35.3: Inverse

Suppose we have vectors $\mathbf{A}, \mathbf{B}$, and $\mathbf{C}$ such that $\mathbf{A} \times \mathbf{B}=\mathbf{C}$. If vectors $\mathbf{B}$ and $\mathbf{C}$ are known, can we solve for vector $\mathbf{A}$ ?

There is no such thing as a "cross division" operation, so we can't do anything similar to $A=C / B$. In fact, there is no unique solution for vector $\mathbf{A}$. There are an infinite number of vectors that can be crossed with $\mathbf{B}$ to yield vector $\mathbf{C}$; the smaller the angle between $\mathbf{A}$ and $\mathbf{B}$, the larger the magnitude $\mathbf{A}$ must have to yield a given vector $\mathbf{C}$.

To solve $\mathbf{A} \times \mathbf{B}=\mathbf{C}$ for vector $\mathbf{A}$, we will need to know vectors $\mathbf{B}$ and $\mathbf{C}$, along with one other piece of information, such as the magnitude of vector $\mathbf{A}$ or the angle $\theta$ between $\mathbf{A}$ and $\mathbf{B}$. Suppose the magnitude $A$ of vector $\mathbf{A}$ is known; then since $|\mathbf{A} \times \mathbf{B}|=A B \sin \theta=C$, we have

$$\sin \theta=\frac{C}{A B}$$

On the other hand, if $\theta$ is known, then

$$A=\frac{C}{B \sin \theta} .$$

In either case, we now know both the magnitude $A$ and the angle $\theta$. Then since $\mathbf{A} \cdot \mathbf{B}=A B \cos \theta$, we can now find the dot product $\mathbf{A} \cdot \mathbf{B}$. Now let's take

$$\mathbf{A} \times \mathbf{B}=\mathbf{C} .$$

Crossing both sides on the right with vector $\mathbf{B}$, we get

$$(\mathbf{A} \times \mathbf{B}) \times \mathbf{B}=\mathbf{C} \times \mathbf{B} .$$

The left-hand side is a vector triple product; applying Eq. (32.3.8), we get

$$\mathbf{B}(\mathbf{A} \cdot \mathbf{B})-B^{2} \mathbf{A}=\mathbf{C} \times \mathbf{B}$$

Solving for vector $\mathbf{A}$, we find

$$\mathbf{A}=\frac{1}{B^{2}}[(\mathbf{B} \times \mathbf{C})+(\mathbf{A} \cdot \mathbf{B}) \mathbf{B}]$$

So if either $A$ or $\theta$ is known, then we can find $\mathbf{A} \cdot \mathbf{B}=A B \cos \theta$; knowing this and $\mathbf{B}$ and $\mathbf{C}$, Eq. (32.25) lets us solve for vector $\mathbf{A}$ (provided $\mathbf{B}, \mathbf{C} \neq \mathbf{0}$ ).