The vectors $\bar{a}$ and $\bar{b}$ are not perpendicular and $\overline{\mathrm{c}}$ and…

The vectors $\bar{a}$ and $\bar{b}$ are not perpendicular and $\overline{\mathrm{c}}$ and $\overline{\mathrm{d}}$ are two vectors satisfying $\overline{\mathrm{b}} \times \overline{\mathrm{c}}=\overline{\mathrm{b}} \times \overline{\mathrm{d}}$ and $\overline{\mathrm{a}} \cdot \overline{\mathrm{d}}=0$, then the vector $\overline{\mathrm{d}}$ is equal to
  1. $\bar{b}+\left(\frac{\bar{b} \cdot \bar{c}}{\bar{a} \cdot \bar{b}}\right) \bar{c}$
  2. $\overline{\mathrm{c}}-\left(\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}}{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}\right) \overline{\mathrm{b}}$
  3. $\bar{b}-\left(\frac{\bar{b} \cdot \bar{c}}{\bar{a} \cdot \bar{b}}\right) \bar{c}$
  4. $\overline{\mathrm{c}}+\left(\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}}{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}\right) \overline{\mathrm{b}}$

Solution

Given, Vectors $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ are not perpendicular $\begin{aligned} \therefore \quad & \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}} \neq 0 \\ & \overline{\mathrm{a}} \cdot \overline{\mathrm{~d}}=0 \\ & \overline{\mathrm{~b}} \times \overline{\mathrm{c}}=\overline{\mathrm{b}} \times \overline{\mathrm{d}} \\ & \Rightarrow \overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{d}}) \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}) \overline{\mathrm{c}}=(\overline{\mathrm{a}} \cdot \overline{\mathrm{~d}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}) \overline{\mathrm{d}} \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}) \overline{\mathrm{d}}=-(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \cdot \overline{\mathrm{b}}+(\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}) \overline{\mathrm{c}} \\ \quad & \Rightarrow \overline{\mathrm{~d}}=\frac{-(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}}{\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}}+\overline{\mathrm{c}} \\ \Rightarrow & \overline{\mathrm{~d}}=\overline{\mathrm{c}}-\left(\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}}{\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}}\right) \overline{\mathrm{b}} \end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 2)

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