The primitive of orientation. Where does it point?
Each bond has a length and a direction.
Look at the bonds. Each bond connects two atoms. Each bond has a length. But each bond also points somewhere.
The O-H bond on the left points in a different direction than the O-H bond on the right.
How do we describe "where a bond points"?
A single number won't do.
"The bond points 0.96 Å" — meaningless. That's a length, not a direction.
"The bond points northeast" — better, but imprecise. And what about 3D?
We need a mathematical object that captures both magnitude (how long) and direction (which way).
That object is the vector.
DIRECTION: "It points."
Before we formalize: you already perceive direction. Which way is up? Point to the door. The wind blows from the west. Direction is a primitive perception. The vector is its mathematical formalization.
A vector is a directed line segment — an arrow with a specific length and direction.
Two vectors are equal if they have the same length and the same direction, regardless of where they are located in space.
These are all the same vector. Location doesn't matter — only length and direction.
The zero vector 0 has magnitude zero. It has no direction (or equivalently, every direction). It is the only vector with |v| = 0.
A scalar is a single number (element of ℝ or ℂ). A vector is an object with magnitude and direction.
| Scalars | Vectors |
|---|---|
| Temperature (300 K) | Velocity (5 m/s northward) |
| Energy | Force |
| Mass | Momentum |
For scalar $c \in \mathbb{R}$ and vector v:
To add v + w:
This is the tip-to-tail or head-to-tail rule.
For vectors u, v, w:
These properties make the set of vectors a vector space over ℝ.
Fix a coordinate system with origin O and axes. In 3D: axes x, y, z, with unit vectors î, ĵ, k̂.
A vector v in 3D can be written:
$\mathbf{v} = v_x \mathbf{\hat{i}} + v_y \mathbf{\hat{j}} + v_z \mathbf{\hat{k}}$
where $v_x$, $v_y$, $v_z$ are the components of v.
Addition: Add component-wise.
$(v_x, v_y, v_z) + (w_x, w_y, w_z) = (v_x + w_x, v_y + w_y, v_z + w_z)$
Scalar multiplication: Multiply each component.
$c(v_x, v_y, v_z) = (cv_x, cv_y, cv_z)$
The magnitude of v = $(v_x, v_y, v_z)$ is:
$|\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$
A unit vector has magnitude 1. Given any nonzero vector v, the unit vector in the direction of v is:
$\mathbf{\hat{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}$
This process is called normalization.
A molecule is specified by the positions of its atoms.
| Atom | x (Å) | y (Å) | z (Å) |
|---|---|---|---|
| O | 0.000 | 0.000 | 0.000 |
| H₁ | 0.958 | 0.000 | 0.000 |
| H₂ | -0.240 | 0.927 | 0.000 |
The bond vector from atom A to atom B is:
$\mathbf{b}_{AB} = \mathbf{r}_B - \mathbf{r}_A$
For atoms at positions r_A and r_B:
$d_{AB} = |\mathbf{r}_B - \mathbf{r}_A| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2}$
This is the Euclidean distance between two points.
We can compute bond vectors and bond lengths. But what about the angle between two bonds?
This requires a new operation: the dot product.
The dot product of v and w is:
$\mathbf{v} \cdot \mathbf{w} = |\mathbf{v}||\mathbf{w}|\cos\theta$
where θ is the angle between the vectors.
Equivalently (in components):
$\mathbf{v} \cdot \mathbf{w} = v_x w_x + v_y w_y + v_z w_z$
Finding the angle:
$\theta = \arccos\left(\frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}||\mathbf{w}|}\right)$
b₁ = (0.958, 0, 0), b₂ = (-0.240, 0.927, 0)
b₁ · b₂ = (0.958)(-0.240) + (0)(0.927) + (0)(0) = -0.230
|b₁| = 0.958, |b₂| = 0.958
$\cos\theta = \frac{-0.230}{0.958 \times 0.958} = -0.251$
$\theta = \arccos(-0.251) = 104.5°$
This is the famous H-O-H bond angle in water.
A dipole moment is a vector quantity.
$\boldsymbol{\mu}_{mol} = \sum_i \boldsymbol{\mu}_i$
CO₂: Each C=O bond is polar, but they point in opposite directions. The vector sum is zero. CO₂ is nonpolar despite having polar bonds.
H₂O: Two O-H bond dipoles at 104.5°. They don't cancel. The vector sum is nonzero. H₂O is polar.
| Operation | Geometric | Algebraic |
|---|---|---|
| Addition | Tip-to-tail | Component-wise |
| Scalar mult | Scale length, possibly reverse | Multiply each component |
| Magnitude | Length of arrow | √(Σvᵢ²) |
| Normalization | Scale to unit length | v/|v| |
| Concept | Vector Interpretation |
|---|---|
| Bond | Displacement vector from atom A to atom B |
| Bond length | Magnitude of bond vector |
| Dipole moment | Vector from δ+ to δ- |
| Molecular dipole | Vector sum of bond dipoles |
| Force | Vector (magnitude and direction) |
Let v = (3, -1, 2) and w = (1, 4, -1).
(a) Compute v + w
(b) Compute v - w
(c) Compute 3v - 2w
(d) Compute |v| and |w|
(e) Find the unit vector in the direction of v
Methane (CH₄) has C at the origin and H atoms at:
(a) Compute all four C-H bond lengths. Are they equal?
(b) Using the dot product, compute the angle between C-H₁ and C-H₂ bonds.
A water molecule has two O-H bond dipoles, each of magnitude 1.5 D, separated by the H-O-H angle of 104.5°.
(a) Set up a coordinate system and write the two bond dipole vectors.
(b) Compute the molecular dipole moment magnitude.