Let $\overline{\mathrm{A}}=2 \hat{i}+\hat{k}, \overline{\mathrm{~B}}=\hat{i}+\hat{j}+\hat{k}$ and…

Let $\overline{\mathrm{A}}=2 \hat{i}+\hat{k}, \overline{\mathrm{~B}}=\hat{i}+\hat{j}+\hat{k}$ and $\overline{\mathrm{C}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$. If a vector $\bar{R}$ satisfies $\bar{R} \times \bar{B}=\bar{C} \times \bar{B}$ and $\overline{\mathrm{R}} \cdot \overline{\mathrm{A}}=0$, then $\overline{\mathrm{R}}$ is given by
  1. $\hat{\mathrm{i}}-8 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
  2. $\hat{i}+8 \hat{j}+2 \hat{k}$
  3. $-\hat{\mathrm{i}}-8 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
  4. $\quad-\hat{i}-8 \hat{j}-2 \hat{k}$

Solution

$\begin{aligned} & \overline{\mathrm{R}} \times \overline{\mathrm{B}}=\overline{\mathrm{C}} \times \overline{\mathrm{B}} \\ & \Rightarrow \overline{\mathrm{A}} \times(\overline{\mathrm{R}} \times \overline{\mathrm{B}})=\overline{\mathrm{A}} \times(\overline{\mathrm{C}} \times \overline{\mathrm{B}}) \\ & \Rightarrow(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}) \overline{\mathrm{R}}-(\overline{\mathrm{A}} \cdot \overline{\mathrm{R}}) \overline{\mathrm{B}}=(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}) \overline{\mathrm{C}}-(\overline{\mathrm{A}} \cdot \overline{\mathrm{C}}) \overline{\mathrm{B}} \\ & \Rightarrow(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}) \overline{\mathrm{R}}-0=(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}) \overline{\mathrm{C}}-(\overline{\mathrm{A}} \cdot \overline{\mathrm{C}}) \overline{\mathrm{B}}\end{aligned}$ $\begin{aligned} & \Rightarrow \overline{\mathrm{R}}=\overline{\mathrm{C}}-\left(\frac{\overline{\mathrm{A}} \cdot \overline{\mathrm{C}}}{\overline{\mathrm{~A}} \cdot \overline{\mathrm{~B}}}\right) \overline{\mathrm{B}} \\ & \overline{\mathrm{~A}} \cdot \overline{\mathrm{C}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}) \\ & =2(4)+0(-3)+1(7) \\ & =15 \\ & \overline{\mathrm{~A}} \cdot \overline{\mathrm{~B}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & =2(1)+0(1)+1(1) \\ & =3 \\ & \therefore \quad \overline{\mathrm{R}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}-\left(\frac{15}{3}\right)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & =4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}-5(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & =-\hat{i}-8 \hat{j}+2 \hat{k} \end{aligned}$

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

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