# What does the dot product of two vectors represent?

I know how to calculate the dot product of two vectors alright. However, it is not clear to me what, exactly, does the dot product represent.

The product of two numbers, $2$ and $3$, we say that it is $2$ added to itself $3$ times or something like that.

But when it comes to vectors $\vec{a} \cdot \vec{b}$, I'm not sure what to say. "It is $\vec{a}$ added to itself $\vec{b}$ times" which doesn't make much sense to me.

• Adding $\vec{a}$ to itself $b$ times ($b$ being a number) is another operation, called the scalar product. The dot product involves two vectors and yields a number.
– user65203
Commented May 22, 2014 at 22:40
• Something not mentioned but of interest is that the dot product is an example of a bilinear function, which can be considered a generalization of multiplication. Commented Mar 15, 2021 at 23:34
• Does this answer your question? Meaning of Dot Products in Regards to Linear Algebra Commented Mar 17, 2021 at 16:58

The dot product tells you what amount of one vector goes in the direction of another. For instance, if you pulled a box 10 meters at an inclined angle, there is a horizontal component and a vertical component to your force vector. So the dot product in this case would give you the amount of force going in the direction of the displacement, or in the direction that the box moved. This is important because work is defined to be force multiplied by displacement, but the force here is defined to be the force in the direction of the displacement.

http://youtu.be/KDHuWxy53uM

• For this interpretation it's important that the vector you are projecting onto has unit length, otherwise you are getting the component of vector 1 along vector 2 scaled by the length of vector 2 Commented May 22, 2014 at 22:52
• Work is a good example of magnitude as well as direction. W = F.D, i.e work is the scalar product of the force and displacement vectors (assuming constant force and direction). Commented May 23, 2014 at 7:37
• This answer is vague. As rVitale points out, your first sentence is only true if the "other" vector is a unit vector. When you say "[pull] a box 10 meters at an inclined angle", you need to be clearer about what the two vectors are: presumably, you mean the force vector and the vector along which the box moves. Commented May 23, 2014 at 9:08
• Obviously it helped the OP. The criticism of the King is blasphemous. Commented May 23, 2014 at 13:05
• Oh this makes sense now. The dot product is zero when they're orthogonal because they don't have any common component! Commented Sep 13, 2015 at 18:08

# Geometric Meaning

As other answers have pointed out, the dot product $$\vec{a} \cdot \vec{b}$$ is related to the angle $$\theta$$ between $$\vec{a}$$ and $$\vec{b}$$ through:

$$\vec a \cdot \vec b = \Vert\vec a\Vert_2 \, \Vert\vec b\Vert_2 \, \cos \theta$$

Assuming that $$a$$ and $$b$$ point into similar directions, i.e., $$\theta \leq 90°$$, we can visualize what this relationship means (skipping the vector arrows and Euclidean norm subscript from now on):

$$p$$ is the vector resulting from an orthogonal projection of $$a$$ onto $$b$$. As the $$\cos$$ is the ratio between the adjacent leg ($$p$$) and the hypotenuse ($$a$$) in the right triangle, i.e.,

$$\cos \theta = \frac{\Vert p \Vert}{\Vert a \Vert},$$

we get for the inner product:

$$a \cdot b = \Vert a \Vert \, \Vert b \Vert \, \frac{\Vert p \Vert}{\Vert a \Vert} = \Vert p \Vert \Vert b \Vert$$

So, the inner product is the length of the vector $$p$$, the projection of $$a$$ onto $$b$$, multiplied by the length of $$b$$. If $$a$$ and $$b$$ point into opposite directions, i.e., $$90° < \theta \leq 180°$$, the dot product will be the negative: $$a \cdot b = - \Vert p \Vert \Vert b \Vert$$

# Derivation

The problem is that the relationship between the dot product and the angle $$\theta$$ is not inherently given. By definition:

$$a \cdot b = \sum_i a_i b_i$$

So, we need to find a link between this and the cosine. From the definition of the dot product, we can see that it scales proportionally with the input vectors, so for non-unit vectors $$u$$ and $$v$$ with the corresponding unit vectors $$\hat{u}$$ and $$\hat{v}$$:

$$u \cdot v = \Vert u \Vert \hat{u} \cdot \Vert v \Vert \hat{v} = \Vert u \Vert \Vert v \Vert \hat{u} \cdot \hat{v}.$$

For simplicity, we will assume $$a$$ and $$b$$ to be unit vectors. Thus, we only need to show

$$a \cdot b = \cos \theta$$

or, by the definition of $$\cos$$, we need to show:

$$a \cdot b = \Vert p \Vert$$

Let's calculate the length of the projection $$p$$ using $$a$$ and $$b$$. We can start by using the Pythagorean theorem:

$$\Vert p \Vert ^2 = \Vert a \Vert ^2 - \Vert c \Vert ^2$$

Because $$a$$ is a unit vector:

$$\Vert p \Vert ^2 = 1 - \Vert c \Vert ^2$$

Now, we need to calculate the length of $$c$$ using the other rectangular triangle. Again, we use the fact that $$b$$ is a unit vector, i.e., $$\Vert b \Vert = 1$$.

\begin{align} \Vert c \Vert ^2 &= \Vert d \Vert ^2 - (\Vert b \Vert - \Vert p \Vert)^2 \\ &= \Vert d \Vert ^2 - (1 - \Vert p \Vert)^2 \\ &= \Vert d \Vert ^2 - 1 + 2 \Vert p \Vert - \Vert p \Vert^2 \end{align}

Now, we can insert this term for $$\Vert c \Vert ^2$$ in the equation above:

\begin{align} \Vert p \Vert ^2 &= 1 - \Vert d \Vert ^2 + 1 - 2 \Vert p \Vert + \Vert p \Vert^2 \\ 0 &= 2 - \Vert d \Vert ^2 - 2 \Vert p \Vert \\ 2 \Vert p \Vert &= 2 - \Vert d \Vert ^2 \end{align}

In the figure, we see that $$\vec a + \vec d = \vec b$$. Therefore, $$\vec d = \vec b - \vec a$$, or:

$$d_i = b_i - a_i$$

Thus, we can express $$\Vert d \Vert^2$$ as:

\begin{align} \Vert d \Vert ^2 &= \sum_i d_i^2 \\ &= \sum_i (b_i - a_i)^2 \\ &= \sum_i b_i^2 - 2 b_i a_i + a_i^2 \\ &= \sum_i b_i^2 - \sum_i 2 b_i a_i + \sum_i a_i^2 \\ &= \Vert b \Vert^2 - \sum_i 2 b_i a_i + \Vert a \Vert^2 \\ &= 1 - \sum_i 2 b_i a_i + 1 \\ &= 2 - 2 \sum_i b_i a_i \end{align}

Finally:

\begin{align} 2 \Vert p \Vert &= 2 - \Vert d \Vert ^2 \\ &= 2 - (2 - 2 \sum_i b_i a_i) \\ &= 2 \sum_i b_i a_i \\ \Vert p \Vert &= \sum_i b_i a_i \end{align}

q.e.d.

• Hello Kilian - thanks for this. I'm not sure how you got the expression for \norm{c}^2 in the second step and how you got the expression for the \norm{p}^2 in the third step. Could you please clarify? Commented Jan 26, 2020 at 17:00
• Same here, very interested in a clarification. Commented Oct 24, 2020 at 10:07
• I updated the equations to make it more clear. Let me know if you have any further questions! Commented Nov 15, 2020 at 12:11
• Thanks for such detailed answer.. helped me to clear many doubts and concepts. Commented Dec 27, 2020 at 8:46
• For me this post would be helped if you outlined what you are going to be proving, before the lenghy proof starts. Alternatively, I would put p is the vector resulting from an orthogonal projection of a onto b in bold. Commented Jan 21, 2021 at 12:00

I think of dot product as the "same-ness" of two vectors. If two vectors are orthogonal (90 degrees on one another) they are 'not at all the same' (dot product =0), and if they are parallel they are 'very much the same'. If you divide their dot product by the product of their magnitude, that is the argument for an arccosine function to find the angle between them. My application for the dot product is finding the angle between two vectors for calculating the force required to pull a cable through two or more pipes with a bend. It's hard to do this in a three dimensional world without knowing how to calculate the dot product. Math makes life really easy :)

It might help to think of multiplication of real numbers in a more geometric fashion. $2$ times $3$ is the length of the interval you get starting with an interval of length $3$ and then stretching the line by a factor of $2$.

For dot product, in addition to this stretching idea, you need another geometric idea, namely projection. Imagine the line $L$ parallel to $\vec b$ through the origin $O$. Now imagine projecting from the tip of the vector $\vec a$, along a line perpendicular to $L$, until hitting $L$ at a point $P$. The dot product $\vec a \cdot \vec b$ is the length of the line segment you get by starting with the line segment $OP$ and then stretching the plane by a factor equal to the length of $\vec b$.

I'm being a little careless about plus and minus signs, but those can be incorporated into this picture too.

• What's the importance of this? What do we get from the product? I mean why we are so interested in finding this dot product? Commented Jul 24, 2018 at 2:26
• Hi @LeeMosher please could you help out with math.stackexchange.com/q/4445701/585488 Commented May 8, 2022 at 8:23
• Wasiq, the importance of this is that when we compute something like work (force in the direction of motion multiplied by the displacement) we can simply write this as $F\cdot d$ where $F$ and $d$ are both vectors. If the vectors are not in the same direction we need to find the component of force that is in the direction of motion. So if the force is being applied at an angle relative to the displacement, $|F|cos(\theta)$ gives us the force in the direction of motion and it is that which gets multiplied by $|d|$, not $|F|$ itself. It's a way of multiplying when the direction matters. Commented May 19 at 2:59

The dot product of two vectors u,v is the area of the parallelogram u,v' where v' is v rotated by 90 degrees.

• This is the most interesting answer to me because it shows a link between cross product and dot product. I will give you the most I can which is one. Commented Jul 13, 2020 at 17:24
• Haha, thank you! Commented Jul 14, 2020 at 19:04
• @Jules You are confusing the dot product with the vector product Commented Jun 14, 2021 at 9:12
• I'm not. The vector (cross) product has a similar interpretation, but without the 90 degree rotation. Commented Jun 15, 2021 at 10:40
• Best answer, but (nitpicking) the dot product is the determinant of $u$ and $v'$, i.e, $$\langle u,v\rangle=\det(u,v').$$ Commented Sep 5, 2023 at 18:54

First of all, if we write $\vec{a} = a \vec{u}$ and $\vec{b} = b \vec{v}$, where $a$ and $b$ are the length of $\vec{a}$ and $\vec{b}$ respectively, then $$\vec{a} \cdot \vec{b} = (a \vec{u})\cdot (b \vec{v}) = ab \,\, \vec{u} \cdot \vec{v};$$ this is a pretty natural property for a product to have.

Now as for $\vec{u} \cdot \vec{v}$, this is equal to $\cos \theta,$ where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$.

As King Squirrel notes, this is also the length of the projection of $\vec{u}$ onto the line through $\vec{v}$, and also the length of the projection of $\vec{v}$ onto the line through $\vec{u}$.

So altogether we get

$$\vec{a} \cdot \vec{b} = a b \, \cos \theta,$$ and it has the interpretation in terms of projecting one vector onto another that King Squirrel discusses.

• does this meaning have any remnant when used over $\Bbb F_p$? Commented Dec 9, 2014 at 7:18

Dot product is the product of magnitudes of 2 vectors with the Cosine of the angle between them. You can take the smaller or the larger angle between the vectors. That is if theta is the angle then you can take (360-theta) as well.

Geometrically, it will also be equal to (read it slowly) the product of “projection” of magnitude of one vector on the other and the magnitude of the 2nd vector.

In Physics, as an example, Mechanical Work is a scalar and a result of dot product of force and displacement vectors. Like-wise, Magnetic flux is the dot product of magnetic field and vector area

Let me try to explain this with an example. Say you wish to find the work done by a force F along X axis over a distance d. However the problem also tells you that the direction of the force is not along the X axis but at an angle of 60 degree with X axis.

Now you know that the work done is the product of force and displacement. But in this case you know that the force is not exactly totally acting in the direction of X axis, since it is inclined at 60 degree. So what you can do is, find what is the contribution of this force in the X direction. Well it turns out with simple trigonometry that it is F Cos60 in direction of X axis. Now you can say that the work done = F Cos 60 X d. This can also be represented as F.d = F d Cos 60. So you see, dot product gives us the magnitude of a certain entity (in this case work) by way of attributing a certain vector (in this case force F) in the direction of the other vector. Here "d" was the other vector along which work was being found

Dot product is a scalar quantity. Watch this video that I have made to understand this better-

What is Dot Product of Vectors

If you have two vectors of lengths $$2$$ and $$3$$.

If they point in the same direction it is reasonable that the product would be $$6$$.

If the one of length $$2$$ is rotated $$180^\circ$$, then it becomes $$-2$$ and the other is still $$3$$. It is reasonable that this product is $$-6$$.

So the range of values is between $$-6$$ and $$6$$.

Half way through the rotation at $$90^\circ$$, the product is halfway between $$-6$$ and $$6$$. The product is $$0$$.

Now

$$2 \cdot 3\cos0^\circ=2 \cdot 3 \cdot 1=6$$

$$2 \cdot 3\cos 90^\circ=2 \cdot 3 \cdot 0=0$$

$$2 \cdot 3\cos180^\circ=2 \cdot 3 \cdot -1=-6$$

Note the length is always positive.

When one calculates A.B, two measurements happen: measurement of how small the angle between them is, and how long A and B are. A.B basically means projection length of A on B, with this length then scaled by the absolute length of B.

One way to think about the interpretation of the dot product is to think how would one maximise or minimise the dot product between two vectors. Let's assume we are trying to maximise the dot product between two vectors that we can modify:

The dot product will be grow larger as the angle between two vector decreases. The dot product A.B will also grow larger as the absolute lengths of A and B increase. This is because as A gets larger, its projected length will be longer, and as B's length gets larger, the scaling of A's projection will grow larger, given that B's absolute length will act as a scaler of A's projection length.

• Please do not post the same answer more than once. If you find that the same answer will address multiple questions, then please flag the questions as duplicates. Commented Mar 17, 2021 at 16:04

These questions are better approached geometrically because they have very well defined applications and real world examples.

For example here we are mentioning the dot product which has a very direct application in calculating work for example.

Work should be force multiplied by the distance moved. This is in simple algebra but when you get to geometry and 3D. Both the force and displacement have directions.

So the dot product helps by projecting the force onto displacement to help you decide how did the force contribute in doing the work.

Three Scenarios:

1 - Force is in direction of displacement: means the force did positive work in moving the object.

2 - Force is perpendicular to displacement: means force did nothing in moving the object in this direction. Zero work.

3 - Force is in the opposite direction of displacement: means force did negative work in moving the object in this direction; slowed it down or stopped it.

When directions are considered, we essentially bring a new dimension to the perception of the entity. (Speed vs Velocity: 5km/h vs 5km/h towards east). Bringing the sense of direction, the question arises, how the entities interact?

In dot product, diagrammatically, what we find is, essentially, the area that is affected by the two entities taken together.

Consider Tetris. You have built a foundation already. Now, a new part is falling and you have the arrow keys to move it around. Two competing vectors, your movement and the falling of the brick/part, will determine how the new part is arranged. The area covered by the falling part would be determined by the dot product of the said vectors.

I don't think the dot product has a very obviously interesting visual interpretation.

I'm not a math teacher, but if I were asked to define the dot product in a course, I'd start by defining the scalar projection first. This has a very intuitive interpretation. And then you'd see immediately that, in order to compute the scalar projection, it's useful to compute the dot product.

So, in a normed vector space $$E$$, let $$u,v \in E$$ two vectors such as $$u, v \neq 0$$.

Formally, the scalar projection of $$v$$ over $$u$$ is the unique scalar $$\alpha$$ such as

$$|| \alpha u ||^2 + || v - \alpha u ||^2 = || v ||^2$$

It might not be true in every normed vector space that such a unique scalar has to exist, but in $$\mathbb{R}^n$$ with the Euclidean norm at least, you can prove that it does exist and that it is indeed unique.

Intuitively/geometrically, you're fixing a point $$P$$ and you're trying to adjust the length and the orientation of a vector that goes in the same direction as $$u$$, (that's $$\alpha u$$), so that the triangle formed by the points: $$P, P+v, P+\alpha u$$ satisfies the Pythagorean property of a right triangle of hypothenuse $$[P, P+v]$$ (that's what the property $$|| \alpha u ||^2 + || v - \alpha u ||^2 = || v ||^2$$ expresses).

If you expand the condition

$$||\alpha u||² + ||v - \alpha u||² = ||v||²$$

using the definition of the Euclidean norm, you'll see that $$\alpha$$ has to be equal to

$$\frac{\sum\limits_{i=1}^{n} u_i v_i}{||u||}$$

(and that this quantity does satisfy the condition).

So geometrically, what is the scalar product of $$u$$ and $$v$$ ? I'd say that it is this quantity that when divided by $$||u||$$ gives you the scalar projection of $$v$$ over $$u$$.