Vectors, Dot & Cross Products — Maths for Quantum Chemistry

Quantities with direction

Dipoles, forces and angular momentum

A vector has both magnitude and direction. We write it in component form $a = (a_{x}, a_{y}, a_{z})$ or as a column matrix, and its magnitude is the Pythagorean length $∣ a ∣ = a_{x}^{2} + a_{y}^{2} + a_{z}^{2}$ . Adding two vectors means adding their components; the resultant points from tail to head of the chain. These ideas are not abstract — every bond dipole in a molecule is a vector, and the molecular dipole moment is literally their vector sum.

Two kinds of multiplication exist for vectors. The dot product $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ is a scalar that measures how much two vectors align; it is zero for perpendicular vectors, and reaches $∣ a ∣∣ b ∣$ when they are parallel. The cross product $a \times b$ is a vector perpendicular to both, with magnitude $∣ a ∣∣ b ∣ sin θ$ equal to the area of the parallelogram they span — it is largest when they are perpendicular and zero when they are parallel.

In a crystal lattice, three basis vectors $a, b, c$ define the unit cell; every lattice point is reached by an integer linear combination $n_{1} a + n_{2} b + n_{3} c$ . Bragg's law and the reciprocal lattice are both vector constructs built directly from these three primitive vectors.

Key formulas at a glance

Magnitude

∣ a ∣ = a_{x}^{2} + a_{y}^{2} + a_{z}^{2}

Dot product

a \cdot b = a_{x} b_{x} + a_{y} b_{y} + a_{z} b_{z} = ∣ a ∣∣ b ∣ cos θ

Cross product (z-comp.)

(a \times b)_{z} = a_{x} b_{y} - a_{y} b_{x}

Resultant dipole

μ_{res} = 2 μ_{bond} cos (θ /2)

Vectors in chemistry

Bond dipoles add as vectors to give the molecular dipole (why CO₂ is non-polar); the dot product lets us extract bond angles from Cartesian coordinates; the cross product gives angular momentum — central to atomic orbital quantum numbers; and lattice translation vectors define crystal structures. Open the vector lab to see all of this live.

A unit vector $\hat{a} = a /∣ a ∣$ has magnitude 1 and carries only direction. Unit vectors are essential for resolving a vector into components along an arbitrary axis (projection) and for writing force-field equations cleanly. The projection of $b$ onto $a$ is:

proj_{a} b = \frac{a \cdot b}{∣ a ∣ ^{2}} a

This is used in NMR to decompose a magnetic field vector into components parallel and perpendicular to a bond axis, and in crystal optics to find how much of an electric field drives a given phonon mode.

Unit vector

\hat{a} = a /∣ a ∣

, so

∣ \hat{a} ∣ = 1

. Normalise before projecting.

Projection

Component of

b

along

\hat{a}

b \cdot \hat{a} = ∣ b ∣ cos θ

Cross product magnitude

∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ

— the area of the parallelogram spanned by

a

and

b

NH₃ dipole sum

Three N–H bond dipoles at 107.8° bond angle plus the lone-pair contribution all add as vectors; the net

μ \approx 1.47 D

points along the symmetry axis.

Common pitfalls

Adding magnitudes instead of components. The net dipole of water is not $1.51 + 1.51 = 3.02 D$ ; the correct result is $2 \times 1.51 cos 52.25° = 1.85 D$ because the x-components cancel. Always decompose before adding.
Using dot instead of cross (or vice versa). The dot product gives a scalar (bond angle, overlap); the cross product gives a vector (torque, angular momentum). If your answer should have a direction, you need the cross product.
Forgetting to normalise. The angle formula $cos θ = a \cdot b / (∣ a ∣∣ b ∣)$ requires dividing by both magnitudes. Omitting either gives an angle that is wrong by an arbitrary scale factor.
Wrong right-hand rule for cross products. $a \times b = - b \times a$ . Swapping the order of a cross product reverses the direction of the torque or angular momentum, which reverses the sense of rotation.

Going deeper

When three bond vectors in a molecule are combined to find the net dipole, the result can be zero even when all individual bonds are polar — because the vector sum cancels. BF₃ (D₃h symmetry, 120° angles) is the classic example: three equivalent B–F bond dipoles arranged at 120° yield a zero net moment. Knowing whether a sum of vectors can cancel depends purely on the symmetry of the arrangement — a key reason why group theory and vectors go hand-in-hand in molecular chemistry.

Vector lab

Adjust two vectors and watch the dot product, cross product, and angle update in real time.

Set two vectors and read their products

a \cdot b = ∣ a ∣∣ b ∣ cos θ

Vector a (orange)

aₓ3

a_y1

Vector b (blue)

bₓ1

b_y3

|a|, |b|3.16, 3.16

dot a·b6.00

angle θ53°

cross (z-component)8.00

The dot product peaks when the vectors align and is zero at 90°. The cross product is largest when they're perpendicular — it measures the area of the parallelogram they span, and points along angular momentum.

Worked example 1Resultant dipole of water from its two O–H bond dipoles

Each O–H bond in water has a bond dipole moment of $μ_{OH} = 1.51 D$ . The H–O–H angle is $104.5°$ . Find the net molecular dipole moment, treating the two bond dipoles as vectors symmetric about the bisector.

Worked example 2Bond angle in a molecule from Cartesian coordinates

In H₂O the oxygen is at the origin, and the two hydrogen atoms are at $r_{1} = (0.757, 0.586, 0) \overset{˚}{A}$ and $r_{2} = (- 0.757, 0.586, 0) \overset{˚}{A}$ . Use the dot product to find the H–O–H bond angle.

Worked example 3Cross product: torque on a bond dipole in an electric field

A bond dipole $μ = (0, 1.51, 0) D$ sits in an electric field $E = (5.0 \times 1 0^{6}, 0, 0) V m^{- 1}$ . The torque is $τ = μ \times E$ . Find its magnitude and the direction it would rotate the dipole.

Worked examples

Pull the vector ideas together across dipole moments, crystal geometry and angular momentum.

Worked example 4Proving CO₂ is non-polar by vector cancellation

Each C=O bond has a dipole of $μ_{CO} = 2.30 D$ directed from C toward O. In linear CO₂ the molecule is O=C=O with a bond angle of 180°. Show that the net molecular dipole is zero.

Worked example 5Lattice translation vector and unit-cell parameter

In a face-centred cubic (FCC) metal the lattice parameter is $a = 3.52 \overset{˚}{A}$ (nickel). The three primitive lattice vectors are $a_{1} = \frac{a}{2} (\hat{j} + \hat{k})$ , $a_{2} = \frac{a}{2} (\hat{i} + \hat{k})$ , and $a_{3} = \frac{a}{2} (\hat{i} + \hat{j})$ . Find the nearest-neighbour distance (length of $a_{1}$ ).

Worked example 6Angular momentum quantum number from the cross product

An electron in a 2p orbital has orbital angular momentum magnitude $∣ L ∣ = l (l + 1) ℏ$ . For $l = 1$ , calculate $∣ L ∣$ in units of $ℏ = 1.055 \times 1 0^{- 34} J s$ . Express classically: if the electron travels in a circle of radius $r = a_{0} = 0.529 \overset{˚}{A}$ with speed $v = ∣ L ∣/ (m_{e} r)$ , what is $v$ ?

Worked example 7Unit vectors and projection: component of a force along a bond axis

An applied force vector $F = (2.0, 3.0, - 1.0) nN$ acts on a molecule. The bond axis is along $d = (1, 1, 1)$ (unnormalised). Find the component of $F$ along the bond axis (the tensile/compressive force on the bond).

Worked example 8Resultant dipole of SO₂ from its two S–O bond dipoles

Each S–O bond in SO₂ has a dipole of $μ_{SO} = 1.63 D$ directed from S toward O. The O–S–O angle is $119.5°$ . Using vector addition, find the net molecular dipole moment.

ChallengeChallenge — net dipole of NH₃ from three N–H bond vectors and the lone pair

Ammonia (NH₃) has C₃v symmetry. Each N–H bond has dipole $μ_{NH} = 1.31 D$ directed from N toward H. The H–N–H angle is $107.8°$ . The lone pair contributes an additional $μ_{lp} \approx 0.70 D$ in the $+ z$ direction (along the C₃ axis, away from H atoms). Sum all four vector contributions and compare with the experimental value of 1.47 D.

Check yourself

Four quick questions on vectors, products, and dipole moments.

Question 1 of 4 · Score 0

A vector differs from an ordinary number because it has both magnitude and:

Choose an answer.