Dot Product and Cross Product: Solving a set of Simultaneous Vector Equations

Question

I was wondering if $\mathbf{A}\times\mathbf{C}=\mathbf{B}$ and $\mathbf{A}\cdot{\mathbf{C}}= p$, is it possible to express C in terms of A, B, and p?

Note: A, B, C are vectors and p is a scalar.

Sorry if this question has been asked before. If so, can anyone give me a link to it?

Answer 1

From Grassmann's “bac min cab”, see https://en.wikipedia.org/wiki/Triple_product#Vector_triple_product, we know $$a\times (a\times c)=a\langle a,c\rangle-c\langle a,a\rangle,$$ hence $$c=\frac{ap-a\times b}{\|a\|^2}.$$

Answer 2

There are two cases to consider. First assume $b \neq 0$. Then since for any three vectors $v_1, v_2, v_3$, we have the identities $$ v_1 \times (v_2 \times v_3) = v_2 \times (v_3 \times v_1) = v_3 \times (v_1 \times v_2), $$ it follows that $$ c \cdot (b \times a) = b \cdot (a \times c) = b \cdot b = \| b \|^2. $$ We have $c \cdot a = p$, and $$ c \cdot b = c \cdot (a \times c) = a \cdot (c \times c) = a \cdot 0 = 0. $$ Now since $\{ b \times a, a, b \}$ is an orthogonal set ($a \times c = b$ implies that $a$ and $b$ are orthogonal, and clearly $b \times a$ is orthogonal to both $a$ and $b$), we can write $$ c = \frac{a \cdot c}{a \cdot a} a + \frac{b \cdot c}{b \cdot b} b + \frac{(b \times a) \cdot c}{(b \times a) \cdot (b \times a)} b \times a = \frac{p}{\|a\|^2}a + \frac{\|b^2\|}{\|b \times a \|^2} b \times a. $$ If you recall that $\|b \times a \| = \| a \| \|b \| \sin \theta$ (where $\theta$ is some geometric angle I do not want to explain here), since $a$ and $b$ are orthogonal $\sin \theta = 1$, and so $$ c = \frac{p a + b \times a}{\| a \|^2}. $$ If $b = 0$, then $a \times c = 0$ and $a \cdot c = p$ implies that $$ a \times \left( c - \frac{pa}{\| a\|^2} \right) = 0, \qquad a \cdot \left( c - \frac{pa}{\| a\|^2} \right) = 0 $$ and so since $a$ is a non-zero vector that is both orthogonal and parallel to $c - \frac{pa}{\| a \|^2}$, we must have $c - \frac{pa}{\|a\|^2} = 0$, i.e. $c = \frac{pa}{\|a\|^2}$. Note that this coincides with the formula coming from the $b \neq 0$ case because $b \times a = 0$ if $b = 0$.

Therefore, $$ c = \frac 1{\|a\|^2} (p a + b \times a). $$ Note that if $a = 0$, your equation gives absolutely no information about $c$, so it is a necessary condition that $a \neq 0$.

Hope that helps,