Let ABC be a triangle of area 15 sqrt{2} and | vedclass

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(A) The area of triangle $ABC$ is given by $	ext{Area} = \frac{1}{2} |\overrightarrow{AB} 	imes \overrightarrow{AC}| = 15 \sqrt{2}$.First,calculate

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$54$

Vedclass · Answer

(A) The area of triangle $ABC$ is given by $	ext{Area} = \frac{1}{2} |\overrightarrow{AB} 	imes \overrightarrow{AC}| = 15 \sqrt{2}$.First,calculate the cross product $\overrightarrow{AB} 	imes \overrightarrow{AC}$:$\overrightarrow{AB} 	imes \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -7 \ 6 & d & -2 \end{vmatrix} = \hat{i}(-4 + 7d) - \hat{j}(-2 + 42) + \hat{k}(d - 12) = (7d - 4)\hat{i} - 40\hat{j} + (d - 12)\hat{k}$.Now,find the magnitude:$|\overrightarrow{AB} 	imes \overrightarrow{AC}|^2 = (7d - 4)^2 + (-40)^2 + (d - 12)^2 = (2 	imes 15 \sqrt{2})^2 = 900 	imes 2 = 1800$.$(49d^2 - 56d + 16) + 1600 + (d^2 - 24d + 144) = 1800$.$50d^2 - 80d + 1760 = 1800 \implies 50d^2 - 80d - 40 = 0 \implies 5d^2 - 8d - 4 = 0$.Solving for $d$:$5d^2 - 10d + 2d - 4 = 0 \implies 5d(d - 2) + 2(d - 2) = 0 \implies (5d + 2)(d - 2) = 0$.Since $d > 0$,we have $d = 2$.Using the vector triangle law $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$,we have $\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB} = (6-1)\hat{i} + (d-2)\hat{j} + (-2 - (-7))\hat{k} = 5\hat{i} + 0\hat{j} + 5\hat{k}$.Now,calculate the squares of the lengths of the sides:$|\overrightarrow{AB}|^2 = 1^2 + 2^2 + (-7)^2 = 1 + 4 + 49 = 54$.$|\overrightarrow{BC}|^2 = 5^2 + 0^2 + 5^2 = 25 + 25 = 50$.$|\overrightarrow{AC}|^2 = 6^2 + 2^2 + (-2)^2 = 36 + 4 + 4 = 44$.The largest side is $\sqrt{54}$,and its square is $54$.

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