The magnitude (length) of $\mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k}$ is:
$$|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$$
Properties:
- $|\mathbf{v}| \geq 0$, and $|\mathbf{v}| = 0 \iff \mathbf{v} = \mathbf{0}$
- $|\lambda\mathbf{v}| = |\lambda|\,|\mathbf{v}|$
- Triangle inequality: $|\mathbf{a} + \mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}|$
Example: Find $|\mathbf{v}|$ for $\mathbf{v} = 3\mathbf{i} - 4\mathbf{j} + 12\mathbf{k}$.
$$|\mathbf{v}| = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$$
A unit vector has magnitude 1. The unit vector in the direction of $\mathbf{v}$ is:
$$\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}$$
Example: Find the unit vector in the direction of $\mathbf{a} = 2\mathbf{i} + 2\mathbf{j} - \mathbf{k}$.
$|\mathbf{a}| = \sqrt{4+4+1} = 3$. Unit vector: $\hat{\mathbf{a}} = \dfrac{1}{3}(2\mathbf{i} + 2\mathbf{j} - \mathbf{k})$.
Standard basis unit vectors:
$$\mathbf{i} = \begin{pmatrix}1\0\0\end{pmatrix}, \quad \mathbf{j} = \begin{pmatrix}0\1\0\end{pmatrix}, \quad \mathbf{k} = \begin{pmatrix}0\0\1\end{pmatrix}$$
The direction of $\mathbf{v}$ is specified by $\hat{\mathbf{v}}$, or equivalently by the angles it makes with the coordinate axes.
Direction cosines: If $\mathbf{v} = (v_1, v_2, v_3)$ with $|\mathbf{v}| = r$:
$$\cos\alpha = \frac{v_1}{r}, \quad \cos\beta = \frac{v_2}{r}, \quad \cos\gamma = \frac{v_3}{r}$$
where $\alpha, \beta, \gamma$ are the angles with the $x$-, $y$-, $z$-axes respectively.
Identity: $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ (since $\hat{\mathbf{v}}$ is a unit vector).
Example: Find the vector of magnitude 6 in the direction of $\mathbf{u} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$.
$|\mathbf{u}| = \sqrt{1+4+4} = 3$. Required vector: \$6 \cdot \hat{\mathbf{u}} = 6 \cdot \dfrac{\mathbf{u}}{3} = 2\mathbf{u} = 2\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}$.
Vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are linearly independent if $\lambda\mathbf{a} + \mu\mathbf{b} + \nu\mathbf{c} = \mathbf{0} \Rightarrow \lambda = \mu = \nu = 0$.
Three 3D vectors are linearly independent iff they are not coplanar (their scalar triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \neq 0$).
KEY TAKEAWAY: The unit vector $\hat{\mathbf{v}}$ encodes direction; the magnitude $|\mathbf{v}|$ encodes size. Together they fully describe any vector.
EXAM TIP: When asked for a vector of a specific magnitude in a given direction, always find the unit vector first, then scale.
COMMON MISTAKE: Computing $|\mathbf{v}|$ as $v_1 + v_2 + v_3$ (sum of components) instead of the square root of the sum of squares.
