# Why does cross product give a vector which is perpendicular to a plane

I was wondering if anyone could give me the intuition behind the cross product of two vectors $\textbf{a}$ and $\textbf{b}$. Why does their cross product $\textbf{n} = \textbf{a} \times \textbf{b}$ give me a vector which is perpendicular to a plane?

I know I can just check this by using dot product but I'm not totally satisfied with "it just works" answer =)

Thank you for any help! =)

• Not just "a" plane; the subspace (plane) of ${\bf R}^3$ generated by $\bf a$ and $\bf b$. If you're using the "morally correct" definition of $\times$; this property is by definition (Q: how do we know bachelors aren't married? A: by definition), the real thing to prove is the computational formula for it, which can be done by noting $\times$ is bilinear and then computing it on the basis vectors. (The "determinant" formula is actually mysterious.)
– anon
Jun 14, 2013 at 6:15
• en.wikipedia.org/wiki/Cross_product Jun 14, 2013 at 6:24

The determinant formula isn't so mysterious. Consider the cross product $\mathbf{v} = \langle a,b,c \rangle \times \langle d,e,f \rangle$ as the formal determinant

$$\det \left(\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ a & b & c \\ d & e & f \end{array} \right)$$

where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are the standard basis vectors. If instead one considers $\mathbf{i}, \mathbf{j}, \mathbf{k}$ as indeterminates and substitutes $x, y, z$ for them, this determinant computes the dot product $\mathbf{v} \cdot \langle x, y, z \rangle$. But letting $\langle x, y, z \rangle$ be $\langle a, b, c \rangle$ or $\langle d, e, f \rangle$ gives a zero determinant, so $\mathbf{v}$ is perpendicular to the latter two vectors, hence to the plane they span, as Omnomnomnom says.

• Is this considered a valid mathematical proof? Easy to understand but is seems like a slight of hand. Apr 30, 2017 at 23:14
• Why do you say that substituting {i,j,k} for v={x,y,z} is the dot product v · {x,y,z}? I calculated that and I didn't get it. Sep 17, 2018 at 4:45
• @Sedumjoy Very late, but the "definition" of the cross product in terms of this determinant is indeed a bit of trickery. You can "properly" define the cross product as simply being the solution (for all $x, y, z$) of $(\langle a,b,c\rangle\times\langle d,e,f\rangle)\cdot\langle x,y,z\rangle=\begin{vmatrix}x&y&z\\a&b&c\\d&e&f\end{vmatrix}.$ The notation with the vectors in the matrix is just a trick to quickly compute the cross product by using certain properties of this definition, and the desired property also comes out quickly.
– HTNW
Sep 1, 2020 at 18:48
• @HTNW ...thank you. The cross product like so many things in maths is so rich in history...that sometimes it pays to be happy with "a bit of trickery" in order to get the job done. Sep 2, 2020 at 20:20

According to Prof. Josiah Willard Gibbs, one of the founders of Vector Analysis, the cross (or skew) product is defined as follows in his 1881 pamphlet Vector Analysis - Arranged for the use of Students in Physics

Definition: The skew product of the vector A into the vector B is the vector quantity C whose direction is the normal upon that side of the plane of A and B on which rotation from A to B through an angle of less than one hundred and eighty degrees appears positive or counter- clockwise ; and whose magnitude is obtained by multiplying the product of the magnitudes of A and B by the sine of the angle from A to B.

This definition, coming from the horse's mouth, as it were, is a geometric definition. Why then, do we in the 21st c. insist on defining it in terms of determinants. The determinant "definition" is a direct consequence of the geometric definition and the distributivity of cross products $$\vec{\mathbf{w}}\times(\vec{\mathbf{u}}+ \vec{\mathbf{v}}) = \vec{\mathbf{w}}\times\vec{\mathbf{u}}+\vec{\mathbf{w}}\times\vec{\mathbf{v}}$$ $$(\vec{\mathbf{u}}+ \vec{\mathbf{v}})\times\vec{\mathbf{w}} = \vec{\mathbf{u}}\times\vec{\mathbf{w}}+\vec{\mathbf{v}}\times\vec{\mathbf{w}}$$ now try to expand $$(a_1\vec{\mathbf{i}}+a_2\vec{\mathbf{j}}+a_3\vec{\mathbf{k}})\times(b_1\vec{\mathbf{i}}+b_2\vec{\mathbf{j}}+b_3\vec{\mathbf{k}})$$ to see that cross products, in fact, distribute over addition, see a geometric proof on page 9 of http://www.math.oregonstate.edu/bridge/papers/dot+cross.pdf

See what happens when you try to take $(a\times b)\cdot a$ or $(a\times b)\cdot b$ (you should get $0$). If a vector is perpendicular to a basis of a plane, then it is perpendicular to that entire plane. So, the cross product of two (linearly independent) vectors, since it is orthogonal to each, is orthogonal to the plane which they span.

Also, while you're trying to develop an intuition for cross products, I highly recommend this video

and, while we're there, might be worth knowing how the angle formula for dot products comes from the law of cosines

Look. I'm not a mathematician, but I have a perspective which can explain why the cross product of two vectors is another vector perpendicular to them. It is not a proof but it will help make that idea fimilar.

One can understand cross product in this way:imagine a line segment that makes colorful marks wherever it moves on a paper. Now make it move in the same direction and for the same distance as another segment, preserving the direction of the first segment. The result would be a parallelogram with area donated by ab sin(θ) where a and b are the lengths of the segments. So we can define cross product of two vectors as "the area of the parallelogram in wich these vectors are adjacent sides".

The first question is: why is it a vector? Imagine a plane containing two vectors a and b and the angle from a to b equals θ, the cross product of a and b equals ||a|| ||b|| sin(θ). That works quite well as a scalar, but the problem comes when you flip the entire plane containing the vectors, which is the same as replacing a for b and vise versa. The angle from a to b becomes -θ, and the cross product becomes ||a|| ||b|| sin(-θ), which equals -||a|| ||b|| sin(θ). Flipping the plane reverses the result. So the cross product can be represented as a vector with its starting point lying on the plane, which points "up or down". If that vector is pointing up, it will point down when you flip the plane and vise versa.

The second question is: why is the cross product perpendicular to the plane? Why doesn't it just point anywhere up or down? That's because when you flip the plane the cross product is completely reversed, which means it's perpendicular to the plane.

Here's another perspective: the cross product is a sort of 'currying' of the volume form (AKA the determinant). That is, we have a function $V(\mathbf{a}, \mathbf{b}, \mathbf{c})$ that determines the volume spanned by three vectors; because this function is linear in each argument then we can fix $\mathbf{a}$ and $\mathbf{b}$ and consider this as a linear function $V_{\mathbf{a}\mathbf{b}}(\mathbf{c})$ that takes vectors to numbers. But we know that every linear function $f(\mathbf{c})$ from vectors to numbers is of the form $\mathbf{v}\cdot\mathbf{c}$ for some vector $\mathbf{v}$, so we can define the cross-product of $\mathbf{a}$ and $\mathbf{b}$ to be this vector $\mathbf{v_{\mathbf{a}\mathbf{b}}}$ such that $V(\mathbf{a}, \mathbf{b}, \mathbf{c}) = \mathbf{v}_{\mathbf{a}\mathbf{b}}\cdot \mathbf{c}$. The cross-product is then orthogonal to $\mathbf{a}$ and $\mathbf{b}$, roughly, because the direction perpendicular to both vectors is the one that maximizes the volume. For a bit more on this approach, see my answer here.

My answer is that this is the case because the cross product is defined to have that property.

• Can you motivate this definition? Jun 14, 2013 at 8:38
• I tell my students --- historically incorrect but in my mind pedagogically sound --- that the cross product was developed to model torques: math.stackexchange.com/questions/388045/… Jun 14, 2013 at 9:46
• @JpMcCarthy I think user1729 meant motivation should be added to the body of the solution. Aug 2, 2013 at 12:28
• This is nonsensical circuit argument. Mar 23, 2015 at 21:38
• @user48672 would you say the definition of a determinant is a nonsensical circular argument? Feb 14, 2018 at 23:17

There are two aspects to this question. First, the definitions of cross and dot products follow directly from the product of quaternions introduced by Hamilton, although he did not give names to these products (Gibbs named them, although he used other names). The fact that the cross product is perpendicular to the vectors being multiplied is directly related to the theory of rotations. The result of the product is a rotation axis.

Second, as described above, it is difficult to speak of a motivation for the definition of cross product. It was a byproduct of the more general concept of product of quaternions. Because of its properties, the cross product became popular when it was shown to be useful to solve physics problem. However, I accidentally found a motivation for the cross product by consideration of the linear combination of two vectors in 3D space. I considered the plane generated by a linear combination of two vectors u and v. When looking for the coefficients in the linear combination required to reach a desired point on the plane, the solution involves the existence of a normal vector n = u × v. This was an unexpected result because the concept of linear combination does not involve any product of vectors.

I discuss all of the preceding in the paper: The linear combination of vectors implies the existence of the cross and dot products, Int. J. Math. Education Sci. Technol., DOI: 10.1080/0020739X.2017.1408149

The paper is available in full here: www.tandfonline.com/eprint/CPN9hQWMMVu7nU2bZebd/full

I would start by something simpler.

What is i x j ?

but why? Well, it's by definition. The cross product of a unit vector in the x-direction (i) and a unit vector in the y-direction (j) is a perpendicular vector in the z-direction (k).

Given the above, one can easily see that: 2i x j = 2k

In 2 dimensions, if you want to multiply a x b

1. Simply breakdown vectors a and b into their i and j components.
2. Multiply the i from the first vector (a) with the j in the second vector (b).
3. Multiply the j from the first vector (a) with the i in the second vector (b).
4. Add them together, remembering that j x i is NEGATIVE k.

so step 2 and 3 work opposite of each other - if you will.

That's it. Now think about this in 3 dimensions remembering that: j x k = i k x i = j and that the reverse cross product gives a negative sign.