The angle between the pair of lines with dire | vedclass

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(C) Let the direction ratios of the two lines be $a_1, b_1, c_1 = 1, 1, 2$ and $a_2, b_2, c_2 = \sqrt{3}-1, -\sqrt{3}-1, 4$. The cosine of the angle $

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let the direction ratios of the two lines be $a_1, b_1, c_1 = 1, 1, 2$ and $a_2, b_2, c_2 = \sqrt{3}-1, -\sqrt{3}-1, 4$. The cosine of the angle $	heta$ between the lines is given by: $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$ Calculating the numerator: $1(\sqrt{3}-1) + 1(-\sqrt{3}-1) + 2(4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6$. Calculating the denominators: $\sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$. $\sqrt{(\sqrt{3}-1)^2 + (-\sqrt{3}-1)^2 + 4^2} = \sqrt{(3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) + 16} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$. Thus,$\cos 	heta = \frac{6}{\sqrt{6} \cdot 2\sqrt{6}} = \frac{6}{2 \cdot 6} = \frac{1}{2}$. Therefore,$	heta = \cos^{-1}(\frac{1}{2}) = 60^\circ$.

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

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