Scalar (dot) product

The scalar product (or dot product) of two vectors is a single number that captures how much they align. It is the simplest of the vector products , and arguably the most useful, because it gives you the angle between two vectors and the component of one along another.

Definition

$\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\,\cos\theta,$ where $\theta \in [0, \pi]$ is the angle between $\vec{a}$ and $\vec{b}$ (when both are placed with the same tail).

Componentwise formula (in any orthonormal basis $\hat{i}, \hat{j}, \hat{k}$): $\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3.$

Why two formulas agree

The componentwise formula is defined this way. To show it equals $|\vec a||\vec b|\cos\theta$, use the law of cosines on the triangle formed by $\vec{a}, \vec{b}, \vec{a} - \vec{b}$: $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}||\vec{b}|\cos\theta.$ Expand the left side componentwise: $|\vec{a} - \vec{b}|^2 = (a_1 - b_1)^2 + \cdots = |\vec{a}|^2 + |\vec{b}|^2 - 2(a_1 b_1 + a_2 b_2 + a_3 b_3)$. Match coefficients: $a_1 b_1 + a_2 b_2 + a_3 b_3 = |\vec{a}||\vec{b}|\cos\theta$.

Key consequences

Perpendicularity test: $\vec{a} \cdot \vec{b} = 0 \iff \vec{a} \perp \vec{b}$ (for non-zero vectors). The reverse is trivial; the forward is from $\cos 90^\circ = 0$.

Magnitude formula: $\vec{a} \cdot \vec{a} = |\vec{a}|^2$.

Angle formula: $\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$.

Projection of $\vec{a}$ on $\vec{b}$: the (signed) length is $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$; the vector projection is $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\vec{b}$.

Component of $\vec{a}$ along the direction $\hat{u}$ (unit vector): $\vec{a}\cdot\hat{u}$.

Cauchy-Schwarz: $|\vec{a}\cdot\vec{b}| \le |\vec{a}|\,|\vec{b}|$, with equality iff $\vec{a}$ and $\vec{b}$ are parallel.

Properties

• Commutative: $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}$.
• Distributive: $\vec{a}\cdot(\vec{b} + \vec{c}) = \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c}$.
• Scalar factors come out: $(\lambda\vec{a})\cdot\vec{b} = \lambda(\vec{a}\cdot\vec{b})$.
• Dot products of basis vectors: $\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1$; cross-pairs $= 0$.

Worked examples

Example 1. $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$, $\vec{b} = -\hat{i} + 4\hat{j} + 2\hat{k}$. Find $\vec{a}\cdot\vec{b}$.

$\vec{a}\cdot\vec{b} = (2)(-1) + (3)(4) + (-1)(2) = -2 + 12 - 2 = 8$.

Example 2. Find the angle between $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - \hat{j}$.

$\vec{a}\cdot\vec{b} = 1 - 1 = 0$. So $\theta = \pi/2$ (perpendicular).

Example 3. $\vec{a} = 3\hat{i} - 4\hat{j} + 0\hat{k}$ and $\vec{b} = -2\hat{j} + \hat{k}$. Find the angle.

$|\vec{a}| = 5$, $|\vec{b}| = \sqrt{5}$. $\vec{a}\cdot\vec{b} = 0 \cdot 0 + (-4)(-2) + 0 \cdot 1 = 8$. So $\cos\theta = \dfrac{8}{5\sqrt 5} = \dfrac{8}{5\sqrt 5}$. (Approximately $0.7155$, $\theta \approx 44.4^\circ$.)

Example 4. Find the projection of $\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ on $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.

$\vec{a}\cdot\vec{b} = 2 + 6 + 2 = 10$. $|\vec{b}| = \sqrt{1 + 4 + 1} = \sqrt{6}$. Projection length: $\dfrac{10}{\sqrt 6}$.

Example 5. Find $\lambda$ so that $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ is perpendicular to $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.

$\vec{a}\cdot\vec{b} = 2 + 2\lambda + 3 = 0$, so $\lambda = -5/2$.

Example 6. Show that $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are perpendicular iff $|\vec{a}| = |\vec{b}|$.

$(\vec{a} + \vec{b})\cdot(\vec{a} - \vec{b}) = \vec{a}\cdot\vec{a} - \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{a} - \vec{b}\cdot\vec{b} = |\vec{a}|^2 - |\vec{b}|^2$. This is zero iff $|\vec{a}| = |\vec{b}|$. (Geometrically: the diagonals of a rhombus are perpendicular.)

Try it yourself

1. $\vec{a} = \hat{i} - \hat{j}$, $\vec{b} = \hat{i} + \hat{j}$. Find $\vec{a}\cdot\vec{b}$.
2. Find the angle between $\vec{a} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.
3. If $\vec{a}\cdot\vec{b} = 0$ and neither vector is zero, what is the angle between them?
4. Find $\lambda$ if $\hat{i} + \hat{j} - \hat{k}$ is perpendicular to $2\hat{i} + \lambda\hat{j} + \hat{k}$.
5. Find the projection of $\vec{a} = 7\hat{i} + \hat{j} - 4\hat{k}$ on $\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}$.
6. If $|\vec{a}| = 3$, $|\vec{b}| = 4$, and the angle between them is $60^\circ$, find $\vec{a}\cdot\vec{b}$.
7. If $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2$, what is the angle between $\vec{a}$ and $\vec{b}$?
8. Show that the diagonals of a square are perpendicular.
9. Show that for any vectors $\vec{a}, \vec{b}, \vec{c}$: $\vec{a}\cdot(\vec{b}+\vec{c}) = \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c}$.
10. Find $|\vec{a} - \vec{b}|$ if $|\vec{a}| = 5$, $|\vec{b}| = 6$, angle between them $\theta$ with $\cos\theta = 1/3$.
11. Find a unit vector perpendicular to both $\hat{i} + \hat{j}$ and $\hat{j} + \hat{k}$. (Try: a vector $\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}$ with $\vec{r}\cdot$ each $= 0$.)
12. Show $(\vec{a} \cdot \vec{b})^2 \le |\vec{a}|^2 |\vec{b}|^2$ (Cauchy-Schwarz).
13. The work done by a force $\vec{F} = 3\hat{i} - \hat{j} + 2\hat{k}$ moving a particle by $\vec{d} = \hat{i} + 2\hat{j} - 3\hat{k}$.
14. The cosine of the angle between the diagonals of the cube $0 \le x, y, z \le 1$.

Pitfalls and tricks

• Dot product is a scalar. Don't write $\vec{a}\cdot\vec{b}$ as a vector.
• Use the angle formula in geometric problems where the angle matters.
• Use the componentwise formula when you have coordinates , it's the fastest path.
• Projection has a sign. Negative projection means $\vec{a}$ points opposite to $\vec{b}$.
• Test perpendicularity via $\vec{a}\cdot\vec{b} = 0$ , far cleaner than computing angles.