The angle between the pair of lines with dire | vedclass

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Question

(C) Let the direction ratios of the two lines be $\vec{a} = (1, 1, 2)$ and $\vec{b} = (\sqrt{3} - 1, -\sqrt{3} - 1, 4)$.The cosine of the angle $	het

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let the direction ratios of the two lines be $\vec{a} = (1, 1, 2)$ and $\vec{b} = (\sqrt{3} - 1, -\sqrt{3} - 1, 4)$.The cosine of the angle $	heta$ between the lines is given by the formula:$\cos 	heta = \frac{|a_1b_1 + a_2b_2 + a_3b_3|}{\sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}}$Calculating the dot product:$a_1b_1 + a_2b_2 + a_3b_3 = 1(\sqrt{3} - 1) + 1(-\sqrt{3} - 1) + 2(4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6$.Calculating the magnitudes:$|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.$|\vec{b}| = \sqrt{(\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2 + 4^2} = \sqrt{(3 + 1 - 2\sqrt{3}) + (3 + 1 + 2\sqrt{3}) + 16} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.Substituting the values:$\cos 	heta = \frac{6}{\sqrt{6} 	imes 2\sqrt{6}} = \frac{6}{2 	imes 6} = \frac{6}{12} = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,we have $	heta = 60^o$.

The angle between the pair of lines with direction ratios $(1, 1, 2)$ and $(\sqrt{3} - 1, -\sqrt{3} - 1, 4)$ is ......... $^o$.

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