Let A be the point of intersection of the lin | vedclass

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(A) The direction vectors of the lines $L_1$ and $L_2$ are $\vec{v_1} = \langle 1, 0, -1 \rangle$ and $\vec{v_2} = \langle 3, 4, 5 \rangle$ respective

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(A) The direction vectors of the lines $L_1$ and $L_2$ are $\vec{v_1} = \langle 1, 0, -1 angle$ and $\vec{v_2} = \langle 3, 4, 5 angle$ respectively.Let $	heta$ be the angle between the lines $L_1$ and $L_2$.$\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|} = \frac{|(1)(3) + (0)(4) + (-1)(5)|}{\sqrt{1^2 + 0^2 + (-1)^2} \sqrt{3^2 + 4^2 + 5^2}} = \frac{|3 + 0 - 5|}{\sqrt{2} \sqrt{9 + 16 + 25}} = \frac{2}{\sqrt{2} \sqrt{50}} = \frac{2}{\sqrt{100}} = \frac{2}{10} = \frac{1}{5}$.Since $\cos 	heta = \frac{1}{5}$,we have $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - (\frac{1}{5})^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5}$.The area of triangle $ABC$ is given by $	ext{Area} = \frac{1}{2} 	imes AB 	imes AC 	imes \sin 	heta$.Given $AB = AC = \sqrt{15}$,we have $	ext{Area} = \frac{1}{2} 	imes \sqrt{15} 	imes \sqrt{15} 	imes \frac{\sqrt{24}}{5} = \frac{1}{2} 	imes 15 	imes \frac{\sqrt{24}}{5} = \frac{3 \sqrt{24}}{2}$.The square of the area is $(\frac{3 \sqrt{24}}{2})^2 = \frac{9 	imes 24}{4} = 9 	imes 6 = 54$.

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