# Visualisation of dot/cross product

If this question is a duplicate / the answer already exists I am very sorry.
I recently started working with vectors in 3 dimensions. My visualisation of the dot product is the following:
Lets say we have the vector v(0,0,1) with the origin O. The dot product between v and the vector with the origin in O and end in P is: positive if P is situated anywhere "above" O and negative otherwise. It's like O would be the centre of a sphere and P would be situated in the upper/lower hemisphere. The radius is infinite. It surely sounds pretty stupid to smarter people but this is how I see it.
Could you point me so something similar regarding the cross product? Thank you very much!

• cross product creates a vector perpendicular to the original two: $\vec{a} \times \vec{b} = \vec{c}$ such that $\vec{c} ⟂ \vec{a}$ and $\vec{c} ⟂ \vec{b}$. – Christopher Marley Jun 2 '18 at 19:17
• @VictorBurlacu Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – gimusi Aug 3 '18 at 21:52

The dot product is the product of the length of one vector and that of the projection of the other onto it, taken algebraically (i.e. negative in the opposite direction). Hence your intuition is correct.

The cross product is a vector with is orthogonal to both input vectors and with length equal to the product of the lengths, times the sine of the angle they form. This is quite coherent because when the two vectors are parallel, the cross product is null, avoiding any indeterminacy in the direction of the "common perpendicular". The "sign" of this vector is conventionally defined by the right-hand rule.

If one of the vectors is unit, you have the following decomposition of a vector in parallel and orthogonal components:

$$\vec p= (\vec p\cdot\vec u)\,\vec u+\vec(p\times \vec u)\times\vec u.$$

Cross product between two vectors

$$\vec A=(x_1,x_2,x_3)\;,\;\;\;\vec B=(y_1,y_2,y_3)$$

is defined as

$$\vec C=\vec A\times \vec B:=\begin{vmatrix}e_1&e_2&e_3\\x_1&x_2&x_3\\y_1&y_2&y_3\end{vmatrix}=(x_2y_3-x_3y_2\,,\,x_3y_1-x_1y_3\,,\,x_1y_2-x_2y_1)$$

and it turns out to be a vector $\vec C$ orthogonal to the plane spanned by $\vec A$ and $\vec B$.

The direction of $\vec C$ is given by the right hand rule.

Lets say we have the vector v(0,0,1) with the origin O. The dot product between v and the vector with the origin in O and end in P is: positive if P is situated anywhere "above" O and negative otherwise. It's like O would be the centre of a sphere and P would be situated in the upper/lower hemisphere. The radius is infinite. It surely sounds pretty stupid to smarter people but this is how I see it.

$$v \cdot OP = \lVert v \rVert \lVert OP \rVert \cos \alpha = \lVert OP \rVert \cos \alpha$$ So indeed the scalar product is positive for $\alpha \in (-\pi/2, \pi/2)$, which is indeed the case if $P$ is in the upper half-plane ($z>0$).

But you leave out the magnitude, which is the length of $OP$ projected in $v$-direction (which is the $z$-direction).

Could you point me so something similar regarding the cross product?

We can add $v \times OP = OQ$ to the scene:

(Large versions here and here)

You can fiddle with the scene here: link

The most striking feature of the vector product is, that $OQ$ is both orthogonal to $v$ and $OP$, it is a normal of the plane spanned by $v$ and $OP$ (rendered in turquoise).

In physics and engineering the vector product is used to describe rotational and it shows up in conjunction with electromagnetic fields.