Precalculus

Unit vectors and i, j notation

A direction alone has no length. How do you write down “this way” as a vector, and then build a vector of any length you like pointing that way?

You need

Magnitude and direction of a vectorPrecalculus

Where you are 4 in Vectors

Trigonometry7

Polar coordinates and vectors10

Vectors4

Component form of a vector Directed line segments Magnitude and direction of a vector

Unit vectors and i, j notation

Why this exists

A vector carries two pieces of information, a length and a direction, and magnitude and direction showed how to read each one off the components \langle a, b \rangle. Often the two pieces arrive separately. Ghada, Jamila and Karim pilot tugboats, and a tug’s compass says which way the bow points while its log says how far it went. The heading has no length of its own. To treat “this way” as a vector at all, you fix its length at the one value that changes nothing when you multiply by it: 1.

A vector of length 1 is a pure direction. Every other vector in that direction is a length times it, and the two unit vectors along the axes, \mathbf{i} and \mathbf{j}, give a second way to write every vector: so much of “one right” plus so much of “one up”.

Definition

A unit vector is a vector with magnitude 1.

For a nonzero vector \mathbf{v} = \langle a, b \rangle, the unit vector in the direction of \mathbf{v} is \mathbf{v} with each component divided by its magnitude:

\frac{\mathbf{v}}{\|\mathbf{v}\|} = \left\langle \frac{a}{\|\mathbf{v}\|},\ \frac{b}{\|\mathbf{v}\|} \right\rangle

The two standard unit vectors point along the positive axes:

\mathbf{i} = \langle 1, 0 \rangle, \qquad \mathbf{j} = \langle 0, 1 \rangle

\mathbf{i}, \mathbf{j} notation. The vector \langle a, b \rangle is also written

a\mathbf{i} + b\mathbf{j}

read “a of \mathbf{i} plus b of \mathbf{j}”: a steps one to the right and b steps one up. For now, a\mathbf{i} + b\mathbf{j} is simply another name for \langle a, b \rangle. Multiplying a vector by a number is a scalar multiple and adding two vectors is vector addition; with those in hand, a\mathbf{i} really is \langle a, 0 \rangle, b\mathbf{j} really is \langle 0, b \rangle, and their sum really is \langle a, b \rangle. A term with a zero coefficient is left out, so \langle 4, 0 \rangle = 4\mathbf{i} and \langle 0, -1 \rangle = -\mathbf{j}.

Intuition

Draw \mathbf{v} from the origin. Its unit vector lies along the same ray but is cut off at length 1, so it ends exactly where the ray crosses the unit circle (Figure 1). That crossing point is (\cos\theta, \sin\theta), where \theta is the direction angle. A unit vector, then, is a point on the unit circle seen as an arrow, and \mathbf{i} and \mathbf{j} are the arrows to the points at 0^\circ and 90^\circ.

Figure 1The unit vector \mathbf{u} lies along \mathbf{v} and ends on the unit circle, so its components are \cos\theta and \sin\theta. The vectors \mathbf{i} and \mathbf{j} are the unit vectors at 0^\circ and 90^\circ.

Dividing both components by the same positive number shrinks the arrow without turning it: the signs of a and b stay, so the quadrant stays, and the ratio \displaystyle \frac{b}{a} stays, so the line stays.

The \mathbf{i}, \mathbf{j} name reads as a set of instructions (Figure 2). 3\mathbf{i} + 2\mathbf{j} says: take three unit steps right, then two unit steps up. A negative coefficient walks the other way, so -\mathbf{i} is one step left.

Figure 2\mathbf{v} = 3\mathbf{i} + 2\mathbf{j} reads as three steps of \mathbf{i} and two steps of \mathbf{j}, the trip \langle 3, 2 \rangle.

Results

Unit vectors are points of the unit circle. For every angle \theta, \langle \cos\theta, \sin\theta \rangle is a unit vector with direction angle \theta, and every unit vector has this form.

The unit vector of a vector. If \mathbf{v} \ne \mathbf{0} has direction angle \theta, then

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

a unit vector pointing the same way as \mathbf{v}.

Every plane vector in \mathbf{i}, \mathbf{j} form. Each \mathbf{v} = \langle a, b \rangle is a\mathbf{i} + b\mathbf{j} for exactly one pair a, b, and \displaystyle \|a\mathbf{i} + b\mathbf{j}\| = \sqrt{a^2 + b^2}.

A given length in a given direction. The vector of magnitude L > 0 and direction angle \theta is

L\cos\theta\,\mathbf{i} + L\sin\theta\,\mathbf{j}

and the vector of magnitude L in the direction of a nonzero \mathbf{w} = \langle p, q \rangle is \displaystyle \left\langle \frac{Lp}{\|\mathbf{w}\|}, \frac{Lq}{\|\mathbf{w}\|} \right\rangle.

Why it is true

Points of the unit circle. By the Pythagorean identity, \displaystyle \sqrt{\cos^2\theta + \sin^2\theta} = 1, and the point (\cos\theta, \sin\theta) lies on the terminal side of \theta, so that is the direction. Conversely, if \langle a, b \rangle is a unit vector then a^2 + b^2 = 1, so (a, b) is on the unit circle. Let \theta be the angle whose terminal side passes through it; the point where that terminal side meets the unit circle is (\cos\theta, \sin\theta) by the definition of cosine and sine, so a = \cos\theta and b = \sin\theta.

The unit vector of a vector. Resolving \mathbf{v} gives a = \|\mathbf{v}\|\cos\theta and b = \|\mathbf{v}\|\sin\theta. Divide each by \|\mathbf{v}\|, which is allowed because \mathbf{v} \ne \mathbf{0} makes it positive: \displaystyle \frac{a}{\|\mathbf{v}\|} = \cos\theta and \displaystyle \frac{b}{\|\mathbf{v}\|} = \sin\theta. By the first result this is a unit vector with the same direction angle \theta as \mathbf{v}.

\mathbf{i}, \mathbf{j} form. a\mathbf{i} + b\mathbf{j} names \langle a, b \rangle, and two vectors in component form are equal exactly when their components match, so the pair is unique. The magnitude is the one for \langle a, b \rangle.

A given length. The vector of magnitude L and direction angle \theta resolves to \langle L\cos\theta, L\sin\theta \rangle. In the direction of \mathbf{w}, \displaystyle \cos\theta = \frac{p}{\|\mathbf{w}\|} and \displaystyle \sin\theta = \frac{q}{\|\mathbf{w}\|} by the second result; substitute.

Worked examples

Example 1

Changing names

(a) Write \langle 5, -7 \rangle in \mathbf{i}, \mathbf{j} form and find its magnitude. (b) Write -3\mathbf{i} + \mathbf{j} and (c) -2\mathbf{j} in component form.

Solution

(a) The first component is the coefficient of \mathbf{i} and the second the coefficient of \mathbf{j}: \langle 5, -7 \rangle = 5\mathbf{i} - 7\mathbf{j}. A negative coefficient is written with a minus sign rather than “+ (-7)\mathbf{j}”. The magnitude is \displaystyle \sqrt{25 + 49} = \sqrt{74}.

(b) A bare \mathbf{j} has coefficient 1, so -3\mathbf{i} + \mathbf{j} = \langle -3, 1 \rangle.

(c) There is no \mathbf{i} term, so the first component is 0: -2\mathbf{j} = \langle 0, -2 \rangle, two units straight down.

Example 2

A unit vector with whole-number magnitude

Find the unit vector in the direction of \mathbf{v} = \langle -8, 15 \rangle.

Solution

First the magnitude: \displaystyle \|\mathbf{v}\| = \sqrt{64 + 225} = \sqrt{289} = 17. Divide each component by it:

\frac{\mathbf{v}}{\|\mathbf{v}\|} = \left\langle -\frac{8}{17}, \frac{15}{17} \right\rangle = -\frac{8}{17}\mathbf{i} + \frac{15}{17}\mathbf{j}

Check. \displaystyle \frac{64}{289} + \frac{225}{289} = \frac{289}{289} = 1, so the length is 1. Both \mathbf{v} and its unit vector have a negative first and positive second component, so both point into quadrant II.

Example 3

A unit vector with a radical

Find the unit vector in the direction of \mathbf{v} = 2\mathbf{i} - 6\mathbf{j}.

Solution

\displaystyle \|\mathbf{v}\| = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}. Dividing,

\begin{aligned} \frac{\mathbf{v}}{\|\mathbf{v}\|} &= \frac{2}{2\sqrt{10}}\mathbf{i} - \frac{6}{2\sqrt{10}}\mathbf{j} = \frac{1}{\sqrt{10}}\mathbf{i} - \frac{3}{\sqrt{10}}\mathbf{j} \\ &= \frac{\sqrt{10}}{10}\mathbf{i} - \frac{3\sqrt{10}}{10}\mathbf{j} \end{aligned}

The last line multiplies top and bottom by \displaystyle \sqrt{10}, which changes no value.

Check. \displaystyle \frac{10}{100} + \frac{90}{100} = 1.

Example 4

From a direction angle

(a) Write the unit vector with direction angle 210^\circ. (b) Write the vector of magnitude 14 in the same direction. Give both in \mathbf{i}, \mathbf{j} form.

Solution

(a) A unit vector is \langle \cos\theta, \sin\theta \rangle, and 210^\circ is 30^\circ past the negative x-axis, in quadrant III where both are negative:

\mathbf{u} = \cos 210^\circ\,\mathbf{i} + \sin 210^\circ\,\mathbf{j} = -\frac{\sqrt{3}}{2}\mathbf{i} - \frac{1}{2}\mathbf{j}

(b) Magnitude 14 multiplies each component of \mathbf{u} by 14: \displaystyle 14\cos 210^\circ\,\mathbf{i} + 14\sin 210^\circ\,\mathbf{j} = -7\sqrt{3}\,\mathbf{i} - 7\mathbf{j}. Check: \displaystyle \sqrt{147 + 49} = \sqrt{196} = 14.

Example 5

A tugboat heads for a buoy

Jamila’s tug is at P(30, 50) on a map in meters, x east and y north. She steers straight toward a buoy at Q(50, 10) and goes 30 meters. Where does she stop, to the nearest hundredth of a meter?

Solution

The heading is the direction of \overrightarrow{PQ} = \langle 20, -40 \rangle, terminal minus initial. Its magnitude is \displaystyle \sqrt{400 + 1600} = \sqrt{2000} = 20\sqrt{5}, so the unit vector is

\left\langle \frac{20}{20\sqrt{5}}, \frac{-40}{20\sqrt{5}} \right\rangle = \left\langle \frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}} \right\rangle

The trip has length 30 in that direction, so multiply each component by 30:

\left\langle \frac{30}{\sqrt{5}}, -\frac{60}{\sqrt{5}} \right\rangle = \langle 6\sqrt{5}, -12\sqrt{5} \rangle \approx \langle 13.42, -26.83 \rangle

The trip starts at P, so its head is (30 + 13.42,\ 50 - 26.83) = (43.42, 23.17). The buoy is \displaystyle 20\sqrt{5} \approx 44.72 meters away, so she is still short of it, as she should be.

Example 6

Unit vectors along a line

Find every unit vector that lies along the line y = 2x, and the direction angle of each, to the nearest tenth of a degree.

Solution

A vector drawn from the origin along y = 2x ends at a point (a, 2a), so it is a\mathbf{i} + 2a\mathbf{j}. Being a unit vector means

\begin{aligned} a^2 + (2a)^2 &= 1 \\ 5a^2 &= 1 \\ a &= \frac{\sqrt{5}}{5} \quad \text{or} \quad a = -\frac{\sqrt{5}}{5} \end{aligned}

So there are two: \displaystyle \frac{\sqrt{5}}{5}\mathbf{i} + \frac{2\sqrt{5}}{5}\mathbf{j}, pointing into quadrant I at \arctan 2 \approx 63.4^\circ, and its opposite \displaystyle -\frac{\sqrt{5}}{5}\mathbf{i} - \frac{2\sqrt{5}}{5}\mathbf{j}, at 63.4^\circ + 180^\circ = 243.4^\circ. A line has two directions, and each has one unit vector.

Boundaries

The zero vector has no unit vector. \displaystyle \frac{\mathbf{0}}{\|\mathbf{0}\|} would divide by 0, and \mathbf{0} has no direction to keep. Every other vector has exactly one unit vector in its direction.

A length must be positive. “Magnitude L in direction \theta” needs L > 0. Using -L gives a vector of length L pointing the opposite way, direction \theta + 180^\circ.

Two letters, two dimensions. \mathbf{i} and \mathbf{j} name every vector in the plane. In space a third unit vector \mathbf{k} along the z-axis is needed; see vectors in space. The letter \mathbf{i} here is unrelated to the imaginary unit i; context and bold type tell them apart.

Where this goes next

Scalar multiples make “multiply each component by 14” an operation of its own, and \displaystyle L\,\frac{\mathbf{v}}{\|\mathbf{v}\|} becomes one product. Vector addition makes a\mathbf{i} + b\mathbf{j} a true sum, which is why \mathbf{i}, \mathbf{j} form is the convenient name when vectors are combined: like terms gather. In forces and velocities a speed times a unit heading is the velocity vector, and in the dot product unit vectors measure the angle between two directions.

Exercises

\mathbf{i} and \mathbf{j} form

Write \langle -6, 13 \rangle in \mathbf{i}, \mathbf{j} form.
Write 7\mathbf{i} - 9\mathbf{j} in component form.
Write \displaystyle \frac{1}{2}\mathbf{i} - 3\mathbf{j} in component form.
Find \|9\mathbf{i} - 40\mathbf{j}\|.
Find \|\mathbf{i} + \mathbf{j}\|.
Write the vector from A(4, -1) to B(-2, 3) in \mathbf{i}, \mathbf{j} form.

The unit vector in the direction of \mathbf{v}

Give each answer exactly, with no radical in a denominator.

\mathbf{v} = \langle -12, -35 \rangle
\mathbf{v} = 20\mathbf{i} - 21\mathbf{j}
\mathbf{v} = \langle 5, 5 \rangle
\mathbf{v} = -2\mathbf{i} + 3\mathbf{j}
\mathbf{v} = \langle 0, -7 \rangle

From a direction angle

Write each vector in \mathbf{i}, \mathbf{j} form, exactly where the angle allows, otherwise to the nearest hundredth.

The unit vector with direction angle 300^\circ.
The unit vector with direction angle \displaystyle \frac{5\pi}{6}.
Magnitude 6, direction angle 135^\circ.
Magnitude 20, direction angle 252^\circ.

A given length along a given vector

Find the vector of magnitude 34 in the direction of \langle 15, -8 \rangle.
Find the vector of magnitude 4 in the direction of 5\mathbf{i} - \mathbf{j}, exactly and to the nearest hundredth.
Karim’s tug is at (-10, 20) on the map, in meters. He steers straight toward a pier corner at (14, 27) and goes 15 meters. Where does he stop?
Ghada’s compass gives her heading as 0.28\mathbf{i} + 0.96\mathbf{j}. Show that this is a unit vector. Starting from (2, -3), she goes 12.5 meters on that heading. Where does she stop?

Unknown parts

Find every k for which \displaystyle \frac{2}{3}\mathbf{i} + k\mathbf{j} is a unit vector.
Find the unit vector pointing opposite to \langle -5, 12 \rangle.

Reasoning

Explain why the zero vector has no unit vector in its direction.
Ghada says \displaystyle \frac{1}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} is a unit vector because \displaystyle \frac{1}{2} + \frac{1}{2} = 1. Is she right? Explain.
Explain why a nonzero vector \mathbf{v} has exactly two unit vectors lying along its line, and name them.
Show that \|a\mathbf{i} + b\mathbf{j}\| \ge |a|, and say when the two are equal.

Answers

-6\mathbf{i} + 13\mathbf{j}. The components become the coefficients.
\langle 7, -9 \rangle. The coefficients become the components.
\displaystyle \left\langle \frac{1}{2}, -3 \right\rangle.
41. \displaystyle \sqrt{81 + 1600} = \sqrt{1681}.
\displaystyle \sqrt{2}. \mathbf{i} + \mathbf{j} = \langle 1, 1 \rangle, and \displaystyle \sqrt{1 + 1} = \sqrt{2}.
-6\mathbf{i} + 4\mathbf{j}. Terminal minus initial: \langle -2 - 4,\ 3 - (-1) \rangle = \langle -6, 4 \rangle.
\displaystyle \left\langle -\frac{12}{37}, -\frac{35}{37} \right\rangle. \displaystyle \|\mathbf{v}\| = \sqrt{144 + 1225} = \sqrt{1369} = 37.
\displaystyle \frac{20}{29}\mathbf{i} - \frac{21}{29}\mathbf{j}. \displaystyle \|\mathbf{v}\| = \sqrt{400 + 441} = \sqrt{841} = 29.
\displaystyle \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle. \displaystyle \|\mathbf{v}\| = \sqrt{50} = 5\sqrt{2}, and \displaystyle \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}.
\displaystyle -\frac{2\sqrt{13}}{13}\mathbf{i} + \frac{3\sqrt{13}}{13}\mathbf{j}. \displaystyle \|\mathbf{v}\| = \sqrt{4 + 9} = \sqrt{13}.
\langle 0, -1 \rangle = -\mathbf{j}. \|\mathbf{v}\| = 7, and \langle 0, -7 \rangle points straight down.
\displaystyle \frac{1}{2}\mathbf{i} - \frac{\sqrt{3}}{2}\mathbf{j}. \displaystyle \cos 300^\circ = \frac{1}{2} and \displaystyle \sin 300^\circ = -\frac{\sqrt{3}}{2}.
\displaystyle -\frac{\sqrt{3}}{2}\mathbf{i} + \frac{1}{2}\mathbf{j}. \displaystyle \cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2} and \displaystyle \sin\frac{5\pi}{6} = \frac{1}{2}.
\displaystyle -3\sqrt{2}\,\mathbf{i} + 3\sqrt{2}\,\mathbf{j}. \displaystyle 6\cos 135^\circ = 6\left(-\frac{\sqrt{2}}{2}\right) and \displaystyle 6\sin 135^\circ = 6 \cdot \frac{\sqrt{2}}{2}.
-6.18\mathbf{i} - 19.02\mathbf{j}. 20\cos 252^\circ \approx -6.180 and 20\sin 252^\circ \approx -19.021.
\langle 30, -16 \rangle. \|\langle 15, -8 \rangle\| = 17, so the unit vector is \displaystyle \left\langle \frac{15}{17}, -\frac{8}{17} \right\rangle; times 34.
\displaystyle \frac{10\sqrt{26}}{13}\mathbf{i} - \frac{2\sqrt{26}}{13}\mathbf{j} \approx 3.92\mathbf{i} - 0.78\mathbf{j}. \displaystyle \|5\mathbf{i} - \mathbf{j}\| = \sqrt{26}, so the vector is \displaystyle \frac{20}{\sqrt{26}}\mathbf{i} - \frac{4}{\sqrt{26}}\mathbf{j}.
(4.4, 24.2). The heading is \langle 24, 7 \rangle with magnitude 25; \displaystyle 15\left\langle \frac{24}{25}, \frac{7}{25} \right\rangle = \langle 14.4, 4.2 \rangle, added to (-10, 20). The corner is 25 meters off, so he is short of it.
(5.5, 9). 0.28^2 + 0.96^2 = 0.0784 + 0.9216 = 1. The trip is \langle 12.5 \cdot 0.28,\ 12.5 \cdot 0.96 \rangle = \langle 3.5, 12 \rangle, added to (2, -3).
\displaystyle k = \frac{\sqrt{5}}{3} or \displaystyle k = -\frac{\sqrt{5}}{3}. \displaystyle \frac{4}{9} + k^2 = 1 gives \displaystyle k^2 = \frac{5}{9}.
\displaystyle \left\langle \frac{5}{13}, -\frac{12}{13} \right\rangle. The unit vector of \langle -5, 12 \rangle is \displaystyle \left\langle -\frac{5}{13}, \frac{12}{13} \right\rangle; the opposite direction flips both signs.
Its magnitude is 0, so \displaystyle \frac{\mathbf{0}}{\|\mathbf{0}\|} divides by zero. Also, a unit vector has length 1 and a direction, and \mathbf{0} has no direction for it to share.
No. Its magnitude is \displaystyle \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{\sqrt{2}}{2} \ne 1. Length comes from the sum of the squares of the components, not the sum of the components.
The line through the origin along \mathbf{v} crosses the unit circle at exactly two points, one on each of its rays, and the unit vectors along the line are the arrows to those points: \displaystyle \frac{\mathbf{v}}{\|\mathbf{v}\|}, pointing the same way as \mathbf{v}, and its opposite, with both components’ signs flipped.
\displaystyle \|a\mathbf{i} + b\mathbf{j}\| = \sqrt{a^2 + b^2} \ge \sqrt{a^2} = |a|, because b^2 \ge 0 and the square root grows with its input. Equality holds exactly when b^2 = 0, that is, when b = 0 and the vector lies along the x-axis.