The angle between the lines frac{x}{1} = frac | vedclass

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Question

(A) The direction ratios of the first line are $\vec{a} = (1, 0, -1)$.The direction ratios of the second line are $\vec{b} = (3, 4, 5)$.The angle $	h

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{1}{5}\right)$

Vedclass · Answer

(A) The direction ratios of the first line are $\vec{a} = (1, 0, -1)$.The direction ratios of the second line are $\vec{b} = (3, 4, 5)$.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 $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.Substituting the values:$\cos 	heta = \frac{|(1)(3) + (0)(4) + (-1)(5)|}{\sqrt{1^2 + 0^2 + (-1)^2} \sqrt{3^2 + 4^2 + 5^2}}$$\cos 	heta = \frac{|3 + 0 - 5|}{\sqrt{1 + 0 + 1} \sqrt{9 + 16 + 25}}$$\cos 	heta = \frac{|-2|}{\sqrt{2} \sqrt{50}} = \frac{2}{\sqrt{2} \cdot 5\sqrt{2}} = \frac{2}{5 \cdot 2} = \frac{1}{5}$.Therefore,$	heta = \cos^{-1}\left(\frac{1}{5}ight)$.

The angle between the lines $\frac{x}{1} = \frac{y}{0} = \frac{z}{-1}$ and $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$ is

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