Vectors

A vector is an arrow with a length and a direction. Just like positions, vectors exist before we measure or describe them. Unlike positions, vectors can mean many different things, such as position vectors, velocities, etc. Vectors are not anchored to particular positions in space, so we can slide a vector around and locate it at any position.

Change:

Two vectors, which may or may not be the same vector. Moving a vector around does not change it: it is still the same vector.

Notation

Some textbooks differentiate between free vectors, which are free to slide around, and bound vectors, which are anchored in space. We will only use free vectors.

We will use the over-arrow notation $ \vec{a} $ for vector quantities. Other common notations include bold $ \boldsymbol{a} $ and under-bars $ \underline{a} $. For unit (length one) vectors we will use an over-hat $ \hat{a} $.

Units

When using coordinates to describe the location of physical positions in the real world, we must always have appropriate units of length for the coordinates. For example, the following coordinates all specify the same position:

$$ \begin{aligned} x &= 2{\rm\ m} & x &= 6.56{\rm\ ft} & r &= 6.32{\rm\ m} & r &= 249{\rm\ in} \\ y &= -6{\rm\ m} & y &= -6 \times 10^6{\rm\ \mu m} & \theta &= 1.25{\rm\ rad} & \theta &= 71.6^\circ. \end{aligned} $$

It is important to distinguish between units and dimensions. The base dimensions include mass (M), length (L), and time (T), and can be combined to produce dimensions such as L/T or L T$ ^{-1} $ for velocity. Each physical quantity such as velocity has just one set of dimensions, and there is no choice in this.

Units are particular ways of measuring dimensional quantities, and include the SI units kilogram (kg), meter (m), and second (s), as well as the customary U.S. units pound (lb) and foot (ft). A given physical quantity can be written in terms of many different units, although all of these must conform to the dimensions of the quantity. For example, velocity has dimension L/T and so it can be written in terms of the units m/s, ft/s, km/h, or many other choices.

Angles are an example of a dimensionless quantity (having dimension 1), but they still have units, typically either radians (rad) or degrees (°).

Did you know?

The SI unit system is the definitive system of measurement used in science and engineering. Other systems, such as U.S. customary units are defined in terms of SI units, so that an inch is defined to be exactly 2.54 cm, for example.

The National Institute of Standards and Technology (NIST) maintains an excellent reference guide to SI units, including precise rules and style conventions for writing units in scientific and engineering work.

Unit vectors

A unit vector is any vector with a length of one. We use the special over-hat notation $ \hat{a} $ to indicate when a vector is a unit vector. Any non-zero vector $ \vec{a} $ gives a unit vector $ \hat{a} $ that specifies the direction of $ \vec{a} $.

Normalization to unit vector. #rvv-eu

$$ \begin{aligned} \hat{a} =\frac{\vec{a}}{a}\end{aligned} $$

If we compute the length of $ \hat{a} $ then we find:

$$ \| \hat{a} \| = \left\| \frac{\vec{a}}{a} \right\| = \frac{\|\vec{a}\|}{a} = \frac{a}{a} = 1, $$

so $ \hat{a} $ is really a unit vector, and it is in the same direction as $ \vec{a} $ as they differ only by a scalar factor.

Any vector can be written as the product of its length and direction:

Vector decomposition into length and direction. #rvv-ei

$$ \begin{aligned} \vec{a} = a\hat{a}\end{aligned} $$

This follows from rearranging #rvv-eu.

Three vectors and their decompositions into lengths and directional unit vectors.

Vectors bases

To describe vectors mathematically, we write them as a combination of basis vectors. An orthonormal basis is a set of two (in 2D) or three (in 3D) basis vectors which are orthogonal (have 90° angles between them) and normal (have length equal to one). We will not be using non-orthogonal or non-normal bases.

Any other vector can be written as a linear combination of the basis vectors:

Components of a vector. #rvv-ec

$$ \vec{a} = a_1 \,\hat{\imath}+ a_2 \,\hat{\jmath} + a_3 \,\hat{k} $$

The numbers $a_1, a_2, a_3$ are called the components of $ \vec{a} $ in the $ \,\hat{\imath}, \hat{\jmath}, \hat{k} $ basis. If we are in 2D then we will only have two components for a vector.

Writing a vector as the sum of scaled basis vectors. The scale factors are the components of the vector. Here $ \vec{a} = 3\hat\imath + 2\hat\jmath $, so the components of $ \vec{a} $ are $a_1 = 3$ and $a_2 = 2$.

We draw the symbol $\odot$ (arrow tip) to indicate a vector coming out of the page, and $\otimes$ (arrow fletching) to indicate an arrow going into the page.

Two standard arrangements of the basis vectors when working in 2D. Either $\hat\jmath$ is the vertical and $ \hat{k} $ is out of the page, or $ \hat{k} $ is the vertical and $\hat\jmath$ is into the page. In both cases $\hat\imath$ is horizontal.

Notation note

Just as for position coordinates, we can write the vector components $3\hat\imath + 2\hat\jmath$ as the ordered list $(3, 2)$ if we know which basis we are using. Because we often will be using several bases simultaneously, we will generally write the components explicitly in the $3\hat\imath + 2\hat\jmath$ form.

Did you know?

The use of the letter $i,j,k$ for basis vectors is due to William Hamilton, who was motivated by thinking of basis vectors as extensions of the complex number $i$. This notation was popularized by the book Vector Analysis: A Text Book for the Use of Students of Mathematics and Physics Founded upon the Lectures of J. Willard Gibbs (1901), by E. B. Wilson. This book also introduced the use of bold letters to represent vectors.

Length of vectors

The length of a vector $ \vec{a} $ is written either $ \| \vec{a} \| $ or just plain $a$. The length can be computed using Pythagorus’ theorem:

Pythagorus' length formula. #rvv-ey

$$ a = \|\vec{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} $$

First we prove Pythagorus' theorem for right-angle triangles. For side lengths $a$ and $b$ and hypotenuse $c$, the fact that $a^2 + b^2 = c^2$ can be seen graphically below, where the gray area is the same before and after the triangles are rotated in the animation:

Pythagorus' theorem immediately gives us vector lengths in 2D. To find the length of a vector in 3D we can use Pythagorus' theorem twice, as shown below. This gives the two right-triangle calculations:

$$ \begin{aligned} \ell^2 &= a_1^2 + a_2^2 \\ a^2 &= \ell^2 + a_3^2 = a_1^2 + a_2^2 + a_3^2. \end{aligned} $$

Click and drag to rotate.

Warning: Length must be computed in a single basis. #rvv-wl

The Pythagorean length formula can only be used if all the components are written in a single orthonormal basis.

Computing the length of a vector using Pythagorus' theorem.

Some common integer vector lengths are $ \vec{a} = 4\hat\imath + 3\hat\jmath $ (length $a = 5$) and $ \vec{b} = 12\hat\imath + 5\hat\jmath $ (length $b = 13$).

Warning: Adding vectors does not add lengths. #rvv-wa

If $ \vec{c} = \vec{a} + \vec{b} $, then $ \|\vec{c}\| \ne \|\vec{a}\| + \|\vec{b}\| $ unless $ \vec{a} $ and $ \vec{b} $ are parallel and in the same direction.

It will always be true, however, that $ \|\vec{c}\| \le \|\vec{a}\| + \|\vec{b}\| $. This fact is known as the triangle inequality, for reasons that should be obvious.

Did you know?

Sets of three integers $a,b,c$ where $a^2 + b^2 = c^2$ are called Pythagorean triples. A long list of such triples is given on the Plimpton 322 clay tablet written by the ancient Babylonians around 1800 BCE, although it is unclear how they generated these numbers. Pythagorean triples lead to complex mathematics, including the curious patterns shown below and Fermat's Last Theorem.

The values of $a$ and $b$ for all Pythagorean triples $a,b,c$ with $a$ and $b$ up to 2000.

Changing bases

To change the basis that a vector is written in, we need to know how the basis vectors are related. We do this by writing one set of basis vectors in terms of the other basis vectors. If we want to change from $ \hat\imath,\hat\jmath $ to $ \hat{u},\hat{v} $, then we need to write $ \hat\imath,\hat\jmath $ in terms of $ \hat{u},\hat{v} $ and then substitute the expressions.

Example: Example: Basis change. #rvv-xn

For example, if we have $ \vec{a} = 3\,\hat{\imath} + 2\,\hat{\jmath} $ and we want to write this in the $ \,hat{u}, ,hat{v} $ basis, then we need to know $ \,\hat{\imath}, \,\hat{\jmath} $ in terms of $ \,\hat{u}, \,\hat{v} $.

From above we see that:

$$ \begin{aligned} \hat{\imath} &= \cos\theta \, \hat{u} - \sin\theta \, \hat{v} = \frac{1}{\sqrt{2}} \,\hat{u} - \frac{1}{\sqrt{2}} \,\hat{v} \\ \hat{\jmath} &= \sin\theta \, \hat{u} + \cos\theta \, \hat{v} = \frac{1}{\sqrt{2}} \,\hat{u} + \frac{1}{\sqrt{2}} \,\hat{v}.\end{aligned} $$

Then we can substitute and re-arrange:

$$ \begin{aligned} \vec{a} &= 3\,\hat{\imath} + 2\,\hat{\jmath} \\ &= 3\left(\frac{1}{\sqrt{2}} \,\hat{u} - \frac{1}{\sqrt{2}} \,\hat{v}\right) + 2\left(\frac{1}{\sqrt{2}} \,\hat{u} + \frac{1}{\sqrt{2}} \,\hat{v}\right) \\ &= \left(\frac{3}{\sqrt{2}} + \frac{2}{\sqrt{2}} \right) \,\hat{u} + \left(-\frac{3}{\sqrt{2}} + \frac{2}{\sqrt{2}} \right) \,\hat{v} \\ &= \frac{5}{\sqrt{2}} \,\hat{u} - \frac{1}{\sqrt{2}} \,\hat{v}.\end{aligned} $$

If we want to convert back the other way then we would need to know $ \,\hat{u}, \,\hat{v} $ in terms of $ \,\hat{\imath}, \,\hat{\jmath} $. We can find this by solving for $ \,\hat{u}, \,\hat{v} $ above, giving:

$$ \begin{aligned} \hat{u} &= \cos\theta \, \hat\imath + \sin\theta \, \hat\jmath = \frac{1}{\sqrt{2}} \,\hat\imath + \frac{1}{\sqrt{2}} \,\hat\jmath \\ \hat{v} &= -\sin\theta \, \hat\imath + \cos\theta \, \hat\jmath = -\frac{1}{\sqrt{2}} \,\hat\imath + \frac{1}{\sqrt{2}} \,\hat\jmath.\end{aligned} $$

We can also write the general expressions for basis change, as below.

Change of basis formulas. #rvv-eg

$$ \begin{aligned}\vec{a} &=a_i \, \hat\imath + a_j \, \hat\jmath + a_k \, \hat{k}& \vec{a} &=a_u \, \hat{u} + a_v \, \hat{v} + a_w \, \hat{w} \\[1em]a_i &= a_u u_i + a_v v_i + a_w w_i& a_u &= a_i i_u + a_j j_u + a_k k_u \\a_j &= a_u u_j + a_v v_j + a_w w_j& a_v &= a_i i_v + a_j j_v + a_k k_v \\a_k &= a_u u_k + a_v v_k + a_w w_k& a_w &= a_i i_w + a_j j_w + a_k k_w\end{aligned} $$

We will derive the first set of equations (the second set are derived similarly). The vector $ \vec{a} $ can be written in both the $ \hat\imath,\hat\jmath,\hat{k} $ and $ \hat{u},\hat{v},\hat{w} $ bases:

$$ \begin{aligned} \vec{a} &= a_i \hat\imath + a_j \hat\jmath + a_k \hat{k} & \vec{a} &= a_u \hat{u} + a_v \hat{v} + a_w \hat{w}. \end{aligned} $$

We can write each $ \hat{u},\hat{v},\hat{w} $ basis vector in terms of the $ \hat\imath,\hat\jmath,\hat{k} $ basis:

$$ \begin{aligned} \hat{u} &= u_i \hat\imath + u_j \hat\jmath + u_k \hat{k} \\ \hat{v} &= v_i \hat\imath + v_j \hat\jmath + v_k \hat{k} \\ \hat{w} &= w_i \hat\imath + w_j \hat\jmath + w_k \hat{k}. \end{aligned} $$

Substituting these expressions into $ \vec{a} $ gives:

$$ \begin{aligned} \vec{a} &= a_u \hat{u} + a_v \hat{v} + a_w \hat{w} \\ &= a_u (u_i \hat\imath + u_j \hat\jmath + u_k \hat{k}) + a_v (v_i \hat\imath + v_j \hat\jmath + v_k \hat{k}) + a_w (w_i \hat\imath + w_j \hat\jmath + w_k \hat{k}) \\ &= (a_u u_i + a_v v_i + a_w w_i) \hat\imath + (a_u u_j + a_v v_j + a_w w_j) \hat\jmath + (a_u u_k + a_v v_k + a_w w_k) \hat{k} \\ &= a_i \hat\imath + a_j \hat\jmath + a_k \hat{k}. \end{aligned} $$

Comparing the last two lines gives the component formulas.

In 2D the change between two orthonormal bases is a rotation by an angle $ \theta $, resulting in the change of basis expression below.

Change of basis formula in 2D. #rvv-e2

$$ \begin{aligned} \vec{a} &= a_i \, \hat\imath + a_j \, \hat\jmath & \vec{a} &= a_u \, \hat{u} + a_v \, \hat{v} \\[1em] a_i &= \cos\theta \, a_u - \sin\theta \, a_v & a_u &= \cos\theta \, a_i + \sin\theta \, a_j \\ a_j &= \sin\theta \, a_u + \cos\theta \, a_v & a_v &= -\sin\theta \, a_i + \cos\theta \, a_j \end{aligned} $$

Elementary geometry gives the relationships between the basis vectors:

$$ \begin{aligned} \hat\imath &= \cos\theta \, \hat{u} - \sin\theta \, \hat{v} & \hat{u} &= \cos\theta \, \hat\imath + \sin\theta \, \hat\jmath \\ \hat\jmath &= \sin\theta \, \hat{u} + \cos\theta \, \hat{v} & \hat{v} &= -\sin\theta \, \hat\imath + \cos\theta \, \hat\jmath. \end{aligned} $$

Thus we have the components:

$$ \begin{aligned} i_u &= \cos\theta & i_v &= -\sin\theta & u_i &= \cos\theta & u_j &= \sin\theta \\ j_u &= \sin\theta & j_v &= \cos\theta & v_i &= -\sin\theta & v_j &= \cos\theta. \end{aligned} $$

Substituting these into #rvv-eg and ignoring the third components gives the desired expressions.

Vector expressions are true no matter which basis we write the vectors in, even if they are written in different bases.

Example: Vector addition in different bases. #rvv-xa

Adding $ \vec{a} $ and $ \vec{b} $ to get the result $ \vec{c} $ is a well-defined operation even before any basis is used, so it cannot depend on the basis chosen. As we see below, we can do the calculation in either the $ \hat\imath,\hat\jmath $ or $ \hat{u}, \hat{v} $ basis.

Show components: none $ \hat\imath,\hat\jmath $ $ \hat{u}, \hat{v} $ mixed

$$ \begin{aligned} \vec{c} &= \vec{a} + \vec{b} \\ &= (3\hat\imath + 2\hat\jmath) + (3\hat\imath - \hat\jmath) \\ &= 6\hat\imath + \hat\jmath \\ \vec{c} &= \vec{a} + \vec{b} \\ &= (3.5\hat{u} - 0.7\hat{v}) + (1.4\hat{u} - 2.8\hat{v}) \\ &= 4.9\hat{u} - 3.5\hat{v} \\ \vec{c} &= \vec{a} + \vec{b} \\ &= (3\hat\imath + 2\hat\jmath) + (1.4\hat{u} - 2.8\hat{v}) \\ &= 3\hat\imath - 2.8\hat{v} + 2\hat\jmath + 1.4\hat{u}. \end{aligned} $$

The component order in the mixed expression is arbitrary.

Example Problem: Cross product in different bases. #rvv-xx

Consider the two vectors shown below and their components in two bases, together with their lengths:

$$ \begin{aligned} \vec{a} &= 3\hat\imath + 2\hat\jmath = 3.5\hat{u} - 0.7\hat{v} & a &= \sqrt{3^2 + 2^2} = 3.6 \\ \vec{b} &= 3\hat\imath - \hat\jmath = 1.4\hat{u} - 2.8\hat{v} & b &= \sqrt{3^2 + 1^2} = 3.2 \end{aligned} $$

Show components: none $ \hat\imath,\hat\jmath $ $ \hat{u}, \hat{v} $

Compute the cross product $ \vec{a} \times \vec{b} $ using: (1) the angle formula #rvv-el; (2) the component formula #rvv-ex with $ \vec{a}, \vec{b} $ both in the $ \hat\imath,\hat\jmath $ basis, both in the $ \hat{u}, \hat{v} $ basis, and with $ \vec{a} $ in the $ \hat\imath,\hat\jmath $ basis and $ \vec{b} $ in the $ \hat{u}, \hat{v} $ basis.

(1) The dot product is $ \vec{a} \cdot \vec{b} = 7 $ and the vector lengths are $a = 3.6$ and $b = 3.2$, so $\cos\theta = 7 / (ab)$ and $\theta \approx 53^\circ$. Now using #rvv-el gives:

$$ \begin{aligned} \vec{a} \times \vec{b} &= a b \sin\theta ( -\hat{k}) \\ &\approx -9 \hat{k}. \end{aligned} $$

(2) Using the component formula #rvv-ex gives:

$$ \begin{aligned} (3\hat\imath + 2\hat\jmath) \times (3\hat\imath - \hat\jmath) &= -3 \hat\imath \times \hat\jmath + 6 \hat\jmath \times \hat\imath \\ &= -3 \hat{k} - 6 \hat{k} \\ &= -9 \hat{k} \\ (3.5 \hat{u} - 0.7 \hat{v}) \times (1.4 \hat{u} - 2.8 \hat{v}) &= - (3.5 \times 2.8) \hat{u} \times \hat{v} - (0.7 \times 1.4) \hat{v} \times \hat{u} \\ &= -10 \hat{k} + \hat{k} \\ &= -9 \hat{k} \\ (3\hat\imath + 2\hat\jmath) \times (1.4\hat{u} - 2.8\hat{v}) &= (3 \times 1.4) \hat\imath \times \hat{u} - (3 \times 2.8) \hat\imath \times \hat{v} \\ &\quad + (2 \times 1.4) \hat\jmath \times \hat{u} - (2 \times 2.8) \hat\jmath \times \hat{v} \\ &= 4.2 \sin 53^\circ \, \hat{k} - 8.5 \sin 143^\circ \, \hat{k} \\ &\quad - 2.8 \sin 53^\circ \, \hat{k} - 5.7 \sin 53^\circ \, \hat{k} \\ &= - 9 \hat{k}. \end{aligned} $$

Example: Dot product is independent of basis. #rvv-xd

Equation #rvv-ed makes it clear that the dot product does not depend on which basis we use to write $ \vec{a} $ and $ \vec{b} $, so long as we use the same orthonormal basis for both of them. This is because the dot product only depends on the lengths and angle between the vectors, which are real physical quantities that don’t change just because we use a different basis.

However, we can also verify directly that the component equation #rvv-es for the dot product does not depend on which basis we use. To keep the algebra short, we will only do this in 2D.

We compute the dot product using #rvv-es in the $\hat\imath,\hat\jmath$ basis and substitute in the change-of-basis expressions #rvv-eg, giving:

$$ \begin{aligned} \vec{a} \cdot \vec{b} &= a_i b_i + a_j b_j \\ &= (u_i a_u + v_i a_v ) (u_i b_u + v_i b_v) + (u_j a_u + v_j a_v) (u_j b_u + v_j b_v) \\ &= (u_i^2 + u_j^2) a_u b_u + (v_i^2 + v_j^2) a_v b_v + (u_i v_i + u_j b_j) (a_u b_v + a_v b_u) \\ &= \| \hat{u} \|^2 a_u b_u + \| \hat{v} \|^2 a_v b_v + (\hat{u} \cdot \hat{v}) (a_u b_v + a_v b_u) \\ &= a_u b_u + a_v b_v. \end{aligned} $$

To get the last line we used the fact that $ \hat{u} $ and $ \hat{v} $ form an orthornormal basis, so that they each have length 1 (that is, $ \|\hat{u}\| = \|\hat{v}\| = 1 $) and they are orthogonal (that is, $ \hat{u} \cdot \hat{v} = 0 $ ).

This then shows that

$$ \begin{aligned} a_i b_i + a_j b_j &= a_u b_u + a_v b_v \end{aligned} $$

and so it doesn’t matter which basis we use to compute $ \vec{a} \cdot \vec{b} $ , so long as we use an orthonormal basis.

Projection and complementary projection

The projection and complementary projection are:

Projection of $\vec{a}$ onto $\vec{b}$. #rvv-ep

$$ \operatorname{Proj}(\vec{a},\vec{b})= (\vec{a} \cdot \hat{b}) \hat{b}= (a \cos\theta) \, \hat{b} $$

Complementary projection of $\vec{a}$ with respect to $\vec{b}$. #rvv-em

$$ \begin{aligned}\operatorname{Comp}(\vec{a}, \vec{b})&= \vec{a} -\operatorname{Proj}(\vec{a}, \vec{b}) =\vec{a} - (\vec{a} \cdot \hat{b}) \hat{b} \\\left\|\operatorname{Comp}(\vec{a}, \vec{b}) \right\|&= a \sin\theta\end{aligned} $$

Adding the projection and the complementary projection of a vector just give the same vector again, as we can see on the figure below.

Projection of $ \vec{a} $ onto $ \vec{b} $ and the complementary projection.

As we see in the diagram above, the complementary projection is orthogonal to the reference vector:

Complementary projection is orthogonal to the reference. #rvv-er

$$ \operatorname{Comp}(\vec{a}, \vec{b}) \cdot \vec{b} = 0 $$

Using the definitions of the complementary projection rvv-em and projection rvv-ep, we compute:

$$ \begin{aligned} \operatorname{Comp}(\vec{a}, \vec{b}) \cdot \vec{b} &= \Big(\vec{a} - (\vec{a} \cdot \hat{b}) \hat{b}\Big) \cdot \vec{b} \\ &= \vec{a} \cdot \vec{b} - (\vec{a} \cdot \hat{b}) (\hat{b} \cdot \vec{b}) \\ &= a b \cos\theta - (a\cos\theta) b \\ &= 0. \end{aligned} $$

Change in length and direction

Two useful derivatives are the rates of change of a vector's length and direction:

Derivative of vector length. #rvc-el

$$ \dot{a} = \dot{\vec{a}} \cdot \hat{a} $$

We start with the dot product expression #rvv-ed for length and differentiate it:

$$ \begin{aligned} a &= \sqrt{\vec{a} \cdot \vec{a}} \\ \frac{d}{dt} a &= \frac{d}{dt} \big( (\vec{a} \cdot \vec{a})^{1/2} \big) \\ \dot{a} &= \frac{1}{2} (\vec{a} \cdot \vec{a})^{-1/2} (\dot{\vec{a}} \cdot \vec{a} + \vec{a} \cdot \dot{\vec{a}}) \\ &= \frac{1}{2\sqrt{a^2}} (2 \dot{\vec{a}} \cdot \vec{a}) \\ &= \dot{\vec{a}} \cdot \hat{a}.\end{aligned} $$

Derivative of vector direction. #rvc-eu

$$ \dot{\hat{a}} = \frac{1}{a}\operatorname{Comp}(\dot{\vec{a}}, \vec{a}) $$

We take the definition #rvv-eu for the unit vector and differentiate it:

$$ \begin{aligned} \hat{a} &= \frac{\vec{a}}{a} \\ \frac{d}{dt} \hat{a} &= \frac{d}{dt}\left(\frac{\vec{a}}{a}\right) \\ \dot{\hat{a}} &= \frac{\dot{\vec{a}} a - \vec{a} \dot{a}}{a^2} \\ &= \frac{\dot{\vec{a}}}{a} - \frac{\dot{\vec{a}} \cdot \hat{a}}{a^2} \vec{a} \\ &= \frac{1}{a} \big( \dot{\vec{a}} - (\dot{\vec{a}} \cdot \hat{a}) \hat{a} \big)\\ &= \frac{1}{a} \operatorname{Comp}(\dot{\vec{a}}, \vec{a}).\end{aligned} $$

Here we observed at the end that we had the expression #rvv-em for the complementary projection of the derivative $ \dot{\vec{a}} $ with respect to $ \vec{a} $ itself.

An immediate consequence of the derivative of direction formula is that the derivative of a unit vector is always orthogonal to the unit vector:

Derivative of unit vector is orthogonal. #rvc-eu2

$$ \dot{\hat{a}} \cdot \hat{a} = 0 $$

From #rvc-eu we know that $ \dot{\hat{a}} $ is in the direction of $ \operatorname{Comp}(\dot{\vec{a}}, \vec{a}) $ , and from #rvv-er we know that this is orthogonal to $ \vec{a} $ (and also $ \hat{a} $).

Recall that we can always write a vector as the product of its length and direction, so $ \vec{a} = a \hat{a} $. This gives the following decomposition of the derivative of $ \vec{a} $.

Vector derivative decomposition. #rvc-em2

$$ \begin{aligned} \dot{\vec{a}} &=\underbrace{\dot{a}\hat{a}}_{\operatorname{Proj}(\dot{\vec{a}},\vec{a})} + \underbrace{a\dot{\hat{a}}}_{\operatorname{Comp}(\dot{\vec{a}},\vec{a})}\end{aligned} $$

Differentiating $ \vec{a} = a \hat{a} $ and substituting in #rvv-el and #rvv-eu gives

$$ \begin{aligned} \dot{\vec{a}} &= \dot{a} \hat{a} + a \dot{\hat{a}} \\ &= ( \dot{\vec{a}} \cdot \hat{a} ) \hat{a} + a \frac{1}{a} \operatorname{Comp}(\dot{\vec{a}}, \hat{a}) \\ &= \operatorname{Proj}(\dot{\vec{a}}, \vec{a}) + \operatorname{Comp}(\dot{\vec{a}}, \vec{a}). \end{aligned} $$

Show:

Vector derivatives can be decomposed into length changes (projection onto $ \vec{a} $) and direction changes (complementary projection). Compare to Figure #rvv-fu.