The acute angle between the lines x = -2 + 2t | vedclass

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Question

(A) The direction ratios of the first line are $a_1 = 2, b_1 = -4, c_1 = 1$. The direction ratios of the second line are $a_2 = -1, b_2 = 2, c_2 = 3$.

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{7}{3\sqrt{6}}\right)$

Vedclass · Answer

(A) The direction ratios of the first line are $a_1 = 2, b_1 = -4, c_1 = 1$. The direction ratios of the second line are $a_2 = -1, b_2 = 2, c_2 = 3$. The cosine of the angle $	heta$ between the two lines is given by: $\cos 	heta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} ight|$ Substituting the values: $\cos 	heta = \left| \frac{(2)(-1) + (-4)(2) + (1)(3)}{\sqrt{2^2 + (-4)^2 + 1^2} \sqrt{(-1)^2 + 2^2 + 3^2}} ight|$ $\cos 	heta = \left| \frac{-2 - 8 + 3}{\sqrt{4 + 16 + 1} \sqrt{1 + 4 + 9}} ight| = \left| \frac{-7}{\sqrt{21} \sqrt{14}} ight|$ $\cos 	heta = \frac{7}{\sqrt{21 	imes 14}} = \frac{7}{\sqrt{3 	imes 7 	imes 2 	imes 7}} = \frac{7}{7\sqrt{6}} = \frac{1}{\sqrt{6}}$ Therefore,$	heta = \cos^{-1}\left(\frac{1}{\sqrt{6}}ight)$.

The acute angle between the lines $x = -2 + 2t, y = 3 - 4t, z = -4 + t$ and $x = -2 - t, y = 3 + 2t, z = -4 + 3t$ is

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