Vector Algebra · Lesson 3 of 9

Vector Components and Direction

“Translate geometric arrows into components, then calculate magnitudes, unit vectors and direction information.”

Learning Objectives

• Distinguish scalar components from vector components in a rectangular basis. • Derive magnitude and perform componentwise vector operations. • Construct unit vectors and vectors of specified magnitude. • Test collinearity and calculate direction ratios and cosines. • Represent every unit direction in the coordinate plane.

Building a vector from three basic directions

The unit vectors $\hat{i}, \hat{j}, \hat{k}$ point along the positive x-, y- and z-axes. They are mutually perpendicular and each has length one. By travelling along these three directions, with signed distances, we can reach any point in space.

For $P (x, y, z)$ , first move $x$ along the x-axis, then $y$ parallel to the y-axis, and finally $z$ parallel to the z-axis. The head-to-tail rule says that the sum of these displacements is $O P$ . Negative coordinates simply require movement in the corresponding negative axis direction.

Rectangular component formLaTeX

r = x \hat{i} + y \hat{j} + z \hat{k}

x, y, z

are scalar components: signed numbers. The arrows

x \hat{i}, y \hat{j}, z \hat{k}

are vector components.

Three perpendicular components— Follow the component route O to Q to R to P; its resultant is OP.

Magnitude and basic operations

The component route has right angles, so Pythagoras applies twice. The projection into the XY-plane has length $x^{2} + y^{2}$ . Combining that length with the perpendicular z-component gives the full length in space.

Magnitude in componentsLaTeX

∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}

For

a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}

, square all three signed components and add before taking the nonnegative square root.

Addition combines movements in the same basic direction. Equality therefore requires each corresponding component to be equal, rather than merely requiring the two lengths to agree. The following component rules express the geometric operations already introduced.

Component operationsLaTeX

a + b a - b λ a = (a_{1} + b_{1}) \hat{i} + (a_{2} + b_{2}) \hat{j} + (a_{3} + b_{3}) \hat{k} = (a_{1} - b_{1}) \hat{i} + (a_{2} - b_{2}) \hat{j} + (a_{3} - b_{3}) \hat{k} = (λ a_{1}) \hat{i} + (λ a_{2}) \hat{j} + (λ a_{3}) \hat{k}

Here

b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}

. The vectors are equal exactly when

a_{1} = b_{1}

a_{2} = b_{2}

and

a_{3} = b_{3}

Example — Equality compared with equal length

Problem
Find $x, y, z$ if $x \hat{i} + 2 \hat{j} + z \hat{k} = 2 \hat{i} + y \hat{j} + \hat{k}$ . Then compare $\hat{i} + 2 \hat{j}$ with $2 \hat{i} + \hat{j}$ .

1.Equate the corresponding components: $x = 2$ , $y = 2$ , $z = 1$ . These three conditions establish vector equality.
2.Both $\hat{i} + 2 \hat{j}$ and $2 \hat{i} + \hat{j}$ have magnitude $5$ .
3.Their corresponding components differ, so they are not equal vectors. Equal length alone does not determine direction.

The component rules also explain how scalar multiplication and addition interact. You can verify each identity by looking at one component at a time. This is a useful alternative to drawing several scaled arrows.

Scalar and distributive lawsLaTeX

k a + m a = (k + m) a, k (m a) = (k m) a, k (a + b) = k a + k b

k, m

are real scalars. These identities hold for all components, hence for the vectors themselves.

Normalising and choosing a magnitude

A unit vector keeps only direction information. Calculate the length of a nonzero vector, then divide every component by that same length. To produce a different required length, multiply the unit vector by that length.

Unit and prescribed-length vectorsLaTeX

\overset{a}{^} = \frac{a _{1} i ^ + a _{2} j ^ + a _{3} k ^}{a _{1}^{2} + a _{2}^{2} + a _{3}^{2}}, v = M \overset{a}{^}

M > 0

gives a vector of magnitude

M

in the same direction. The opposite direction is

- M \overset{a}{^}

Example — A unit vector in space

Problem
Find a unit vector along $2 \hat{i} + 3 \hat{j} + \hat{k}$ .

1.The magnitude is $2^{2} + 3^{2} + 1^{2} = 14$ .
2.Divide all three components by it: $\overset{a}{^} = (2 \hat{i} + 3 \hat{j} + \hat{k}) / 14$ .
3.Checking its squared magnitude gives $(4 + 9 + 1) /14 = 1$ . Its components retain the original signs and proportions.

Example — A vector with a required length

Problem
Construct a vector of magnitude 7 along $\hat{i} - 2 \hat{j}$ .

1.The given vector has magnitude $1 + 4 = 5$ .
2.The unit direction is $(\hat{i} - 2 \hat{j}) / 5$ .
3.Multiplying by 7 gives $(7 \hat{i} - 14 \hat{j}) / 5$ . Its length is 7, and the positive multiplier preserves direction.

Example — Normalising a resultant

Problem
Find the unit vector along the sum of $2 \hat{i} + 2 \hat{j} - 5 \hat{k}$ and $2 \hat{i} + \hat{j} + 3 \hat{k}$ .

1.Add first: $s = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}$ .
2.Its magnitude is $16 + 9 + 4 = 29$ .
3.The unit vector is $(4 \hat{i} + 3 \hat{j} - 2 \hat{k}) / 29$ . Normalising each addend separately would generally change the resultant direction.

Collinearity and direction information

Two nonzero vectors are collinear exactly when one is a scalar multiple of the other. A positive multiple gives the same direction and a negative multiple gives the opposite direction. In components this means one common scale factor must work in all three positions.

Component test for collinearityLaTeX

b_{1} = λ a_{1}, b_{2} = λ a_{2}, b_{3} = λ a_{3}

This is the safe form of the test. Ratios such as

b_{1} / a_{1}

should only be formed when their denominators are nonzero.

Example — Opposite but collinear

Problem
Show that $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $- 4 \hat{i} + 6 \hat{j} - 8 \hat{k}$ are collinear.

1.The second vector is $- 2$ times the first in every component.
2.Therefore they are collinear and oppositely directed.
3.Their magnitudes have ratio 1:2, so they are not equal vectors and are not negatives of one another with equal length.

The scalar components provide direction ratios. Dividing them by the vector magnitude gives direction cosines, precisely the components of the unit vector along it. A common positive rescaling of direction ratios preserves the same orientation; a negative rescaling describes the opposite orientation.

Direction from componentsLaTeX

l = \frac{a _{1}}{∣ a ∣}, m = \frac{a _{2}}{∣ a ∣}, n = \frac{a _{3}}{∣ a ∣}, \overset{a}{^} = l \hat{i} + m \hat{j} + n \hat{k}

The direction-cosine triple has squared sum one. The component triple generally does not.

Example — Ratios and cosines

Problem
Find direction ratios and cosines for $\hat{i} + \hat{j} - 2 \hat{k}$ .

1.One direction-ratio triple is $(1, 1, - 2)$ .
2.The magnitude is $1 + 1 + 4 = 6$ .
3.The direction cosines are $(1/ 6, 1/ 6, - 2/ 6)$ . Their squared sum is $(1 + 1 + 4) /6 = 1$ .

All unit directions in a plane

A vector in the XY-plane has no z-component. For a unit vector, its endpoint lies on the unit circle. As its direction angle varies, the endpoint passes through every point of that circle, including directions in all four quadrants.

Unit vectors in the coordinate planeLaTeX

u = cos θ \hat{i} + sin θ \hat{j}, 0 \leq θ < 2 π

The magnitude is one because

cos^{2} θ + sin^{2} θ = 1

. The angle is measured from the positive x-axis.

Unit directions trace a circle— The horizontal and vertical components are cos θ and sin θ.

Common mistake

Do not divide by a zero component when testing collinearity. Compare all components using a common multiplier instead. Also, a direction-ratio triple need not have length one; a direction-cosine triple must.

Quick check

For $a = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ , the scalar y-component is

Quick check

What is $∣ \hat{i} + 2 \hat{j} + 2 \hat{k} ∣$ ?

Quick check

A unit vector along $3 \hat{i} + 4 \hat{j}$ is

Quick check

Which is collinear with $\hat{i} - 2 \hat{j}$ ?

Quick check

The unit vector making angle $π /6$ with the positive x-axis in the XY-plane is

Practice Problems

For $a = 2 \hat{i} - 7 \hat{j} - 3 \hat{k}$ , find its scalar components, vector components and magnitude.
Find $x, y$ when $2 \hat{i} + 3 \hat{j} = x \hat{i} + y \hat{j}$ . Give two unequal vectors with the same magnitude.
Find the sum of $\hat{i} - 2 \hat{j} + \hat{k}$ , $- 2 \hat{i} + 4 \hat{j} + 5 \hat{k}$ and $\hat{i} - 6 \hat{j} - 7 \hat{k}$ .
Find a vector of magnitude 8 in the direction of $5 \hat{i} - \hat{j} + 2 \hat{k}$ .
Find the direction cosines of $\hat{i} + 2 \hat{j} + 3 \hat{k}$ . Explain why $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the positive axes.
If a unit vector makes angles $π /3$ , $π /4$ and an acute angle $γ$ with the positive axes, determine $γ$ and its components.
Find all real values of $x$ for which $x (\hat{i} + \hat{j} + \hat{k})$ is a unit vector. Explain the two signs.

Key Takeaways

Components turn geometric vector operations into arithmetic on signed numbers. Scalar components are numbers; vector components include the corresponding basis vector. Normalise the required vector only after forming it. Collinearity requires one common component multiplier; direction cosines are normalised components. Every unit vector in the XY-plane has cosine and sine as its two components.

Previous · Lesson 2

Adding and Scaling Vectors

Next · Lesson 4

Vectors Between Points and the Section Formula

Building a vector from three basic directions

Magnitude and basic operations

Normalising and choosing a magnitude

Collinearity and direction information

All unit directions in a plane

Quiz

Practice Problems

Key Takeaways