$\left[\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right]\,\left[\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right]\,\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right]$

$a_2{b}_3-a_3{b}_2$

$\= a_2{c}_3-a_3{c}_2$

$-a_1{b}_3\+a_3{b}_1$

$\= -a_1{c}_3\+a_3{c}_1$

$a_1{b}_2-a_2{b}_1$

$\= a_1{c}_2-a_2{c}_1$

Initialize

$⟨a_1\,a_2\,a_3⟩$$\stackrel{\text{assign to a name}}{\to }$$A$

$⟨b_1\,b_2\,b_3⟩$$\stackrel{\text{assign to a name}}{\to }$$B$

$⟨c_1\,c_2\,c_3⟩$$\stackrel{\text{assign to a name}}{\to }$$C$

Equate corresponding components of the vectors $\mathbf{A}\times \mathbf{B}$ and $\mathbf{A}\times \mathbf{C}$ 

$\mathbf{A}\times \mathbf{B}\,\mathbf{A}\times \mathbf{C}$

$\stackrel{\text{equate}}{\to }$

$\left[a_2{b}_3-a_3{b}_2\=a_2{c}_3-a_3{c}_2\,-a_1{b}_3\+a_3{b}_1\=-a_1{c}_3\+a_3{c}_1\,a_1{b}_2-a_2{b}_1\=a_1{c}_2-a_2{c}_1\right]$

$\stackrel{\text{solve (specified)}}{\to }$

$\left\{b_2\=\frac{a_1{c}_2\+a_2{b}_1-a_2{c}_1}{a_1}\,b_3\=\frac{a_1{c}_3\+a_3{b}_1-a_3{c}_1}{a_1}\right\}$

$\stackrel{\text{assign to a name}}{\to }$

$S_1$

Obtain the vector B that corresponds to this solution

$\genfrac{}{}{0ex}{}{\mathbf{B}}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{S_1}$ = $$

${\mathbf{B}}_1\=\left[\begin{array}{c}{b}_1\\ \frac{a_2}{a_1}\(b_1-c_1\)\+c_2\\ \frac{a_3}{a_1}\(b_1-c_1\)\+c_3\end{array}\right]$

Cyclically permute the indices in ${\mathbf{B}}_1$ to obtain alternate forms of the vector B

${\mathbf{B}}_2\=\left[\begin{array}{c}\frac{a_1}{a_2}\(b_2-c_2\)\+c_1\\ b_2\\ \frac{a_3}{a_2}\(b_2-c_2\)\+c_3\end{array}\right]$

${\mathbf{B}}_3\=\left[\begin{array}{c}\frac{a_1}{a_3}\(b_3-c_3\)\+c_1\\ \frac{a_2}{a_3}\(b_3-c_3\)\+c_2\\ b_3\end{array}\right]$

Initialize

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{MultivariateCalculus}\right)\:$

$\mathbf{A}\,\mathbf{B}\,\mathbf{C}≔⟨a_1\,a_2\,a_3⟩\,⟨b_1\,b_2\,b_3⟩\,⟨c_1\,c_2\,c_3⟩\:$

Equate corresponding components of the vectors $\mathbf{A}\times \mathbf{B}$ and $\mathbf{A}\times \mathbf{C}$ 

$q≔\mathrm{Equate}\left(\mathrm{CrossProduct}\left(\mathbf{A}\,\mathbf{B}\right)\,\mathrm{CrossProduct}\left(\mathbf{A}\,\mathbf{C}\right)\right)$

$\left[a_2{b}_3-a_3{b}_2\=a_2{c}_3-a_3{c}_2\,-a_1{b}_3\+a_3{b}_1\=-a_1{c}_3\+a_3{c}_1\,a_1{b}_2-a_2{b}_1\=a_1{c}_2-a_2{c}_1\right]$

Solve for B under the assumptions that $a_1\ne 0$, $a_2\ne 0$, or $a_3\ne 0$ 

$$\begin{gathered} S_1≔\mathrm{solve}\left(q\,\left\{b_2\,b_3\right\}\right)\; \\ {S}_2≔\mathrm{solve}\left(q\,\left\{b_1\,b_3\right\}\right)\; \\ {S}_3≔\mathrm{solve}\left(q\,\left\{b_1\,b_2\right\}\right) \end{gathered}$$

$\left\{b_2\=\frac{a_1{c}_2\+a_2{b}_1-a_2{c}_1}{a_1}\,b_3\=\frac{a_1{c}_3\+a_3{b}_1-a_3{c}_1}{a_1}\right\}$

Obtain the vector B that corresponds to each solution

$\mathrm{eval}\left(\mathbf{B}\,S_1\right)$ = $$

${\mathbf{B}}_1\=\left[\begin{array}{c}{b}_1\\ \frac{a_2}{a_1}\(b_1-c_1\)\+c_2\\ \frac{a_3}{a_1}\(b_1-c_1\)\+c_3\end{array}\right]$

$\mathrm{eval}\left(\mathbf{B}\,S_2\right)$ = $$

${\mathbf{B}}_2\=\left[\begin{array}{c}\frac{a_1}{a_2}\(b_2-c_2\)\+c_1\\ b_2\\ \frac{a_3}{a_2}\(b_2-c_2\)\+c_3\end{array}\right]$

$\mathrm{eval}\left(\mathbf{B}\,S_3\right)$ = $$

${\mathbf{B}}_3\=\left[\begin{array}{c}\frac{a_1}{a_3}\(b_3-c_3\)\+c_1\\ \frac{a_2}{a_3}\(b_3-c_3\)\+c_2\\ b_3\end{array}\right]$