The angle between the straight lines frac{x - | vedclass

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(B) The direction ratios of the first line are $\vec{a_1} = (2, 5, -3)$.The direction ratios of the second line are $\vec{a_2} = (-1, 8, 4)$.The angle

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{26}{9\sqrt{38}}\right)$

Vedclass · Answer

(B) The direction ratios of the first line are $\vec{a_1} = (2, 5, -3)$.The direction ratios of the second line are $\vec{a_2} = (-1, 8, 4)$.The angle $	heta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.$\cos 	heta = \frac{|2(-1) + 5(8) + (-3)(4)|}{\sqrt{2^2 + 5^2 + (-3)^2} \sqrt{(-1)^2 + 8^2 + 4^2}}$$\cos 	heta = \frac{|-2 + 40 - 12|}{\sqrt{4 + 25 + 9} \sqrt{1 + 64 + 16}}$$\cos 	heta = \frac{|26|}{\sqrt{38} \sqrt{81}} = \frac{26}{9\sqrt{38}}$.Therefore,$	heta = \cos^{-1}\left(\frac{26}{9\sqrt{38}}ight)$.

The angle between the straight lines $\frac{x - 2}{2} = \frac{y - 1}{5} = \frac{z + 3}{-3}$ and $\frac{x + 1}{-1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ is

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