What do the three lines with direction ratios | vedclass

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(A) Let the direction ratios of the three lines be $L_1 = (1, 1, 2)$,$L_2 = (\sqrt{3}-1, -\sqrt{3}-1, 4)$,and $L_3 = (-\sqrt{3}-1, \sqrt{3}-1, 4)$.To

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Equilateral triangle

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(A) Let the direction ratios of the three lines be $L_1 = (1, 1, 2)$,$L_2 = (\sqrt{3}-1, -\sqrt{3}-1, 4)$,and $L_3 = (-\sqrt{3}-1, \sqrt{3}-1, 4)$.To find the angles between these lines,we use the formula $\cos 	heta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$.First,calculate the magnitudes: $|L_1| = \sqrt{1^2+1^2+2^2} = \sqrt{6}$.$|L_2| = \sqrt{(\sqrt{3}-1)^2 + (-\sqrt{3}-1)^2 + 4^2} = \sqrt{(3+1-2\sqrt{3}) + (3+1+2\sqrt{3}) + 16} = \sqrt{24} = 2\sqrt{6}$.$|L_3| = \sqrt{(-\sqrt{3}-1)^2 + (\sqrt{3}-1)^2 + 4^2} = \sqrt{(3+1+2\sqrt{3}) + (3+1-2\sqrt{3}) + 16} = \sqrt{24} = 2\sqrt{6}$.Now,calculate the dot products:$L_1 \cdot L_2 = 1(\sqrt{3}-1) + 1(-\sqrt{3}-1) + 2(4) = \sqrt{3}-1-\sqrt{3}-1+8 = 6$.$L_1 \cdot L_3 = 1(-\sqrt{3}-1) + 1(\sqrt{3}-1) + 2(4) = -\sqrt{3}-1+\sqrt{3}-1+8 = 6$.$L_2 \cdot L_3 = (\sqrt{3}-1)(-\sqrt{3}-1) + (-\sqrt{3}-1)(\sqrt{3}-1) + 4(4) = -(3-1) - (3-1) + 16 = -2 - 2 + 16 = 12$.Calculate the cosines of the angles:$\cos 	heta_{12} = \frac{6}{\sqrt{6} \cdot 2\sqrt{6}} = \frac{6}{12} = \frac{1}{2} \implies 	heta_{12} = 60^\circ$.$\cos 	heta_{13} = \frac{6}{\sqrt{6} \cdot 2\sqrt{6}} = \frac{6}{12} = \frac{1}{2} \implies 	heta_{13} = 60^\circ$.$\cos 	heta_{23} = \frac{12}{2\sqrt{6} \cdot 2\sqrt{6}} = \frac{12}{24} = \frac{1}{2} \implies 	heta_{23} = 60^\circ$.Since all angles are $60^\circ$,the lines form an equilateral triangle.

What do the three lines with direction ratios $(1, 1, 2)$,$(\sqrt{3}-1, -\sqrt{3}-1, 4)$,and $(-\sqrt{3}-1, \sqrt{3}-1, 4)$ represent?

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