Let for a triangle ABC,overline{AB} = -2hat{i | vedclass

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(A) Given that $\overline{AB} + \overline{BC} + \overline{CA} = \vec{0}$,we have $\overline{BC} = -\overline{AB} - \overline{CA}$.$\overline{BC} = -(-

Vedclass · Accepted Answer

$60$

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(A) Given that $\overline{AB} + \overline{BC} + \overline{CA} = \vec{0}$,we have $\overline{BC} = -\overline{AB} - \overline{CA}$.$\overline{BC} = -(-2\hat{i} + \hat{j} + 3\hat{k}) - (4\hat{i} + 3\hat{j} + \delta\hat{k}) = -2\hat{i} - 4\hat{j} - (3 + \delta)\hat{k}$.Since $\overline{CB} = -\overline{BC}$,we get $\overline{CB} = 2\hat{i} + 4\hat{j} + (3 + \delta)\hat{k}$.Comparing with $\overline{CB} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$,we have $\alpha = 2$,$\beta = 4$,and $\gamma = 3 + \delta$.The area of triangle $ABC$ is given by $\frac{1}{2} |\overline{AC} 	imes \overline{AB}| = 5\sqrt{6}$.$\overline{AC} = -\overline{CA} = -4\hat{i} - 3\hat{j} - \delta\hat{k}$.$\overline{AC} 	imes \overline{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -4 & -3 & -\delta \ -2 & 1 & 3 \end{vmatrix} = \hat{i}(-9 + \delta) - \hat{j}(-12 - 2\delta) + \hat{k}(-4 - 6) = (\delta - 9)\hat{i} + (2\delta + 12)\hat{j} - 10\hat{k}$.$|\overline{AC} 	imes \overline{AB}|^2 = (\delta - 9)^2 + (2\delta + 12)^2 + (-10)^2 = (2\sqrt{5\sqrt{6}})^2 = 600$.$(\delta^2 - 18\delta + 81) + (4\delta^2 + 48\delta + 144) + 100 = 600 \Rightarrow 5\delta^2 + 30\delta - 275 = 0 \Rightarrow \delta^2 + 6\delta - 55 = 0$.$(\delta + 11)(\delta - 5) = 0$. Since $\delta > 0$,we have $\delta = 5$.Then $\gamma = 3 + 5 = 8$,so $\overline{CB} = 2\hat{i} + 4\hat{j} + 8\hat{k}$ and $\overline{CA} = 4\hat{i} + 3\hat{j} + 5\hat{k}$.$\overline{CB} \cdot \overline{CA} = (2)(4) + (4)(3) + (8)(5) = 8 + 12 + 40 = 60$.

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