$\overline{\mathrm{r}}=-4 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ is a linear combination of…

$\overline{\mathrm{r}}=-4 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ is a linear combination of the vector $\bar{a}=-\hat{i}-4 \hat{j}+3 \hat{k}$ and $\bar{b}=-8 \hat{i}-\hat{j}+3 \hat{k}$, then
  1. $\overline{\mathrm{r}}=\frac{-4}{3} \overline{\mathrm{a}}+\frac{2}{3} \overline{\mathrm{b}}$
  2. $\overline{\mathrm{r}}=\frac{4}{3} \overline{\mathrm{a}}+\frac{2}{3} \overline{\mathrm{b}}$
  3. $\overline{\mathrm{r}}=\frac{-1}{3} \overline{\mathrm{a}}+\frac{2}{3} \overline{\mathrm{b}}$
  4. $\overline{\mathrm{r}}=\frac{1}{3} \overline{\mathrm{a}}-\frac{1}{3} \overline{\mathrm{b}}$

Solution

Let $r=x \bar{a}+y \bar{b}$ $\begin{aligned} & \therefore-4 \hat{i}-6 \hat{j}-2 \hat{k}=x(-\hat{i}+4 \hat{j}+3 \hat{k})+y(-8 \hat{i}-\hat{j}+3 \hat{k}) \\ & =(-x-8 y) \hat{i}+(4 x-y) \hat{j}+(3 x+3 y) \hat{k} \\ & \therefore \quad-x-8 y \quad \ldots(1) \\ & 4 x-y \quad=-4 \quad \ldots(2) \\ & 3 x+3 y \quad=-2 \quad \ldots(3) \end{aligned}$ Solving (1) and (2), we get $\mathrm{y}=\frac{2}{3}$ and $\mathrm{x}=\frac{-4}{3}$

Asked in: MHT CET 2021 (20 Sep Shift 2)

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