If C is a given non-zero scalar and \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{B}}\) are given non-zero vectors such that \(\overline{\mathrm{A}}\), is perpendicular to \(\overline{\mathrm{B}}\). If vector \(\overline{\mathrm{X}}\) is such that \(\overline{\mathrm{A}} \cdot \overline{\mathrm{X}}=\mathrm{C}\) and \(\overline{\mathrm{A}} \times \overline{\mathrm{X}}=\overline{\mathrm{B}}\) then \(\overline{\mathrm{X}}\) is given by

\(\begin{aligned} & \overline{\mathrm{A}} \times \overline{\mathrm{X}}=\overline{\mathrm{B}} \\ & \Rightarrow \overline{\mathrm{A}} \times(\overline{\mathrm{A}} \times \overline{\mathrm{X}})=\overline{\mathrm{A}} \times \overline{\mathrm{B}} \\ & \Rightarrow(\overline{\mathrm{A}} \cdot \overline{\mathrm{X}}) \overline{\mathrm{A}}-(\overline{\mathrm{A}} \cdot \overline{\mathrm{A}}) \overline{\mathrm{X}}=\overline{\mathrm{A}} \times \overline{\mathrm{B}} \\ & \Rightarrow \mathrm{CA}-|\overline{\mathrm{A}}|^2 \overline{\mathrm{X}}=\overline{\mathrm{A}} \times \overline{\mathrm{B}} \quad \ldots[\because \overline{\mathrm{A}} \cdot \overline{\mathrm{X}}=\mathrm{C}] \\ & \Rightarrow \overline{\mathrm{X}}=\frac{\mathrm{CA}-\overline{\mathrm{A}} \times \overline{\mathrm{B}}}{|\overline{\mathrm{A}}|^2}\end{aligned}\)