If $\alpha,\ \beta$ are roots of the equation $x^{2}+2x+5=0$ and $\vec{a}=(\alpha+\beta)\hat{i}+\alpha\beta\hat{j} \vec{b}=\alpha\beta\hat{i}+(\alpha+\beta)\hat{j}+(\alpha^{2}+\beta^{2})\hat{k}$ then $\vec{a}\times\vec{b}=$
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$\hat{i}+12\hat{j}+12\hat{k}$
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$-30\hat{i}+12\hat{j}-4\hat{k}$
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$-30\hat{i}-12\hat{j}-21\hat{k}$
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$\hat{i}-12\hat{j}+29\hat{k}$
C
Correct answer
Explanation
For x^2 + 2x + 5 = 0, alpha + beta = -2 and alpha * beta = 5. Then alpha^2 + beta^2 = (alpha + beta)^2 - 2(alpha * beta) = 4 - 10 = -6. Thus, a = -2i + 5j and b = 5i - 2j - 6k. The cross product a x b is (-2i + 5j + 0k) x (5i - 2j - 6k) = (-30i - 12j - 21k).