Vectors: components, unit vectors, dot and cross product

Vectors are the language of this course. Velocity and acceleration in kinematics, force in dynamics, work in energy, collisions in momentum: all are written with vectors. The midterm has 4-5 direct questions, and all of them are mechanical operations: take components, add, multiply. If you know the formula, they will not get away.

Golden rules:

1. Always add vectors by components. Adding magnitudes ($6 + 8 = 14$) is correct only for vectors in the same direction.
2. Do not trust the $\tan^{-1}$ result, check the quadrant. If $A_x < 0$, add $180^\circ$.
3. The dot product gives a number, the cross product gives a vector.
4. Order matters in the cross product: $\vec B\times\vec A = -\vec A\times\vec B$.
5. Keep the calculator in DEG mode. In RAD mode $\cos35$ comes out negative.

Scalars and vectors

A scalar consists only of a number and a unit: mass, time, temperature, speed, energy. A vector carries magnitude and direction: displacement, velocity, acceleration, force.

Notation: $\vec A$ is the vector itself, $A$ or $|\vec A|$ is its magnitude (always $\ge 0$).

Geometrically, two vectors are added head-to-tail: $\vec R = \vec A + \vec B$ is the arrow from the start of $\vec A$ to the tip of $\vec B$. Subtraction is adding the opposite: $\vec A - \vec B = \vec A + (-\vec B)$. $-\vec B$ is a vector of the same magnitude in the opposite direction.

Components

The projections of a vector onto the $x$ and $y$ axes. If the angle $\theta$ is measured counterclockwise from the +x axis:

$A_x = A\cos\theta \qquad A_y = A\sin\theta$

The reverse, from components to magnitude and direction:

$A = \sqrt{A_x^2 + A_y^2} \qquad \theta = \tan^{-1}\!\left(\frac{A_y}{A_x}\right)$

Example 1. $\vec A$ is 20.0 m, in the direction $150^\circ$. $A_x = 20.0\cos150^\circ = -17.3$ m, $A_y = 20.0\sin150^\circ = 10.0$ m. The signs come out right on their own because we gave the angle from +x.

Quadrant check

The $\tan^{-1}$ function of the calculator gives results only between $-90^\circ$ and $+90^\circ$. So:

PITFALL · the “north of west” angle. The question may not give the angle from +x. “40° north of west” means 40° from west (−x) toward north: from +x it is $180^\circ - 40^\circ = 140^\circ$. “30° east of north” is 30° from north toward east: from +x it is $60^\circ$. Draw first, then take components.

Unit vectors

$\hat i, \hat j, \hat k$ are unitless vectors of magnitude 1 in the $x, y, z$ directions respectively. Every vector is written with them:

$\vec A = A_x\,\hat i + A_y\,\hat j + A_z\,\hat k$

The unit vector in the direction of any $\vec A$: $\hat A = \vec A / A$.

Addition, subtraction, multiplication by a scalar

These are done component by component:

$\vec A \pm \vec B = (A_x \pm B_x)\,\hat i + (A_y \pm B_y)\,\hat j + (A_z \pm B_z)\,\hat k \qquad c\vec A = cA_x\,\hat i + cA_y\,\hat j + cA_z\,\hat k$

Example 2. $\vec A$: 30.0 m, $60.0^\circ$. $\vec B$: 20.0 m, $200^\circ$. $\vec R = \vec A + \vec B$?

Ax = 30.0 cos 60.0° = 15.0       Ay = 30.0 sin 60.0° = 26.0
Bx = 20.0 cos 200°  = -18.8      By = 20.0 sin 200°  = -6.84
Rx = 15.0 - 18.8    = -3.79      Ry = 26.0 - 6.84    = 19.1
R  = √(3.79² + 19.1²) = 19.5 m
tan⁻¹(19.1 / -3.79) = -78.8°  (calculator)
Rx < 0, Ry > 0 → quadrant II → θ = -78.8° + 180° = 101°

Common-sense check: $R$ should be between $|A - B| = 10$ and $A + B = 50$. 19.5 m fits.

Dot product (scalar product)

It produces a number from two vectors:

$\vec A\cdot\vec B = AB\cos\theta = A_xB_x + A_yB_y + A_zB_z$

Here $\theta$ is the angle between the two vectors ($0$ to $180^\circ$). Unit vector rules: $\hat i\cdot\hat i = \hat j\cdot\hat j = \hat k\cdot\hat k = 1$, and different pairs are zero.

Uses:

• The angle between two vectors: $\cos\theta = \dfrac{\vec A\cdot\vec B}{AB}$.
• Perpendicularity test: $\vec A\cdot\vec B = 0 \Leftrightarrow \vec A \perp \vec B$.
• Projection: the component of $\vec A$ along $\vec B$ is $\vec A\cdot\vec B / B$.
• Later, work: $W = \vec F\cdot\Delta\vec r$ (week 10).

Example 3. $\vec a = 2\,\hat i + 3\,\hat j - \hat k$, $\vec b = \hat i - 2\,\hat j + 2\,\hat k$. What is the angle between them?

a·b = (2)(1) + (3)(-2) + (-1)(2) = 2 - 6 - 2 = -6
|a| = √(4 + 9 + 1) = √14 = 3.742      |b| = √(1 + 4 + 4) = 3
cos θ = -6 / (3.742 × 3) = -0.5345    →   θ = 122.3°

Sign information: if the dot product is negative, the angle is larger than $90^\circ$. Eliminate the acute angles in the options immediately.

Cross product (vector product)

It produces a vector from two vectors. Its magnitude is:

$|\vec A\times\vec B| = AB\sin\theta$

This magnitude is the area of the parallelogram formed by $\vec A$ and $\vec B$. Its direction is perpendicular to both vectors and is found with the right-hand rule: as the fingers of the right hand curl from $\vec A$ to $\vec B$, the thumb points in the direction of $\vec A\times\vec B$.

The component formula:

$\vec A\times\vec B = (A_yB_z - A_zB_y)\,\hat i + (A_zB_x - A_xB_z)\,\hat j + (A_xB_y - A_yB_x)\,\hat k$

Instead of memorizing, use the cycle: $x \to y \to z \to x$. For the $\hat i$ component, “y z minus z y”, for $\hat j$ “z x minus x z”, for $\hat k$ “x y minus y x”.

Unit vectors: $\hat i\times\hat j = \hat k$, $\hat j\times\hat k = \hat i$, $\hat k\times\hat i = \hat j$. In reverse order a minus appears ($\hat j\times\hat i = -\hat k$). The product with the same vector is zero: $\hat i\times\hat i = 0$.

Example 4. $\vec A = 2\,\hat i + \hat j$, $\vec B = \hat i - 3\,\hat j + 2\,\hat k$. $\vec A = (2, 1, 0)$, $\vec B = (1, -3, 2)$.

î :  AyBz - AzBy = (1)(2) - (0)(-3)  =  2
ĵ :  AzBx - AxBz = (0)(1) - (2)(2)   = -4
k̂ :  AxBy - AyBx = (2)(-3) - (1)(1)  = -7
A × B = 2î - 4ĵ - 7k̂
check: A·(A×B) = (2)(2) + (1)(-4) + 0 = 0   (perpendicular, correct)

TIP · the perpendicularity check. After finding the cross product, take the dot product of the result with $\vec A$. If it is not zero, a sign has slipped. This test cannot catch $\vec B\times\vec A$ (it is perpendicular too), so check the order separately.

Common mistakes