If C is a given non-zero scalar and $\overline{\mathrm{A}}$ and $\overline{\mathrm{B}}$ are given non-zero…

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
  1. $\frac{C \bar{A}+\bar{A} \times \bar{B}}{|\overline{\mathrm{~A}}|^2}$
  2. $\frac{\mathrm{C} \overline{\mathrm{A}} \times \overline{\mathrm{B}}}{|\overline{\mathrm{A}}|^2}$
  3. $\frac{\mathrm{C} \overline{\mathrm{A}}-\overline{\mathrm{A}} \times \overline{\mathrm{B}}}{|\overline{\mathrm{A}}|^2}$
  4. $\frac{C \bar{A}+\bar{B}}{|\overline{\mathrm{~A}}|^2}$

Solution

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

Asked in: MHT CET 2024 (02 May Shift 2)

Practice more Vectors questions on Aicharya