If $\overline{\mathrm{a}}=\mathrm{m} \overline{\mathrm{b}}+\mathrm{n} \overline{\mathrm{c}}$, where…

If $\overline{\mathrm{a}}=\mathrm{m} \overline{\mathrm{b}}+\mathrm{n} \overline{\mathrm{c}}$, where $\overline{\mathrm{a}}=4 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}-18 \hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$, $\overline{\mathrm{c}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$, then $\mathrm{m}+\mathrm{n}=$
  1. $1$
  2. $2$
  3. $3$
  4. $-1$

Solution

Given: $\begin{aligned} & \overline{\mathrm{a}}=4 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}-18 \hat{\mathrm{k}} \\ & \overline{\mathrm{b}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overline{\mathrm{c}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}} \end{aligned}$ Also, $\bar{a}=m \bar{b}+n \bar{c}$ $\begin{aligned} & \Rightarrow 4 \hat{i}+13 \hat{j}-18 \hat{k}=m(\hat{i}-2 \hat{j}+3 \hat{k})+n(2 \hat{i}+3 \hat{j}-4 \hat{k}) \\ & \Rightarrow 4 \hat{i}+13 \hat{j}-18 \hat{k} \\ & =(m+2 n) \hat{i}+(-2 m+3 n) \hat{j}+(3 m-4 n) \hat{k} \end{aligned}$ Comparing, we get $m+2 n=4 \text { and }-2 m+3 n=13$ Solving above equations, we get $\begin{aligned} & \mathrm{m}=-2 \text { and } \mathrm{n}=3 \\ \therefore \quad & \mathrm{m}+\mathrm{n}=-2+3=1 \end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 2)

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