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TS EAMCET 20-July-2022 Shift 2 Solved Paper

Section: Mathematics

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Question : 30 of 160

Marks: +1, -0

If

\alpha, \beta, \gamma

are real numbers such that

\left(\frac{7}{3}+\beta\right) \overline{i}-\overline{j}+(\alpha+\gamma) \overline{k}=\frac{5}{3}(\alpha \overline{i}+\overline{j}-\overline{k})

+\beta(2 \overline{j}+\overline{k})+(\overline{i}+\gamma \overline{j}+3 \overline{k})

5\alpha-9\beta+13\gamma=

4
12
0
15

Solution:

\left(\frac{7}{3}+\beta\right) \hat{i}-\hat{j}+(\alpha+\gamma) \hat{k}

=\frac{5}{3}(\alpha \hat{i}+\hat{j}-\hat{k})+\beta(2\hat{j}+\hat{k})+(\hat{i}+\gamma \hat{j}+3\hat{k})

\Rightarrow (7+3\beta)\hat{i}-3\hat{j}+(3\alpha+3\gamma)\hat{k}

=5(\alpha \hat{i}+\hat{j}-\hat{k})+\beta(6\hat{j}+3\hat{k})+(3\hat{i}+3\gamma\hat{j}+9\hat{k})

\Rightarrow (7+3\beta)\hat{i}-3\hat{j}+(3\alpha+3\gamma)\hat{k}

=(5\alpha+3)\hat{i}+(5+6\beta+3\gamma)\hat{j}+(-5+3\beta+9)\hat{k}

On comparing corresponding components,

7+3\beta=5\alpha+3

\Rightarrow 3\beta=5\alpha-4

....(i)

-3=5+6\beta+3\gamma

\Rightarrow 3\gamma=-8-6\beta

\Rightarrow 3\gamma=-10\alpha

.....(ii)

3\alpha+3\gamma=-5+3\beta+9

3\alpha+3\gamma=-5+3\beta+9

\Rightarrow 3\alpha=4+3\beta-3\gamma

....(iii)

On putting the values from Eqs. (i) and (ii) into Eq. (iii), we get

3\alpha=4+(5\alpha-4)-(-10\alpha)

\Rightarrow 3\alpha=5\alpha+10\alpha

\Rightarrow \alpha=0

From Eq. (i),

3\beta=0-4

\Rightarrow \beta=\frac{-4}{3}

From Eq. (ii),

3\gamma=0

\Rightarrow \gamma=0

\therefore 5\alpha-9\beta+13\gamma=0-9\left(\frac{-4}{3}\right)+0=12

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