1
$\begingroup$

If I have two vectors $\vec{b}$ and $\vec{v}$, and I know that $$ \vec{b} \times \vec{v} = \vec{c} $$ and $$ \vec{b}\cdot\vec{v} = \lambda $$ can I find the $\vec{v}$ vector in terms of the $\vec{c}$ vector, $\vec{b}$ vector, and $\lambda$? I have been struggling with for quite sometime. And I hope it's time I asked for help.

$\endgroup$
  • $\begingroup$ Welcome to Physics SE, and don't worry about the equations. In general SE uses Mathjax, see physics.stackexchange.com/help/notation. There is also a certain policy on homework questions, meta.physics.stackexchange.com/questions/714/…, meaning it is expected that you show whatever work you have done on the question so far, and what point exactly is unclear. $\endgroup$ – udrv Nov 29 '15 at 3:58
  • $\begingroup$ while this question has received some answers, it is clearly not about mainstream physics (although the equations may be the result of it) and is more suitable on Math.SE. I am voting to close this question. $\endgroup$ – nluigi Nov 29 '15 at 8:01
3
$\begingroup$

Yes. If $\bf B \times V=C$ and ${\bf B \cdot V}=\lambda$, the BAC-CAB rule tells you:

$\bf B\times C=B\times(B\times V)=B(B\cdot V)-V(B\cdot B)$

So

${\bf B\times C}={\bf B}\lambda-{\bf V}B^2$

and

$${\bf V}=\frac{{\bf B}\lambda-{\bf B\times C}}{B^2}$$

$\endgroup$
  • 4
    $\begingroup$ Assuming $\bf B \perp C$, or there is no solution, and $\bf B \neq 0$, or there are infinitely many. $\endgroup$ – Conifold Nov 29 '15 at 4:16
  • $\begingroup$ @Conifold The only constraint is $B^2\neq 0$, as is easily gleaned from the answer. $\endgroup$ – user12029 Sep 7 '16 at 8:30
  • $\begingroup$ No, you also need $\bf B$ and $\bf C$ to be orthogonal, which is necessary to satisfy the first equation, and it does not follow from $\bf B \neq 0$ or the formula. Applying the formula will give $\bf V$ parallel to $\bf B$, implying that $\bf B \times V=0$ rather than $\bf C$. In other words, without this constraint the formula is wrong. $\endgroup$ – Conifold Sep 10 '16 at 21:53
  • $\begingroup$ @Conifold OK, I guess that is fair to emphasize. But the post is 100% correct with no caveats. The op said "I know that ..." and the post starts with "if". $\endgroup$ – user12029 Sep 11 '16 at 1:24
1
$\begingroup$

Not always, but it in almost all cases you can. To find the conditions whereunder it can be done and the inversion formula, reason as below where I give a more "visual" analysis than the other answers. It's also worth mentioning that the consideration of your question leads to one conception of the skew field $\mathbb{H}$ of quaternions (which I'll say more about at the end of the answer).

We assume we know $\vec{A}, \vec{C}=\vec{A}\times\vec{B}$ and $\Delta=\vec{A}\cdot\vec{B}$.

Case 1: $\vec{A}=\vec{0}$:

In this case, if, suposedly $\vec{A}\times\vec{B}\neq\vec{0}$ or $\vec{A}\cdot\vec{B}\neq{0}$ then there can be no vector $\vec{B}$ solving the equations, since $\vec{A}\times$ and $\vec{A}\cdot$ yields the zero vector and naught, respectively. If $\vec{A}\cdot\vec{B}={0}$ and $\vec{A}\cdot\vec{B}=0$ then every vector $\vec{B}$ is clearly a solution.

Case 2: $\vec{A}\neq\vec{0};\,\vec{A}\times\vec{B}=\vec{0}$:

$\vec{A},\,\vec{B}$ must be collinear and $\vec{B}=\lambda\vec{A};\,\lambda\in\mathbb{R}$. Then $\lambda = \vec{A}\cdot\vec{B}/|\vec{A}|^2$ and we are done: $\vec{B}$ has a unique solution and $\lambda$ defines it.

Case 3: $\vec{A}\neq\vec{0};\,\vec{A}\times\vec{B}\neq\vec{0}$ and $(\vec{A}\times\vec{B})\cdot\vec{A}\neq0$:

There is no solution, since there is no vector $\vec{B}$ such that $(\vec{A}\times\vec{B})\cdot\vec{A}\neq0$.

Case 4: $\vec{A}\neq\vec{0};\,\vec{A}\times\vec{B}\neq\vec{0}$ (therefore $\vec{B}\neq\vec{0}$)and $(\vec{A}\times\vec{B})\cdot\vec{A}=0$:

$\vec{B}$ is orthogonal to $\vec{C}\neq\vec{0}$. Also $\vec{A}\times\vec{C}\neq\vec{0}$ (since $\vec{C}\neq\vec{0}$ and $\vec{C}$ is orthogonal to $\vec{A}$), therefore, the plane of all vectors orthogonal to $\vec{C}$ (which contains $\vec{B}$) is spanned by $\vec{A}$ and $\vec{C}\times\vec{A}$. Therefore:

$$\vec{B} = b_1\,\vec{A}+b_2\,\vec{A}\times\vec{C}\tag{1}$$

so now we form $\vec{A}\cdot\vec{B}$ to find:

$$b_1=\frac{\vec{A}\cdot\vec{B}}{|\vec{A}|^2}\tag{2}$$

and form $(\vec{A}\times\vec{C})\cdot\vec{B}=(\vec{A}\times\vec{B})\cdot\vec{C}=|\vec{C}|^2$ (which is known) to find:

$$b_2 = \frac{|\vec{C}|^2}{|\vec{A}\times\vec{C}|^2}\tag{3}$$

and we have a unique solution defined by $b_1,\,b_2$.

This is of course the exact result you get using the cookbook formula, but the above steps visualize the Euclidean geometry more fully and evocatively, in my opinion.


More on Quaternions:

It's worth mentioning that one way to define the quaternions is to consider the set

$$\mathbb{H}=\{(x,\,\vec{X})|\;x\in\mathbb{R},\,\vec{X}\in\mathbb{R}^3\}\tag{3}$$

of ordered pairs of real numbers and three dimensional real vectors and to define multiplication of entities in this set by:

$$(x,\,\vec{X})\cdot(y,\,\vec{Y})\stackrel{def}{=}(x\,y - \vec{X}\cdot\vec{Y},\,x\,\vec{Y}+y\,\vec{X}+\vec{X}\times\vec{Y})\tag{4}$$

It's not hard, albeit a little tedious, to check that the product defined in (4) is associative and that every nonzero element of $\mathbb{H}$ is invertible because, by (4), $(x,\,\vec{X})\cdot(x,\,-\vec{X}) = (x^2+|\vec{X}|^2,\,\vec{0})$, whence, if $x^2+|\vec{X}|^2\neq 0$:

$$(x,\,\vec{X})^{-1} = \frac{1}{x^2+|\vec{X}|^2}\,(x,\,-\vec{X})\tag{5}$$

It is also easy to check that (3) and (4) do indeed define the quaternions and the product group structure for nonzero quaternions. Now your question can be formulated: given $\mathbf{a}=(0,\,\vec{A})$ and $\mathbf{c}=(\Delta,\,\vec{C})$, find $\mathbf{b}$ such that $\mathbf{a}\,\mathbf{b} = (\Delta,\,\vec{C})$. Byt the group property, $\mathbf{b}\in\mathbb{H}$ exists iff $|\vec{A}|\neq0$ and the solution is:

$$\mathbf{b} = \frac{1}{|\vec{A}|^2}\,(0,\,-\vec{A})\,(\Delta,\,\vec{C}) = \frac{1}{|\vec{A}|^2} (\vec{A}\cdot\vec{C},\,-\Delta\,\vec{A} -\vec{A}\times\vec{C})\tag{6}$$

This quaternion is a pure vector only if $\vec{A}\cdot\vec{C}=0$, in which case our solution is given by (6). Thus we readily recover the required solution, as well as neatly understanding all the different cases, through the quaternion multiplicative group laws.

$\endgroup$
0
$\begingroup$

Notice, we have $\vec b\times \vec v=\vec c$ or $ \vec v\times \vec b=-\vec c$ & $\vec b\cdot \vec v= \lambda$ now, $$\vec b\times \vec c=-\vec b\times(\vec v\times \vec b )$$$$\vec b\times \vec c=-\vec v(\vec b\cdot \vec b)+\vec b(\vec b\cdot \vec v)$$ $$|\vec b|^2\vec v=\lambda\vec b-\vec b\times \vec c$$ $$\vec v=\frac{\lambda\vec b-\vec b\times \vec c}{|\vec b|^2}$$ $\ \ \ \forall\ \ \ \ |\vec b|\ne 0$ & $\lambda\vec b\ne \vec b\times \vec c$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.