0quadrant IV\theta = \alpha + 360^\circ
a = 0,\ b > 0straight up\theta = 90^\circ
a = 0,\ b < 0straight down\theta = 270^\circ
a = 0,\ b = 0nowhere: the zero vectorno direction angle

The middle row adds 180^\circ because a left-pointing vector lies on the opposite ray to the one the key returns. The third row’s ray is the right one already, at a negative angle, and adding 360^\circ names the same ray inside [0^\circ, 360^\circ). The signs of a and b are exactly the signs of \cos\theta and \sin\theta, so this is pinning down a quadrant from two signs.

From magnitude and direction back to components. Solve the cosine and sine equations above for a and b:

\mathbf{v} = \langle \|\mathbf{v}\|\cos\theta,\ \|\mathbf{v}\|\sin\theta \rangle

This is called resolving the vector into components. No quadrant fixing is needed in this direction: cosine and sine already carry the right signs for every \theta. It is the same computation that turns a polar point into x and y in converting polar to rectangular, and navigation with bearings does it with angles measured clockwise from north instead.

Worked examples

Example 1

Magnitude and direction in quadrant I

Find the magnitude and direction angle of \mathbf{v} = \langle 7, 24 \rangle.

Solution
\begin{aligned} \|\mathbf{v}\| &= \sqrt{7^2 + 24^2} = \sqrt{49 + 576} \\ &= \sqrt{625} = 25 \end{aligned}

Both components are positive, so \mathbf{v} points into quadrant I, the one quadrant where the arctangent key answers directly:

\theta = \arctan\frac{24}{7} \approx 73.7^\circ

Check. 25\cos 73.7^\circ \approx 7.0 and 25\sin 73.7^\circ \approx 24.0.

Example 2

Two vectors the tangent cannot tell apart

Find the magnitude and direction angle of \displaystyle \mathbf{u} = \langle -3, 3\sqrt{3} \rangle and of \displaystyle \mathbf{w} = \langle 3, -3\sqrt{3} \rangle.

Solution

Squaring removes the signs, so both have the same magnitude:

\|\mathbf{u}\| = \|\mathbf{w}\| = \sqrt{9 + 27} = \sqrt{36} = 6

Both also have the same ratio, \displaystyle \frac{3\sqrt{3}}{-3} = \frac{-3\sqrt{3}}{3} = -\sqrt{3}, so the key gives \displaystyle \arctan(-\sqrt{3}) = -60^\circ for both. That cannot be both answers: the vectors point opposite ways along one line (Figure 2).

x y 120° −60°
\mathbf{u}
\mathbf{w}
Figure 2Example 2. Both vectors have \displaystyle \frac{b}{a} = -\sqrt{3}, and the arctangent key gives -60^\circ for both. That is the ray of \mathbf{w}; \mathbf{u} lies on the opposite ray, at 120^\circ.

\mathbf{u} has a < 0 and b > 0, so it points into quadrant II, on the ray opposite the one the key found: \theta_{\mathbf{u}} = -60^\circ + 180^\circ = 120^\circ. \mathbf{w} has a > 0 and b < 0, quadrant IV, the key’s own ray at a negative angle: \theta_{\mathbf{w}} = -60^\circ + 360^\circ = 300^\circ.

Check. 6\cos 120^\circ = 6\left(-\tfrac{1}{2}\right) = -3 and \displaystyle 6\sin 120^\circ = 6 \cdot \tfrac{\sqrt{3}}{2} = 3\sqrt{3}, the components of \mathbf{u}.

Example 3

A logged run between two buoys

Emi’s logger records a run from a buoy at P(15, 40) to a buoy at Q(-55, -20). How far did she travel, to the nearest hundredth of a meter, and what was her direction angle, to the nearest tenth of a degree?

Solution

The run is the vector from P to Q, terminal minus initial:

\overrightarrow{PQ} = \langle -55 - 15,\ -20 - 40 \rangle = \langle -70, -60 \rangle

Its magnitude is \displaystyle \sqrt{4900 + 3600} = \sqrt{8500} = 10\sqrt{85} \approx 92.20 meters. Both components are negative, so she drove into quadrant III, southwest of where she started. The key gives \displaystyle \arctan\frac{-60}{-70} = \arctan\frac{6}{7} \approx 40.6^\circ, a ray pointing up and to the right, the opposite of her heading. Adding 180^\circ:

\theta \approx 40.6^\circ + 180^\circ = 220.6^\circ

Where the buoys sit made no difference: only the components entered.

Example 4

Components from a magnitude and a direction

Write each vector in component form: (a) \|\mathbf{v}\| = 12 and \theta = 150^\circ; (b) \|\mathbf{v}\| = 9 and \theta = 270^\circ; (c) \|\mathbf{v}\| = 10 and \displaystyle \theta = \frac{5\pi}{4}.

Solution

Each is \langle \|\mathbf{v}\|\cos\theta,\ \|\mathbf{v}\|\sin\theta \rangle, with the exact values of cosine and sine.

(a) \displaystyle \cos 150^\circ = -\frac{\sqrt{3}}{2} and \displaystyle \sin 150^\circ = \frac{1}{2}, so \displaystyle \mathbf{v} = \langle -6\sqrt{3}, 6 \rangle. The negative first component appeared on its own, from cosine. Check: \displaystyle \sqrt{108 + 36} = \sqrt{144} = 12.

(b) \cos 270^\circ = 0 and \sin 270^\circ = -1, so \mathbf{v} = \langle 0, -9 \rangle, straight down.

(c) The angle is in radians, and the formula does not care which unit names the angle. Here \displaystyle \cos\frac{5\pi}{4} = \sin\frac{5\pi}{4} = -\frac{\sqrt{2}}{2}, so \displaystyle \mathbf{v} = \langle -5\sqrt{2}, -5\sqrt{2} \rangle.

Example 5

Two runs from a course card

From the launch ramp, Lotta drives 150 meters at direction angle 64^\circ. Juho drives 80 meters in a direction 35^\circ south of west. Resolve each run into components, to the nearest hundredth of a meter.

x y 64° 35° Lotta Juho O
Figure 3Example 5. Lotta’s run has direction angle 64^\circ. Juho’s is described as 35^\circ south of west, which is the direction angle 180^\circ + 35^\circ = 215^\circ.
Solution

Lotta’s angle is already a direction angle:

\langle 150\cos 64^\circ,\ 150\sin 64^\circ \rangle \approx \langle 65.76, 134.82 \rangle

so she ends about 65.76 meters east and 134.82 meters north of the ramp.

“35^\circ south of west” means: face west, then turn 35^\circ toward south. West is 180^\circ, and turning from west toward south is counterclockwise, so the direction angle is 215^\circ:

\langle 80\cos 215^\circ,\ 80\sin 215^\circ \rangle \approx \langle -65.53, -45.89 \rangle

Check with the triangle in Figure 3. Its legs are 80\cos 35^\circ \approx 65.53 (west) and 80\sin 35^\circ \approx 45.89 (south), and attaching the signs by hand, negative for west and for south, gives the same vector.

Example 6

A missing component

The vector \langle a, 8 \rangle has magnitude 17 and points into quadrant II. Find a and the direction angle, to the nearest tenth of a degree.

Solution

The magnitude gives an equation for a:

\begin{aligned} a^2 + 8^2 &= 17^2 \\ a^2 &= 225 \\ a &= 15 \ \text{or}\ a = -15 \end{aligned}

Two vectors, \langle 15, 8 \rangle and \langle -15, 8 \rangle, have length 17 and this second component. Quadrant II needs a < 0, so a = -15. Then a < 0 calls for 180^\circ:

\theta = \arctan\frac{8}{-15} + 180^\circ \approx -28.1^\circ + 180^\circ = 151.9^\circ
Example 7

Magnitude and a tangent

A vector has magnitude 26, and its direction angle \theta has \displaystyle \tan\theta = \frac{5}{12} but does not lie in quadrant I. Find the vector.

Solution

A positive tangent means \theta is in quadrant I or III, so here it is in quadrant III, where cosine and sine are both negative. A right triangle with legs 5 and 12 has long side \displaystyle \sqrt{25 + 144} = 13, so \displaystyle |\cos\theta| = \frac{12}{13} and \displaystyle |\sin\theta| = \frac{5}{13}. With the quadrant III signs,

\begin{aligned} \mathbf{v} &= \left\langle 26\left(-\tfrac{12}{13}\right),\ 26\left(-\tfrac{5}{13}\right) \right\rangle \\ &= \langle -24, -10 \rangle \end{aligned}

Check. \displaystyle \sqrt{576 + 100} = \sqrt{676} = 26, and \displaystyle \frac{-10}{-24} = \frac{5}{12}. The tangent alone allowed \langle 24, 10 \rangle as well; the quadrant chose between them. The angle itself is \displaystyle \arctan\frac{5}{12} + 180^\circ \approx 202.6^\circ.

Boundaries

The zero vector has no direction. \langle 0, 0 \rangle has magnitude 0, and its drawing is a single point, so no ray and no direction angle. Every other vector has exactly one direction angle in [0^\circ, 360^\circ).

\displaystyle \arctan\frac{b}{a} is a direction angle only in quadrant I. For a < 0 it is off by 180^\circ, and for a vector in quadrant IV it is negative. When a = 0 the ratio does not exist, and the direction is 90^\circ or 270^\circ, read from the sign of b.

The angle convention must be stated. Direction angles here start at the positive x-axis and turn counterclockwise. A bearing starts at north and turns clockwise, and “35^\circ south of west” starts at west. Convert to a direction angle before using \langle \|\mathbf{v}\|\cos\theta, \|\mathbf{v}\|\sin\theta \rangle.

Magnitude is never negative, and it is 0 only for the zero vector. A magnitude and an angle name one vector; a magnitude alone, or a tangent alone, leaves several.

This is the plane. In space a vector has three components and its direction needs more than one angle; see vectors in space.

Where this goes next

Dividing a vector by its magnitude gives a vector of length 1 in the same direction, and the two unit vectors along the axes are the \mathbf{i} and \mathbf{j} of i and j notation. Stretching a vector multiplies its magnitude and keeps or reverses its direction, a scalar multiple. Resolving is what makes vector addition practical: two runs given as lengths and angles are resolved, added component by component, and turned back into a length and an angle. With pushes and speeds in place of runs, that is the work of forces and velocities as vectors. And \|\mathbf{u}\|\|\mathbf{v}\|\cos\theta reappears as the dot product.

Exercises

Magnitude

  1. Find \|\langle -9, 40 \rangle\|.
  2. Find \|\langle 8, -4 \rangle\| exactly.
  3. Find \|\langle 2.4, -1 \rangle\|.
  4. Find the magnitude of the vector from A(-2, 7) to B(4, -1).

Direction angle

Give each magnitude exactly, and each direction angle exactly or to the nearest tenth of a degree.

  1. \displaystyle \langle 5, 5\sqrt{3} \rangle
  2. \langle -5, 5 \rangle
  3. \langle -3, -5 \rangle
  4. \langle 3, -7 \rangle
  5. \langle 0, -6 \rangle
  6. \langle -11, 0 \rangle

Components from magnitude and direction

Write each vector in component form, exactly where the angle allows, otherwise to the nearest hundredth.

  1. \|\mathbf{v}\| = 14, \theta = 60^\circ
  2. \|\mathbf{v}\| = 8, \theta = 225^\circ
  3. \|\mathbf{v}\| = 4, \displaystyle \theta = \frac{11\pi}{6}
  4. \|\mathbf{v}\| = 40, \theta = 317^\circ
  5. \|\mathbf{v}\| = 2.5, \theta = 100^\circ

On the lake

  1. Emi drives 120 meters in a direction 20^\circ north of west. Give her direction angle and resolve the run into components, to the nearest hundredth of a meter.
  2. Lotta’s logger shows her at (-45, -60), in meters from the launch ramp. How far is she from the ramp, and what is the direction angle of the trip from the ramp to her, to the nearest tenth of a degree?
  3. Juho drives straight from a buoy at (30, -10) to a buoy at (-20, 30). Find the length of the run, exactly and to the nearest hundredth of a meter, and its direction angle to the nearest tenth of a degree.

Unknown parts

  1. The vector \langle a, -20 \rangle has magnitude 29 and points into quadrant IV. Find a and the direction angle, to the nearest tenth of a degree.
  2. A vector has magnitude 15, and its direction angle \theta lies in quadrant III with \displaystyle \cos\theta = -\frac{3}{5}. Find the vector.
  3. Find every vector with magnitude 10 and horizontal component 6, with its direction angle to the nearest tenth of a degree.

Reasoning

  1. Juho reports the direction angle of \langle -4, -1 \rangle as \displaystyle \arctan\frac{-1}{-4} \approx 14.0^\circ. Explain what went wrong and give the correct angle.
  2. Show that the vector \langle r\cos\theta,\ r\sin\theta \rangle has magnitude r for every angle \theta and every r \ge 0.
  3. Explain why \langle a, b \rangle and \langle -a, -b \rangle, for \langle a, b \rangle \ne \langle 0, 0 \rangle, have the same magnitude and direction angles that differ by 180^\circ.
  4. Explain why a nonzero vector has exactly one direction angle in [0^\circ, 360^\circ), and why the zero vector has none.
Answers
  1. 41. \displaystyle \sqrt{81 + 1600} = \sqrt{1681} = 41.
  2. \displaystyle 4\sqrt{5} \approx 8.94. \displaystyle \sqrt{64 + 16} = \sqrt{80} = \sqrt{16 \cdot 5}.
  3. 2.6. \displaystyle \sqrt{5.76 + 1} = \sqrt{6.76}.
  4. 10. The vector is \langle 4 - (-2),\ -1 - 7 \rangle = \langle 6, -8 \rangle, and \displaystyle \sqrt{36 + 64} = 10.
  5. 10 and 60^\circ. \displaystyle \sqrt{25 + 75} = 10; quadrant I, \displaystyle \arctan\sqrt{3} = 60^\circ.
  6. \displaystyle 5\sqrt{2} and 135^\circ. \displaystyle \sqrt{50}; a < 0, so \arctan(-1) + 180^\circ = -45^\circ + 180^\circ.
  7. \displaystyle \sqrt{34} \approx 5.83 and 239.0^\circ. Quadrant III: \displaystyle \arctan\frac{5}{3} + 180^\circ \approx 59.0^\circ + 180^\circ.
  8. \displaystyle \sqrt{58} \approx 7.62 and 293.2^\circ. Quadrant IV: \displaystyle \arctan\left(-\frac{7}{3}\right) + 360^\circ \approx -66.8^\circ + 360^\circ.
  9. 6 and 270^\circ. a = 0 and b < 0: straight down.
  10. 11 and 180^\circ. a < 0 with b = 0: \arctan 0 + 180^\circ.
  11. \displaystyle \langle 7, 7\sqrt{3} \rangle. \displaystyle 14 \cdot \frac{1}{2} and \displaystyle 14 \cdot \frac{\sqrt{3}}{2}.
  12. \displaystyle \langle -4\sqrt{2}, -4\sqrt{2} \rangle. \displaystyle \cos 225^\circ = \sin 225^\circ = -\frac{\sqrt{2}}{2}.
  13. \displaystyle \langle 2\sqrt{3}, -2 \rangle. \displaystyle \cos\frac{11\pi}{6} = \frac{\sqrt{3}}{2} and \displaystyle \sin\frac{11\pi}{6} = -\frac{1}{2}.
  14. \langle 29.25, -27.28 \rangle. 40\cos 317^\circ \approx 29.254 and 40\sin 317^\circ \approx -27.280.
  15. \langle -0.43, 2.46 \rangle. 2.5\cos 100^\circ \approx -0.434 and 2.5\sin 100^\circ \approx 2.462.
  16. \theta = 160^\circ and \langle -112.76, 41.04 \rangle. Face west, 180^\circ, and turn 20^\circ toward north, which is clockwise: 180^\circ - 20^\circ. Then 120\cos 160^\circ \approx -112.76 and 120\sin 160^\circ \approx 41.04.
  17. 75 meters at 233.1^\circ. \displaystyle \sqrt{2025 + 3600} = \sqrt{5625} = 75; quadrant III, \displaystyle \arctan\frac{4}{3} + 180^\circ \approx 53.1^\circ + 180^\circ.
  18. \displaystyle 10\sqrt{41} \approx 64.03 meters at 141.3^\circ. The run is \langle -20 - 30,\ 30 - (-10) \rangle = \langle -50, 40 \rangle, length \displaystyle \sqrt{4100}; a < 0, so \arctan(-0.8) + 180^\circ \approx -38.7^\circ + 180^\circ.
  19. a = 21 and \theta \approx 316.4^\circ. a^2 = 841 - 400 = 441, and quadrant IV needs a > 0; then \displaystyle \arctan\left(-\frac{20}{21}\right) + 360^\circ \approx -43.6^\circ + 360^\circ.
  20. \langle -9, -12 \rangle. In quadrant III, \displaystyle \sin\theta = -\sqrt{1 - \frac{9}{25}} = -\frac{4}{5}, so \displaystyle \langle 15\left(-\frac{3}{5}\right),\ 15\left(-\frac{4}{5}\right) \rangle.
  21. \langle 6, 8 \rangle at 53.1^\circ and \langle 6, -8 \rangle at 306.9^\circ. 36 + b^2 = 100 gives b = 8 or b = -8; then \displaystyle \arctan\frac{4}{3} \approx 53.1^\circ, and \displaystyle \arctan\left(-\frac{4}{3}\right) + 360^\circ \approx 306.9^\circ.
  22. The key returns a ray pointing right, but \langle -4, -1 \rangle points left, into quadrant III. The ray at 14.0^\circ belongs to \langle 4, 1 \rangle, the opposite vector. The correct angle is \displaystyle \arctan\frac{1}{4} + 180^\circ \approx 194.0^\circ.
  23. The magnitude is \displaystyle \sqrt{r^2\cos^2\theta + r^2\sin^2\theta}, which equals \displaystyle \sqrt{r^2(\cos^2\theta + \sin^2\theta)}, then \displaystyle \sqrt{r^2}, then r, by the Pythagorean identity and because r \ge 0.
  24. (-a)^2 + (-b)^2 = a^2 + b^2, so the magnitudes match. The heads (a, b) and (-a, -b) are reflections of each other through O, so the two position drawings lie on one line through O on opposite rays, and opposite rays differ by a half turn, 180^\circ.
  25. A nonzero vector’s position drawing is one ray from O, and as \theta runs once around from 0^\circ up to (not including) 360^\circ the terminal side passes every ray exactly once. The zero vector’s drawing is the single point O, which lies on every ray and points along none.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0