Dot Product and Angle Between Vectors

The dot product (or scalar product) is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number (a scalar). It is a powerful tool in both algebra and geometry.

Calculating the Dot Product

For two 2D vectors $\vec{u} = \langle u_1, u_2 \rangle$ and $\vec{v} = \langle v_1, v_2 \rangle$, the dot product is calculated by multiplying corresponding components and adding the results:

$\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2$

(This extends naturally to 3D vectors: $u_1 v_1 + u_2 v_2 + u_3 v_3$)

Finding the Angle Between Vectors

The dot product connects algebra and geometry. Geometrically, it is defined as:

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

where $\theta$ is the angle between the two vectors, and $|\vec{u}|$ and $|\vec{v}|$ are their magnitudes (lengths). Rearranging this gives a reliable formula for finding the angle between any two vectors:

$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$

Example: Find the angle between $\vec{u} = \langle 2, 3 \rangle$ and $\vec{v} = \langle -1, 4 \rangle$.

1. Find the dot product: $\vec{u} \cdot \vec{v} = (2)(-1) + (3)(4) = -2 + 12 = 10$
2. Find the magnitudes: $|\vec{u}| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$ $|\vec{v}| = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}$
3. Calculate the angle: $\cos \theta = \frac{10}{\sqrt{13}\sqrt{17}} = \frac{10}{\sqrt{221}}$ $\theta = \arccos\left(\frac{10}{\sqrt{221}}\right) \approx 47.7^\circ$

Orthogonal (Perpendicular) Vectors

If two non-zero vectors are perpendicular, the angle between them is $90^\circ$. Since $\cos(90^\circ) = 0$, their dot product must be zero.

Rule: Two vectors $\vec{u}$ and $\vec{v}$ are orthogonal if and only if: $\vec{u} \cdot \vec{v} = 0$

Vector Projection

The vector projection of $\vec{u}$ onto $\vec{v}$ gives the component (or "shadow") of $\vec{u}$ that lies perfectly along the direction of $\vec{v}$. The formula is:

$\text{proj}_{\vec{v}} \vec{u} = \left( \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2} \right) \vec{v}$

Example: Find the projection of $\vec{u} = \langle 3, 4 \rangle$ onto $\vec{v} = \langle 1, 0 \rangle$.

1. Dot product: $\vec{u} \cdot \vec{v} = (3)(1) + (4)(0) = 3$
2. Magnitude squared of $\vec{v}$: $|\vec{v}|^2 = 1^2 + 0^2 = 1$
3. Projection: $\text{proj}_{\vec{v}} \vec{u} = \left( \frac{3}{1} \right) \langle 1, 0 \rangle = \langle 3, 0 \rangle$

This means the component of $\langle 3, 4 \rangle$ acting purely in the horizontal direction (along $\langle 1, 0 \rangle$) is exactly $\langle 3, 0 \rangle$.