# What is the direction of dot product and cross product of vector A and B? [closed]

What is the direction of the dot product and the cross product of vectors A and B?

• No better, concise answer than what Red Act has provided! – Swami Sep 26 '14 at 6:09

## 4 Answers

The dot product of two vectors doesn't have a direction, because it's a scalar (single number), not a vector.

The cross product $A\times B$ is perpendicular to both $A$ and $B$. If on your right hand, your thumb is sticking up and pointing in the direction of $A$, and your index finger is pointing in the direction of $B$, then your palm is facing in the direction of $A\times B$

The result of a dot product is, as Red Act said, a scalar, it must not have a direction. The result of a cross product must be a vector. It must have a direction. Curl your right hand fingers - moving them from $A$ towards $B$ via the shorter angle, your thumb gives the direction of the result of the cross product.

As with the other answers by Swami and Red ACt, the scalar product has no direction, the vector product is orthogonal to the plane defined by the two vectors (and nought if the two vectors are colinear).

All this seems a bit arbitrary at first meeting: you're clearly a bit bewildered by this and this is quite normal. So, although the following is a bit further to your question's requirements, I think it may help: it would certainly have helped me when I first came across these ideas.

The scalar product is a scalar and is precisely the further information, other than the cross product itself, that you need to "invert" the cross product

The scalar and vector product together are working parts of a more intuitive whole: an invertible vector product. Suppose we have vectors $X, \,Y$, then the two products are the scalar $z=X\cdot Y$ and the vector $Z=X\times Y$. Now, suppose we know $z$, $Z$ and $X$: we can reconstruct from this information $Y$ as follows: we have the identity $X\times(X\times Y)=X\cdot Y X - |X|^2 Y$ so that

$$Y = \frac{1}{|X|^2}(z X - X\times Z)\tag{1}$$

i.e. we have retrieved $Y$, and we can do this as long as $X$ is nonzero so that $|X|^2\neq0$. The vector product $X\times Y$ is unchanged by the addition of any scalar multiple of $Y$ to $X$, i.e. $X\times Y = (X + \alpha\,Y)\times Y$. So, once again, we see why:

The scalar product is a scalar and is precisely the further information you need to "invert" the cross product

This futher information is $\alpha$ in the expression $X\times Y = (X + \alpha\,Y)\times Y$, which the vector product annihilates. So the scalar product's $X \cdot Y$ role is to fetch the component of $X$ in the direction of $Y$, i.e. the part annihilated by the vector product. If instead of 3 dimensional vectors, we think about four dimensional entities of the form $\chi = (x,\,X)$ where $x$ is a scalar and $X$ a three dimensional vector, we can define a new, generalised product on these entities using the scalar and vector product:

$$(x,\,X) \, (y,\,Y) \stackrel{def}{=} (x\,y - X\cdot Y,\,x\,Y + y\,X + X\times Y)$$

then, with a bit of fiddling, you can show that the inverse of the entity $(x,\,X)$ is indeed:

$$(x,\,X)^{-1} = \frac{1}{x^2+|X|^2}\,(x,\,-X)\tag{3}$$

and so the inverse exists and is unique as long as the "length" $x^2 + |X|^2$ of the entity $(x,\,X)$ is not nought. What you've now defined is in fact the skew field of quaternions and historically these were the first entities to be used for 3D spatial vector analysis before Oliver Heaviside and Josiah Gibbs "pruned" them back to the separate scalar and vector product. Sometimes calculations are easier with quaternions (I certainly find the angular momentum of a falling cat easier to think about, as in my article here) or, more generally in Clifford Algebras, but it sounds as though it might be some time before you meet those. The case of "inverting the scalar and vector product" given in (1) is a special case of (2) and (3) when there are no scalar parts to the entities, i.e. when we deal with "pure vectors" (a.k.a. pure imaginary quaternions) of the form $(0,\,X)$ and $(0,\,Y)$.

To complete the given answers you have to keep in mind that the number of the cross product is the multiplication of the numbers of thus vectors and the sinus of the angle between them. If the angle is zero than there does not exist a direction of the cross product.