Find the angle between the two lines frac{x+1 | vedclass

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Question

(A) The direction ratios of the first line are $\vec{b_1} = (2, 3, 6)$.The direction ratios of the second line are $\vec{b_2} = (10, 2, -11)$.The angl

Vedclass · Accepted Answer

$\cos ^{-1}\left(\frac{8}{21}\right)$

Vedclass · Answer

(A) The direction ratios of the first line are $\vec{b_1} = (2, 3, 6)$.The direction ratios of the second line are $\vec{b_2} = (10, 2, -11)$.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 $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.Calculating the dot product: $a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(10) + (3)(2) + (6)(-11) = 20 + 6 - 66 = -40$.Calculating the magnitudes: $|\vec{b_1}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.$|\vec{b_2}| = \sqrt{10^2 + 2^2 + (-11)^2} = \sqrt{100 + 4 + 121} = \sqrt{225} = 15$.Thus,$\cos 	heta = \frac{|-40|}{7 	imes 15} = \frac{40}{105} = \frac{8}{21}$.Therefore,$	heta = \cos ^{-1}\left(\frac{8}{21}ight)$.

Find the angle between the two lines $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and $\frac{x-1}{10}=\frac{y+3}{2}=\frac{z+4}{-11}$.

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