If $\alpha , \beta$ are roots of the equation $x ^ { 2 } + 2 x + 5 = 0$ and $\overline { a } = ( \alpha + \beta ) \overline { i } + \alpha \beta \overline { j }$ ,$\overline { b } = \alpha \beta \overline { i } + ( \alpha + \beta ) \overline { j } + \left( \alpha ^ { 2 } + \beta ^ { 2 } \right) \overline { k }$ than $\overline { a } \times \overline { b } = $
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$\overline { i } + 12 \overline { j } + 12 \overline { k }$
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$- 30 \vec { i } + 12 \vec { j } - 5 \vec { k }$
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$- 30 \overline { i } - 12 \overline { j } - 21 \overline { k }$
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$\overline { i } - 12 \overline { j } + 29 \overline { k }$
C
Correct answer
Explanation
x^2+2x+5=0. alpha+beta = -2, alpha*beta = 5. alpha^2+beta^2 = (alpha+beta)^2 - 2*alpha*beta = 4 - 10 = -6. a = -2i + 5j. b = 5i - 2j - 6k. a x b = (-2i + 5j + 0k) x (5i - 2j - 6k). Determinant: i(5*-6 - 0) - j(-2*-6 - 0) + k(-2*-2 - 5*5) = -30i - 12j - 21k.