Unit Vectors
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 2.1: “Vectors in the Plane” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/2-1-vectors-in-the-plane |
| Textbook used in class | Stewart, Calculus, Section 12.2: “Vectors” (Example 7) |
Opening Scenario
To specify a direction without committing to any particular length, use a unit vector: a vector of magnitude exactly $1$. Every nonzero vector determines a unique unit vector pointing the same way. This normalization step appears constantly in physics (normal forces, unit normals to surfaces) and in computer graphics (normalizing camera directions).
Quick Reference
Unit vector. A vector $\hat{\mathbf{u}}$ is a unit vector if $|\hat{\mathbf{u}}| = 1$.
Normalizing. The unit vector in the direction of $\mathbf{a} \neq \mathbf{0}$ is \[ \hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}. \]
Standard basis vectors: \[ \mathbf{i} = \langle 1, 0, 0\rangle, \quad \mathbf{j} = \langle 0, 1, 0\rangle, \quad \mathbf{k} = \langle 0, 0, 1\rangle. \] Each is a unit vector along one coordinate axis.
Writing any vector in terms of $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$: $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$.
Key Concepts
1. Finding a Unit Vector
Example 1. Find the unit vector in the direction of $\mathbf{a} = \langle 2, -1, 2\rangle$. (Stewart 12.2, Example 7.)
$|\mathbf{a}| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
\[ \hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{1}{3}\langle 2, -1, 2\rangle = \left\langle \frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right\rangle. \]
Verify: $|\hat{\mathbf{a}}| = \sqrt{4/9 + 1/9 + 4/9} = \sqrt{9/9} = 1$. Correct.
2. Decomposing a Vector Into Magnitude Times Direction
Every nonzero vector $\mathbf{a}$ factors as \[ \mathbf{a} = |\mathbf{a}|\,\hat{\mathbf{a}}. \] This separates the magnitude (scalar) from the direction (unit vector). This decomposition appears when you write a force as “magnitude times direction,” a velocity as “speed times direction,” etc.
dividing a vector by its magnitude squared. The unit vector is $\mathbf{a}/|\mathbf{a}|$, not $\mathbf{a}/|\mathbf{a}|^2$. Dividing by $|\mathbf{a}|^2$ gives a vector of magnitude $1/|\mathbf{a}|$, not $1$. Always check that $|\hat{\mathbf{a}}| = 1$ after normalizing.
Common Errors Summary
| Error | Correction |
|---|---|
| Dividing each component by the square of the magnitude | Divide by $|\mathbf{a}|$ (the magnitude itself), not $|\mathbf{a}|^2$ |
| Normalizing the zero vector | $\mathbf{0}$ has no unit vector (magnitude is $0$, so division is undefined) |
Common Misconceptions
dividing a vector by its squared magnitude produces a unit vector.
This is the multiplicative-not-additive error. The unit vector in the direction of $\mathbf{a}$ is $\mathbf{a}/|\mathbf{a}|$, which has magnitude $|\mathbf{a}|/|\mathbf{a}| = 1$. Dividing by $|\mathbf{a}|^2$ instead gives a vector of magnitude $1/|\mathbf{a}|$, not $1$. The normalization step divides by the first power of the magnitude. Verification by computing the magnitude of the result is the quickest check.
Leveled Practice
Problem 1. Find the unit vector in the direction of $\mathbf{v} = \langle -3, 0, 4\rangle$.
Show answer
$|\mathbf{v}| = \sqrt{9 + 0 + 16} = 5$. $\hat{\mathbf{v}} = \langle -3/5, 0, 4/5\rangle$.
Problem 2. Find a vector of length $10$ in the direction of $\mathbf{a} = \langle 1, 1, 1\rangle$.
Show answer
$|\mathbf{a}| = \sqrt{3}$. $\hat{\mathbf{a}} = \langle 1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}\rangle$. Desired vector: $10\hat{\mathbf{a}} = \langle 10/\sqrt{3}, 10/\sqrt{3}, 10/\sqrt{3}\rangle$.