Dot product: the 3 formulas (coordinates, norm×cos, projection)

Three formulas, one result

Pick the fastest based on what you’re given.

1. From coordinates

$\vec{u} \cdot \vec{v} = x_u x_v + y_u y_v$

The most useful in practice.

2. From norms and angle

$\vec{u} \cdot \vec{v} = \|\vec{u}\| \times \|\vec{v}\| \times \cos(\theta)$

where $\theta$ is the angle between the vectors.

3. From projection (less common)

If $\vec{u'}$ is the orthogonal projection of $\vec{u}$ onto $\vec{v}$: $\vec{u} \cdot \vec{v} = \vec{u'} \cdot \vec{v}$

Key property: orthogonality

$\vec{u} \perp \vec{v} \iff \vec{u} \cdot \vec{v} = 0$

This is the perpendicularity test.

Example 1 (coordinates)

$\vec{u}(3, 4)$ and $\vec{v}(-2, 5)$. $\vec{u} \cdot \vec{v} = 3 \times (-2) + 4 \times 5 = -6 + 20 = 14$.

Example 2 (orthogonality)

Are $\vec{u}(2, 3)$ and $\vec{v}(6, -4)$ orthogonal? $\vec{u} \cdot \vec{v} = 12 - 12 = 0$. Yes.

Vector norm (bonus)

$\|\vec{u}\|^2 = \vec{u} \cdot \vec{u} = x_u^2 + y_u^2$

So $\|\vec{u}(3, 4)\| = 5$.

Common mistakes

1. Confusing vector and point coordinates: $\vec{AB} = (x_B - x_A, y_B - y_A)$.
2. Sign of cos(θ): obtuse angle → cos < 0 → negative dot product.
3. Norm vs dot product: $\|\vec{u}\| = \sqrt{\vec{u} \cdot \vec{u}}$, not $\vec{u} \cdot \vec{u}$ directly.

?À toi de jouer

What is the dot product of u(5, −2) and v(3, 4)?

A.7B.23C.−8D.20

Valider