For example, $a \times b = c$

If you only know $a$ and $c$, what method can you use to find $b$?

  • $\begingroup$ It's a set of linear equations. So, you use linear algebra to solve. $\endgroup$ Apr 12, 2011 at 19:13
  • 8
    $\begingroup$ $b$ is not uniquely determined by the knowledge of $a$ and $c$. Do you want to add some additional conditions on $b$ or are you interested in all possible solutions? $\endgroup$
    – Fabian
    Apr 12, 2011 at 19:13
  • 1
    $\begingroup$ Maybe this will help. math.stackexchange.com/questions/4277293/… $\endgroup$
    – Alex Alex
    Oct 15, 2021 at 8:44

5 Answers 5


As Fabian wrote, $b$ is not uniquely determined by $a$ and $c$. Moreover, there is no solution unless $a$ and $c$ are orthogonal. If $a$ and $c$ are orthogonal, then the solutions are $(c \times a)/(a\, . a) + t a$ for arbitrary scalars $t$.

  • 1
    $\begingroup$ To spell this out in words I understand, $b$ lies in the plane orthogonal to $c$ (as must $a$). The component of $b$ orthogonal to $a$ is $|c|/|a|$ in the direction consistent with the right-hand rule. $\endgroup$
    – Henry
    Apr 13, 2011 at 0:12
  • $\begingroup$ To arrive at the result you can use the vector triple product. $\endgroup$
    – Oscillon
    Aug 25, 2021 at 21:16
  • $\begingroup$ Hey, I know this is really old now, but I was wondering if you could actually write out the derivation? Oscillon mentions that one can see the result using a vector triple product, but I’m not sure how to actually do the math to get to that answer. Any help is appreciated in advance. $\endgroup$ Oct 13, 2021 at 16:29

The name "product" for the cross product is unfortunate. It really should not be thought of as a product in the ordinary sense; for example, it is not even associative. Thus one should not expect it to have properties analogous to the properties of ordinary multiplication.

What the cross product really is is a Lie bracket.

  • 2
    $\begingroup$ Anything called a "product" has to be associative? Then what should we call the Cartesian product of sets, seeing as $(A\times B)\times C\ne A\times(B\times C)$? To say nothing of the inner product of vectors, another misnomer. $\endgroup$
    – bof
    Oct 4, 2014 at 5:59

Alright, let's look for an inverse here:

The cross product is restricted to $\mathbb{R}^3$ (assuming reals--but that isn't really a constraint here, I don't think). So we have:

$$ \langle x_1, y_1, z_1\rangle\times\langle x_2, y_2, z_2\rangle = \langle y_1z_2 - z_1y_2, z_1x_2 - x_1z_2, x_1y_2 - y_1x_2\rangle $$

Then we have:

\begin{align} \langle x_3, y_3, z_3\rangle\times\left(\langle x_1, y_1, z_1\rangle\times\langle x_2, y_2, z_2\rangle\right) = & \langle y_3(x_1y_2 - y_1x_2) - z_3(z_1x_2 - x_1z_2),\\ &z_3(y_1z_2 - z_1y_2) - x_3(x_1y_2 - y_1x_2),\\ & x_3(z_1x_2 - x_1z_2) - y_3(y_1z_2 - z_1y_2)\rangle \end{align}

Now this looks very complicated (and it is--if you need to solve it "as is"). Instead we can imagine that we already know $\vec{n}' = \vec{v}_1\times\vec{v_2}$. Then this becomes:

\begin{align} \langle x_3, y_3, z_3\rangle\times\vec{n}' = & \langle y_3n'_z - z_3n'_y, z_3n'_x - x_3n'_z, x_3n'_y - y_3n'_x\rangle \end{align}

Now we are trying to find $\langle x_3, y_3, z_3\rangle$ which is the inverse. So we set this equal to the second argument of the original cross product, we have a set of linear equations for three unknowns ($x_3, y_3, z_3$):

$$ x_2 = y_3n'_z - z_3n'_y \\ y_2 = z_3n'_x - x_3n'_z \\ z_2 = x_3n'_y - y_3n'_x $$

The matrix gives:

$$ \begin{pmatrix} 0 & n_z' & -n_y' \\ -n_z' & 0 &n_x' \\ n_y' & -n_x' & 0 \end{pmatrix} \times \begin{pmatrix}x_3 \\ y_3 \\ z_3\end{pmatrix} = \begin{pmatrix}x_2 \\ y_2 \\ z_2\end{pmatrix} $$

There is no solution when the determinate of this matrix is zero--which is always the case: $n_z'x_x'n_y'- n_y'n_z'n_x' = 0$. This means that unless $\langle x_2, y_2, z_2\rangle$ is $\vec{0}$ there is no solution (since a matrix that has a determinant equal to zero only has a solution when the RHS is zero--in which case it has infinite solutions).

Edit (in regards to comments)

While it's not strictly true that just because the RHS is non-zero (and the matrix is degenerate) there won't be a solution, it is true "in general". Meaning that you can only find solutions in very special cases. This still suggests that there is not an inverse to the cross product (except in very special cases).

  • $\begingroup$ A matrix equation $A\vec x=\vec b$, with $\det A=0$, can have solutions with $\vec b\neq\vec 0$ . Example: $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix}$ , solutions $x=2$, $y\in\mathbb R$ . $\endgroup$
    – mr_e_man
    May 7, 2018 at 6:41
  • $\begingroup$ @mr_e_man Touche. That's an interesting observation. I'm struggling to find generalizations for this case though. I mean clearly solutions exist when the the RHS is in the image of the matrix transformation (in this case that's any vector $\langle x', 0\rangle$). $\endgroup$
    – Jared
    May 8, 2018 at 2:36

By the definition of $A\times B=C$, the angle $(A,C)=(B,C)=90$ degrees. Assume that $\vec{A}$ and $\vec{C}$ are known. We know that $\vec{B}$ lies in the plane with normal vector $\vec{C}$, which $\vec{A}$ also lies in. Now, due to this, and the fact that $\vec{A}$, $\vec{B}$ and $\vec{C}$ is a right-hand system; the conditions for $\vec{B}$ to satisfy $\vec{A}\times \vec{B}=\vec{C}$ is: 1. The length of $\vec{B}$ and the positive angle $Y$ between $\vec{A}$ and $\vec{B}$ satisfy $|\vec{B}|sin(Y)=\dfrac{|\vec{C}|}{|\vec{A}|}$. 2. The angle $Y$ is positive and between 0 and $\pi$ (radians). The condition for $Y$ comes from how we define a right-hand system. Thus, there are infinitely many solutions if nothing else is specified.

If however $|\vec{B}|$ is known, then there will be only two possibilities for Y: $Y_1$ and $Y_2$, where $sin(Y_1)=sin(Y_2)=\dfrac{|\vec{C}|}{(|\vec{A}|*|\vec{B}|)}$, and $Y_2=\pi - Y_1$. This comes from the fact that $sin(v)=sin(\pi-v)$ for any $v$.

One can reason the same way to solve for $\vec{A}$ if only $\vec{B}$ and $\vec{C}$ are known.

  • $\begingroup$ Btw +1 here :) but you can learn Mathjax and improve your answer. $\endgroup$
    – Fawad
    Feb 14, 2017 at 15:02
  • $\begingroup$ I will, but i haven't gotten to that yet 😄 $\endgroup$
    – user357524
    Feb 14, 2017 at 15:05

Okay, so if we look at the geometric interpretation of the cross product it should be pretty intuitive why it is not invertible. All the answers already given are good, but I hope to give a simpler explanation.

Motivation of dot and cross products

As we know, there are two widely used operations that resemble multiplication with vectors.

There is the dot product (scalar) which is used when we only care about parallel components of the vectors — for example when calculating work done $W = \mathbf{\vec F} \cdot \mathbf{\vec s}$, we only care about how far an object is moved in the direction of the force.

Then there is the cross product (vector), which is used when we only care about perpendicular components of the vectors — for example when calculating torque on a door being opened $\vec{\boldsymbol \tau} = \mathbf{\vec F} \times \mathbf{\vec r}$, we only care about the component of the force applied perpendicular to the door. As a bonus, the cross product tells us the plane containing the two vectors in the form of its normal.

Why the cross product is non-invertible

Since we can interpret the cross product $\mathbf{\vec u} = \mathbf{\vec v} \times \mathbf{\vec w}$ as the product of the magnitudes of $\mathbf{\vec v}$ and the perpendicular component of $\mathbf{\vec w}$, the cross product only "cares" about the perpendicular component of $\mathbf{\vec w}$ and therefore disregards the component that is parallel to $\mathbf{\vec v}$. Therefore, we can change the parallel component of $\mathbf{\vec w}$ however we please without affecting the final cross-product $\mathbf{\vec u}$. Hence, in most cases, there exist infinite vectors $\mathbf{\vec w}$ along the line parallel to $\mathbf{\vec v}$ which will yield the same result $\mathbf{\vec u}$.

Note that a very similar argument can be made with the dot product disregarding the perpendicular component.


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