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Edit note: I've adopt an advice to use $\oplus$, which is equivalent to $+$ for vector. And I do distinguish it from that because a scalar, in normal case, can't be added to a vector.(special case)

Both $\hat i$ and $\hat j$ are unit vectors. Assume that: $$\vec A = A_{x} \hat{i} \oplus A_{y} \hat{j}\\ \vec B = B_{x} \hat{i} \oplus B_{y} \hat{j}\\$$

One of my book computes the dot product like this: $$Since \quad \hat i \cdot \hat i = \hat j \cdot \hat j = 1 \quad and \quad \hat i \cdot \hat j = 0\\ \begin{align} \vec A \cdot \vec B &=(A_{x} \hat{i} \oplus A_{y} \hat{j}) \cdot (B_{x} \hat{i} \oplus B_{y} \hat{j})\\ &= A_{x} \hat i \cdot B_{x} \hat i + A_{x} \hat i \cdot B_{y} \hat j\\ &\quad + A_{y}\hat j\cdot B_{x}\hat i+A_{y}\hat j\cdot B_{y}\hat j\\ &= A_{x} B_{x} + 0\\ &\quad + 0 + A_{y} B_{y}\\ &= A_{x} B_{x} + A_{y} B_{y}\\ \end{align}$$

Why the $\oplus$, which is for vector, in the parentheses become the $+$, which is for scalar? Is this related to axioms of vector space?


Edit2: Definition from my book:

We define $\vec A \cdot \vec B$ to be the magnitude of $\vec A$ multiplied by the component of $\vec B$ in the direction of $\vec A$. Expressed as an equation,

$\vec A \cdot \vec B = AB\cos{\phi} = |\vec A||\vec B|\cos{\phi}$

As above, expand the dot product for the two parentheses, and some terms become $0$,

$\vec A \cdot \vec B = A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z}$

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  • $\begingroup$ The definition of the dot product of two vectors $A$ and $B$ is $$A \cdot B = \sum_{i} A_{i} B_{i}$$ Notice that the definition is a sum. I'm not sure what you mean when you say there is a bold $+$ and a 'normal' $+$. $\endgroup$ – mattos Jan 17 '17 at 15:18
  • $\begingroup$ @Mattos But my physics textbook explain it not by definition. it first states that something like $\hat i\cdot\hat i=1$, and $\hat i\cdot\hat j=0$, and then expand the dot product. $\endgroup$ – Postal Model Jan 17 '17 at 15:27
  • $\begingroup$ In the brackets in the second line you have (sum of) vectors. The $\cdot$ is a bilinear map between two vectors giving you a scalar, so if you sum two images of this map you are summing numbers. That's why you "bold +" becomes a simple + $\endgroup$ – Oscar Jan 17 '17 at 15:45
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    $\begingroup$ It's used for a few different things, such as the binary XOR operation, or direct sum in group theory. It isn't normally used for vector addition; I suggested it because it stands out clearly from the regular $+$ symbol, but maybe some people will grumble about it. A less ostentatious alternative would be something like $+'$ or $\tilde +$, etc. $\endgroup$ – Théophile Jan 17 '17 at 16:31
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    $\begingroup$ Actually, a scalar can be added to a vector, much as an imaginary number can be added to a real. You just get something new that is neither a scalar, nor a vector. See Geometric Algebra, for example: slehar.wordpress.com/2014/03/18/… $\endgroup$ – prokaryoticeukaryote Jan 17 '17 at 17:28
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If I understand the question correctly, you are asking about the following identity, which expresses the distributivity of the scalar product over vector addition:

$$\vec{v} \cdot (\vec{w} \oplus \vec{u}) = (\vec{v}\cdot\vec{w})+(\vec{v}\cdot\vec{u})$$

In the above equation, the operator $\oplus$ denotes vector addition, while the operator $+$ denotes scalar addition.

The reason the operator needs to change from vector addition to scalar addition is because the quantities in parentheses on the right-hand side of the equation are not vectors. They are scalars. Indeed if you tried to write $\oplus$ on the right-hand side the resulting expression would not make sense.

As for why it's true, in order to answer that question we would need to know how your book defines the dot product or inner product. In an abstract vector space, an inner product is by definition a pairing that takes two vectors and returns a scalar, and that satisfies various properties, of which the identity above is one. In more elementary treatments, the dot product may be specified by simply defining the dot products of the basis vectors $\hat{i}$ and $\hat{j}$ and extending that definition to the entire space via the distributive property above; in such a development, the distributive property may or may not be stated explicitly, but it really should be.

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  • $\begingroup$ A Physics book is not a Math book, and a college Physics book probably assumes you have already seen dot products so all it needs to do is review the topic, not cover it rigorously. $\endgroup$ – mweiss Jan 17 '17 at 17:45
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There are two equivalent ways to define the dot product. One is with the formula that sums the products of the coordinates. The other starts from the prescribed values for the basis vectors $i$ and $j$ and the assumption that the dot product will be linear in each of its arguments separately.

If necessary, I can flesh this answer out with formulas, but the words carry the idea.

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The dot product is a bilinear product, that is, for any $\alpha,\beta\in\mathbb R$ and any $u,v,w \in V$ where $V$ is the vector space, you have $$(\alpha u + \beta v)\cdot w = \alpha (u\cdot w) + \beta (v\cdot w)$$ and analogously for the second argument. Note that this is part of the definition of the dot product. Also note that on the left hand side we've got an addition of vectors, while on the right hand side we've got an addition of scalars.

Using this, we find \begin{align} a\cdot b &= (a_x\hat i + a_y\hat j)\cdot b\,\,\,\,\\ &= a_x(\hat i\cdot b) + a_y(\hat j\cdot b)\\ &= a_x(\hat i\cdot (b_x\hat i + b_y\hat j)) + a_y(\hat j\cdot (b_x\hat i + b_y\hat j))\\ &= a_x (b_x(\hat i\cdot \hat i)+b_y(\hat i\cdot \hat j)) + a_y (b_x(\hat j\cdot \hat i)+b_y(\hat j\cdot \hat j))\\ &= a_x (1b_x + 0b_y) + a_y (0b_x + 1b_y)\\ &= a_x b_x + a_y b_y \end{align}

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  • $\begingroup$ Where could I find the definition you gave? I found wiki use the sum definition. $\endgroup$ – Postal Model Jan 17 '17 at 16:34
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    $\begingroup$ Look at the Wikipedia page for inner product space (the inner product is a generalization of the dot product). Note that for your case, you can simple ignore the overbars in the definition (they describe complex conjugation, which for real numbers just doesn't do anything). $\endgroup$ – celtschk Jan 17 '17 at 16:46
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The dot product comes from the Law of Cosines, which, in turn, can be derived using basic geometry and trigonometry. The Law of Cosines gives us the relationship between angles (A, B, and C) and the length of sides of an arbitrary triangle:

enter image description here

It states: $c^2 = a^2 + b^2 - 2ab\cos{C}$

If we treat the vertex at C as the origin, and the lines extending from it as vectors, as depticted above, then we can immediately see that we can obtain information about the angle between the vectors of a and b if we know the length of c, which is just the magnitude of b subtracted from a.

So, really, we have

\begin{equation} \left|\vec{a} - \vec{b}\right|^2 = a^2 + b^2 - 2ab\cos{C} \end{equation}

Which can be rewritten as \begin{equation} \left(a_x - b_x\right)^2 + \left(a_y - b_y\right)^2 = a_x^2 + a_y^2 + b_x^2 + b_y^2 - 2ab\cos{C} \end{equation}

We are then left with

\begin{equation} ab\cos{C} = a_xb_x + a_yb_y \end{equation}

This inspires us to define \begin{equation} \vec{a}\cdot \vec{b} = a_xb_x + a_yb_y \end{equation}

This is known as the dot product. It is an operation that takes two vectors, and produces a scalar and tells us what the angle between two vectors is.

Going back to your question, notice that if a and b are perpendicular, for example, if $\vec{a} = \hat{i}$ and $\vec{b} = \hat{j}$, then the dot product is 0. Likewise, if they are the same vector, then the dot product just becomes its squared magnitude. The last observation to make is that the dot product distributes over addition of vectors so that

\begin{equation} \left(\vec{a} + \vec{b}\right)\cdot \vec{c} = \vec{a}\cdot\vec{c} + \vec{b}\cdot \vec{c} \end{equation}

This enables us, then, to define dot product in a different way, without explicitly relying on defining it in terms of components, which is what your book does. In fact, this is the first hint that vectors exhibit a rich algebra, which enables us to encapsulate their geometric properties without directly appealing to vector components.

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  • $\begingroup$ I need to digest the last part you pointed out. Thanks for detail and inspiring explanation! $\endgroup$ – Postal Model Jan 17 '17 at 16:59
  • $\begingroup$ @N1ng While I see the value of defining the dot product, as your book does, without appealing to components, it is not the best way, pedagogically, to introduce it as a concept, and certainly one of the the worst ways, in my opinion, to motivate it. I feel like this is an example of mathematicians' preoccupation with "elegance" getting in the way of clarity. $\endgroup$ – prokaryoticeukaryote Jan 17 '17 at 17:14

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