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TS EAMCET 19-July-2022 Shift 1 Solved Paper

Section: Mathematics

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Question : 30 of 160

Marks: +1, -0

If

(\alpha, \beta, \gamma)

is a triad of real numbers satisfying

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

\alpha^{2} - \beta^{2} + \gamma^{2} =

23
31
40
-6

Solution: 👈: Video Solution

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

+ \gamma (2 \;\hat{i} - \;\hat{j} + \;\hat{k})

\Rightarrow \;\; 1 = \alpha + \beta + 2 \gamma

......(i)

-2 = \alpha + 2 \beta - \gamma

......(ii)

5 = \alpha + 3 \beta + \gamma

.......(iii)

From Eqs. (i) and (ii), we get

- \beta + 3 \gamma = 3

.....(iv)

From Eqs. (ii) and (iii), we get

\beta + 2 \gamma = 7

......(v)

From Eqs. (iv) and (v), we get

\gamma = 2, \text{ and } \beta = 3

\alpha = -6

\alpha^2 - \beta^2 + \gamma^2 = 36 - 9 + 4 = 31

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