If $\mathbf a \perp \mathbf b$, then $|\mathbf a \times \mathbf b|=|\mathbf a||\mathbf b|$.

The result can be interpreted as saying that the length of the vector product equals the area of the rectangle that can be formed by $\mathbf a$ and $\mathbf b$.

Is there any intuition (geometric or otherwise) for this?

Definitions used:

Let $\mathbf{a}=(a_1,a_2,a_3)$ and $\mathbf{b}=(b_1,b_2,b_3)$. Then

  • $\mathbf{a}\times\mathbf{b}=(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1)$,
  • $\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3$,
  • $|\mathbf{a}|=\sqrt{a_1^2 +a_2^2 +a_3^2}$,
  • $\mathbf a \perp \mathbf b$ if $\mathbf{a}\cdot\mathbf{b}=0$.
  • $\begingroup$ Have you tried watching 3blue1brown’s Linear Algebra series on YouTube? $\endgroup$ Jul 12, 2023 at 6:08
  • $\begingroup$ @bananapeel22: No. If there is a specific segment of a specific video that answers this question, please direct me to that. $\endgroup$
    – user986614
    Jul 12, 2023 at 6:16

1 Answer 1


It is not much about Intuition. It is the Direct Consequence of the Definition.

Consider this Definition :
$$A \times B = ||A|| \cdot ||B|| \cdot \sin (\theta) \hat{n}$$
where $\hat{n}$ is the Unit vector Perpendicular to $A$ & $B$.

When $\theta = \pi/2$ ( vectors are Perpendicular , making a rectangle ) , then $\sin (\theta) = 1$.

[[ With that Definition , we can get the vector Components when $A=(a_1,a_2,a_3)$ & $B=(b_1,b_2,b_3)$ , which will match what you have given. ]]

Now , Parallelogram with Sides $||A||$ & $||B||$ with angle $\theta$ will have that Same Area.


The Height of the triangle (Purple line) is $||B|| \cdot \sin (\theta)$
Area of 1 triangle (Blue) is $||A|| \cdot ||B|| \cdot \sin (\theta) / 2$
Area of Parallelogram (made of the 2 triangles) is $2 ||A|| \cdot ||B|| \cdot \sin (\theta) / 2$

That matches the vector Product Definition.

It is not much about Intuition. It is the Direct Consequence of the Definition.


Definition used here is Common.
Check Page 3 : https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-vectorprod-2009-1.pdf :


Page 7 then gives the "Conversion" from that Definition to this formula :


More Details here : https://en.wikipedia.org/wiki/Cross_product

  • $\begingroup$ That is not the definition I gave. $\endgroup$
    – user986614
    Jul 12, 2023 at 6:28
  • $\begingroup$ I have stated [[ in the Part within Square Brackets like this ]] that "my" Definition will give "your" Definition when we write out the Components ( though that Calculation is simple manipulation or terms ) :: My Definition is not actually mine , it is listed here : en.wikipedia.org/wiki/Cross_product#Definition : In Case you want to use that Definition to Derive your "formula" , you can refer to this Section : en.wikipedia.org/wiki/Cross_product#Computing : It uses the rule of Sarrus to get the Components. $\endgroup$
    – Prem
    Jul 12, 2023 at 8:06
  • $\begingroup$ Added a new Section to my Answer , @user24096 , It might make things a little easier !! $\endgroup$
    – Prem
    Jul 12, 2023 at 8:24

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