Let overrightarrow{a}=hat{i}+5hat{j}+alphahat | vedclass

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(D) Given $\vec{a} \cdot \vec{b} = 0$,we have $(1)(1) + (5)(3) + (\alpha)(\beta) = 0$,which implies $1 + 15 + \alpha\beta = 0$,so $\alpha\beta = -16$.

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(D) Given $\vec{a} \cdot \vec{b} = 0$,we have $(1)(1) + (5)(3) + (\alpha)(\beta) = 0$,which implies $1 + 15 + \alpha\beta = 0$,so $\alpha\beta = -16$. Next,calculate $\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 3 & \beta \ -1 & 2 & -3 \end{vmatrix} = \hat{i}(-9 - 2\beta) - \hat{j}(-3 + \beta) + \hat{k}(2 + 3) = (-9 - 2\beta)\hat{i} + (3 - \beta)\hat{j} + 5\hat{k}$. Given $|\vec{b} 	imes \vec{c}| = 5\sqrt{3}$,so $|\vec{b} 	imes \vec{c}|^2 = 75$. $(-9 - 2\beta)^2 + (3 - \beta)^2 + 5^2 = 75$ $(81 + 36\beta + 4\beta^2) + (9 - 6\beta + \beta^2) + 25 = 75$ $5\beta^2 + 30\beta + 115 = 75$ $5\beta^2 + 30\beta + 40 = 0 \Rightarrow \beta^2 + 6\beta + 8 = 0$. Solving for $\beta$,$(\beta + 4)(\beta + 2) = 0$,so $\beta = -4$ or $\beta = -2$. If $\beta = -4$,then $\alpha = 4$. If $\beta = -2$,then $\alpha = 8$. Now,$|\vec{a}|^2 = 1^2 + 5^2 + \alpha^2 = 26 + \alpha^2$. For $\alpha = 4$,$|\vec{a}|^2 = 26 + 16 = 42$. For $\alpha = 8$,$|\vec{a}|^2 = 26 + 64 = 90$. The greatest value is $90$.

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