Find the angle between the lines whose direct | vedclass

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(C) Let the direction ratios of the two lines be $a_1, b_1, c_1$ and $a_2, b_2, c_2$. Given: $(a_1, b_1, c_1) = (1, 1, 2)$ and $(a_2, b_2, c_2) = (\sq

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let the direction ratios of the two lines be $a_1, b_1, c_1$ and $a_2, b_2, c_2$. Given: $(a_1, b_1, c_1) = (1, 1, 2)$ and $(a_2, b_2, c_2) = (\sqrt{3}-1, -\sqrt{3}-1, 4)$. The formula for the angle $	heta$ between two lines is given by: $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$. First,calculate the dot product: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 1(\sqrt{3}-1) + 1(-\sqrt{3}-1) + 2(4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6$. Next,calculate the magnitudes: $\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1+1+4} = \sqrt{6}$. $\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{(\sqrt{3}-1)^2 + (-\sqrt{3}-1)^2 + 4^2} = \sqrt{(3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) + 16} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$. Now,substitute these values into the formula: $\cos 	heta = \frac{6}{\sqrt{6} 	imes 2\sqrt{6}} = \frac{6}{2 	imes 6} = \frac{6}{12} = \frac{1}{2}$. Since $\cos 	heta = \frac{1}{2}$,we have $	heta = 60^{\circ}$.

Find the angle between the lines whose direction ratios are $1, 1, 2$ and $\sqrt{3}-1, -\sqrt{3}-1, 4$. (in $^{\circ}$)

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