Q70 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics
If $\alpha, \beta, \gamma$ are real numbers such that $\left(\frac{7}{3}+\beta\right) \bar{i}-\bar{j}+(\alpha+\gamma) \bar{k}=\frac{5}{3}(\alpha \bar{i}+\bar{j}-\bar{k})+\beta(2 \bar{j}+\bar{k})+(\bar{i}+\gamma \bar{j}+3 \bar{k}), \text { then } 5 \alpha-9 \beta+13 \gamma=$
- A4
- Correct answerB12
- C0
- D15
Solution
$\begin{aligned}<br />& \therefore\left(\frac{7}{3}+\beta\right) \hat{\mathrm{i}}-\hat{\mathrm{j}}+(\alpha+\gamma) \hat{\mathrm{k}}=\frac{5}{3}(\alpha \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\<br />& +\mathrm{B}(2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+(\hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\<br />& \Rightarrow\left(\frac{7}{3}+\beta\right) \hat{\mathrm{i}}-\hat{\mathrm{j}}+(\alpha+\gamma) \hat{\mathrm{k}}=\left(\frac{5}{3} \alpha+1\right) \hat{\mathrm{i}} \\<br />& +\left(\frac{5}{3}+2 \beta+\gamma\right) \hat{\mathrm{j}}+\left(\frac{-5}{3}+\beta+3\right) \hat{\mathrm{k}}<br />\end{aligned}$
comparing both sides, we get
$\begin{aligned}<br />& \frac{7}{3}+\beta=\frac{5}{3} \alpha+1 \Rightarrow 5 \alpha-3 \beta=4 \\<br />& -1=\frac{5}{3}+2 \beta+\gamma \Rightarrow 2 \beta+\gamma=\frac{-8}{3} \\<br />& \alpha+\gamma=\frac{-5}{3}+\beta+3 \Rightarrow \alpha-\beta+\gamma=\frac{4}{3}<br />\end{aligned}$
solving equations (i), (ii), and (iii), we get
$\begin{aligned}<br />& \alpha=0, \beta=\frac{-4}{3}, \gamma=0 \\<br />& \therefore 5 \alpha-9 \beta+13 \gamma=12<br />\end{aligned}$
comparing both sides, we get
$\begin{aligned}<br />& \frac{7}{3}+\beta=\frac{5}{3} \alpha+1 \Rightarrow 5 \alpha-3 \beta=4 \\<br />& -1=\frac{5}{3}+2 \beta+\gamma \Rightarrow 2 \beta+\gamma=\frac{-8}{3} \\<br />& \alpha+\gamma=\frac{-5}{3}+\beta+3 \Rightarrow \alpha-\beta+\gamma=\frac{4}{3}<br />\end{aligned}$
solving equations (i), (ii), and (iii), we get
$\begin{aligned}<br />& \alpha=0, \beta=\frac{-4}{3}, \gamma=0 \\<br />& \therefore 5 \alpha-9 \beta+13 \gamma=12<br />\end{aligned}$
Hence, the correct answer is option: 'B'.