If $\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{a}} \cdot…

If $\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=1$ and $\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$, then $\overline{\mathrm{b}}$ is
  1. $\hat{i}-\hat{j}+\hat{k}$
  2. $2 \hat{j}-\hat{k}$
  3. $\hat{\mathrm{i}}$
  4. $2 \hat{\mathrm{i}}$

Solution

Given, $\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$ $\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}=1$
Let $\overline{\mathrm{b}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}$ $\begin{aligned} & \overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & 1 \\ x & y & \mathrm{z} \end{array}\right| \\ & \Rightarrow \hat{\mathrm{j}}-\hat{\mathrm{k}}=(\mathrm{z}-y) \hat{\mathrm{i}}-\hat{\mathrm{j}}(\mathrm{z}-x)+\hat{\mathrm{k}}(y-x) \\ & \Rightarrow \mathrm{z}-y=0 ...(i)\\ & \mathrm{z}-x=-1...(ii) \end{aligned}$ $\begin{array}{ll} & y-x=-1 ...(iii)\\ \therefore \quad & \text { Also, } \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}=1 \\ \therefore \quad & x+y+\mathrm{z}=1 ...(iv)\end{array}$
Solving (i), (ii), (iii) and (iv), we get $\begin{aligned} & x=1, y=0, \mathrm{z}=0 \\ \therefore \quad & \overline{\mathrm{~b}}=\hat{\mathrm{i}} \end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 1)

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