Linear AlgebraLinear AlgebraCollege

Vectors

Vector arithmetic, dot and cross products, unit and orthogonal vectors.

A vector has both size and direction and is written in components such as $\mathbf{v} = \langle 3, 4 \rangle$ . Vectors add tip to tail, component by component, and scalar multiplication stretches or shrinks them. The magnitude $|\mathbf{v}| = \sqrt{3^2 + 4^2} = 5$ is the length.

The dot product $\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3$ measures how aligned two vectors are and equals $|\mathbf{u}||\mathbf{v}|\cos\theta$ , so a zero dot product means the vectors are perpendicular. In three dimensions the cross product gives a vector perpendicular to both, with length equal to the area of the parallelogram they span.

Key terms

Dot product: Sum of products of components; zero exactly when nonzero vectors are orthogonal.
$\mathbf{u}\cdot\mathbf{v} = \sum u_i v_i$
Unit vector: A vector divided by its length.
$\hat{\mathbf{v}} = \frac{\mathbf{v}}{\lVert\mathbf{v}\rVert}$

Key ideas

Unit vector: $\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}$ .
Angle between vectors: $\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$ .
Projection of $\mathbf{u}$ onto $\mathbf{v}$ : $\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2}\mathbf{v}$ .
Cross product is anti-commutative: $\mathbf{u} \times \mathbf{v} = -\mathbf{v} \times \mathbf{u}$ .

Worked example

Find the angle between $\mathbf{u} = \langle 1, 2, 2 \rangle$ and $\mathbf{v} = \langle 2, 0, 1 \rangle$ .

Dot product: $2 + 0 + 2 = 4$ .
Magnitudes: $|\mathbf{u}| = 3$ , $|\mathbf{v}| = \sqrt{5}$ .
$\cos\theta = \frac{4}{3\sqrt{5}} \approx 0.596$ .
$\theta \approx 53.4^\circ$ .

Answer: about $53.4^\circ$

Common mistakes

Multiplying magnitudes instead of components for the dot product.
Forgetting the square root in the magnitude.
Expecting the dot product to be a vector.

Practice questions

Try each one, then open the answer. Numerical answers on this page are verified by an automated calculation check.

1.Find $|\langle 6, 8 \rangle|$ .
Show answer
$10$
2.Find $\langle 2, -1, 3 \rangle \cdot \langle 4, 5, 1 \rangle$ .
Show answer
$6$
3.For which $k$ is $\langle k, 2 \rangle$ perpendicular to $\langle 3, -6 \rangle$ ?
Show answer
$k = 4$

Vectors

Key terms

Key ideas

Worked example

Common mistakes

Practice questions

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