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The formula for the cross product is : $$ a \times b = \vert a \vert \vert b \vert \sin(\theta) n $$ One of the purposes of the cross product is to get a vector value that is at right angles to both the given vectors.

  • If we are finding such a vector (which is perpendicular/at a right angle to both given vectors) then why do we have to use $\text{'}n\text{'}$ as an input of the cross product calculation, where $n$ = a unit vector at right angles to both $a$ and $b$.

  • Can we find a vector that is at right angles to both given vectors without knowing $\text{'}n\text{'}$?

  • and if we have to use $\text{'}n\text{'}$ , how we would calculate the value of $n$ when we only have vector $a$ and vector $b$?
Björn Friedrich
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user2259784
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  • $\vec n$ is a unit vector pointing in the direction of vector perpendicular to $\vec a$ and $\vec b$. There are two values for $\vec n$. We use $\vec n$ because $|a||b|\sin \theta$ is scalar and $a \times b$ is vector. To find $\vec n$, use right hand rule. I don't get second question, please clarify. –  Mar 02 '17 at 19:30
  • thank you for the answer. If we just talk about vector direction (no magnitude) - what I am trying to understand is that if we already have a vector perpendicular to vector a and vector b i.e. 'n' then we can just use that value as Cross-product because that's the direction Cross product is going to give us anyway? Or am I missing something? – user2259784 Mar 02 '17 at 19:44
  • Yes you can use that value. But you need to be careful as there are two vectors perpendicular to given two vectors and cross product gives only one vector not two. –  Mar 02 '17 at 19:49
  • @user2259784 The definition you mention (plus the bit you didn't mention about $n$ being the right-handed normal as opposed to the left-handed normal) uniquely defines the cross product of two vectors in $\Bbb R^3$. So the important question I have in response to your question is, "OK, you don't like this definition. How would you suggest that we define the cross product?" –  Mar 02 '17 at 21:21

4 Answers4

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That is one way to define it, but the usual way to compute $c = a \times b$ is using coordinates: $$ \eqalign{c_1 &= a_2 b_3 - a_3 b_2\cr c_2 &= a_3 b_1 - a_1 b_3\cr c_3 &= a_1 b_2 - a_2 b_1\cr}$$ So no, you don't need to know $n$ beforehand, you can get it as a result of computing the cross product.

Robert Israel
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Your main arguments against the definition $a\times b := |a||b|\sin(\theta)\hat n$ seem to be (1) that to use this definition, we have to be able to find the right-handed normal vector $\hat n$ first. I don't see that as much of a problem. There are several known methods of finding such a normal vector. And (2) that you seem to believe the cross product is only useful for finding the normal vector. I disagree. A non-exhaustive list of some of the other useful applications of the cross product includes

  • finding the area of the parallelogram with sides as the vectors.
  • describing various physical quantities including torque and the magnetic field.
  • together with the dot product in the form of the scalar triple product, finding the volume of a parallelopiped.
  • describing (/modelling) the Lie algebra $\mathfrak{so}(3)$.

Now here are some arguments for that particular definition of the cross product (BTW, there are other definitions):

  1. It's entirely coordinate-independent. So, when using this definition, not only will not need to worry about having to change our coordinates at some point in the middle of working through a problem, but it makes the notation far more compact than say the formulas in user247327's and Robert Israel's posts.
  2. It's a very geometric definition. It's of the form (scalar) times (unit vector) so we know immediately upon looking at that formula that the length of the cross product is $|a||b|\sin(\theta)$ and the direction that it points in is the right handed normal direction. Again, compare this to the formulas given in user247327's and Robert Israel's posts. It is not at all clear just looking at those definitions what the length or direction of $a\times b$ is.
  3. It suggests a connection with the dot product as defined by $a\cdot b :=|a||b|\cos(\theta)$. For instance, Lagrange's identity becomes very clear: $$(a\cdot b)^2 + |a\times b|^2 = |a|^2|b|^2\cos^2(\theta) + |a|^2|b|^2\sin^2(\theta) = |a|^2|b|^2\big(\cos^2(\theta) + \sin^2(\theta)\big) = |a|^2|b|^2 \\ \bbox[5px,border:2px solid red] {(a\cdot b)^2 + |a\times b|^2 = |a|^2|b|^2}$$

Postscript:

Note that the definition of an object is not always going to be the best way to calculate it in practice. Now that we have this definition, we can show that it implies the formulas in user247327's and Robert Israel's posts under mild assumptions. So if we want to put some numbers to it, there's no reason not to use those formulas instead.

Compare this to the Riemann integral. The definition of the Riemann integral is pretty complicated, right? But luckily we rarely have to use that definition to actually evaluate an integral -- we use the fundamental theorem of calculus.

It's the same thing for the cross product. The definition might be a little more difficult to use, but for the reasons above (and some more complicated reasons I didn't get into) it's a pretty good definition for proving theorems and other things we might need to do with a definition. But it's completely fine to use some other method to actually calculate it.

  • "Your main arguments against the definition a×b:=|a||b|sin(θ)n̂ a×b:=|a||b|sin⁡(θ)n^ seem to be (1) that to use this definition, we have to be able to find the right-handed normal vector n̂ n^ first. I don't see that as much of a problem...."

    I admire visual explanations which a×b:=|a||b|sin(θ)n̂ this formula does but I feel that formula for finding n̂ should also be the part of this formula if not at least it should be well explained, not many people talk about how to calculate n̂. Can you please give an example of calculating normal vector from two given vectors?

     – user2259784 Mar 02 '17 at 22:12
  • While it may not always be the fastest method, you can just use the naive approach: $a$ and $b$ are perpendicular to $n$ iff $$a\cdot n = 0\qquad \text{and}\qquad b \cdot n = 0$$ So, given two vectors $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$, just find any nonzero solution to the system of equations $$\begin{cases} a_1n_1 + a_2n_2 + a_3n_3 = 0 \ b_1n_1 + b_2n_2 + b_3n_3 = 0\end{cases}$$ –  Mar 02 '17 at 22:15
  • "(2) that you seem to believe the cross product is only useful for finding the normal vector. I disagree. A non-exhaustive list of some of the other useful applications of the cross product includes."

    Thank you for giving examples of the application of Cross product. I am coming from computer graphics/shader background and so far finding perpendicular vector is the only application of Cross product I have come across, I will keep looking for more :)

     – user2259784 Mar 02 '17 at 22:16
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Cross product gives you a vector. Thus, since $\|a\|\|b\| \sin( \theta)$ is a scalar, it would not give you a vector. So, to get a vector you need to multiply that scalar by a unit vector $n$ whose length is one and whose direction is determined by the right hand rule.

Pawel
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"One of the purposes of Cross product is to get a vector value that is at the right angle to both the given vectors. If we are finding such vector (which is perpendicular/at the right angle to both given vectors) then why we have to use 'n' as an input of Cross product calculation where n = unit vector at right angles to both a and b."

We don't use n to find the Cross Product. Instead we use the formula Robert Israel gave, which can be written as a "symbolic" determinant: $\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b3\end{array}\right|$

"Can we find out a vector that is at the right angle to both given vectors without knowing 'n'?"

Yes, by using Robert Israel's formula. While it is useful to know the definition you give in order to use the cross product, no one actually calculates the cross product that way. They use that formula instead.

"and if we have to use 'n', how we would find its value at the first place when we only have a and b?"

No, we don't "have to use 'n'"!

user247327
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  • Thank you, that clarifies the idea! So there is no real world application of this formula: a × b = |a| |b| sin(θ) n ? – user2259784 Mar 02 '17 at 20:19
  • @user2259784 You can find the angle between two vectors using this formula and the value of determinant. It is also the area of parallogram formed by $\vec a$ and $\vec b$ as two adjacent sides (but this can done be done with components). –  Mar 02 '17 at 20:38