The angle between two lines frac{x+3}{2}=frac | vedclass

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(D) The given lines are $\frac{x+3}{2}=\frac{y}{-3}=\frac{z+5}{-6}$ and $\frac{x-1}{10}=\frac{y+1}{-2}=\frac{z-3}{11}$. The direction ratios of the fi

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

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(D) The given lines are $\frac{x+3}{2}=\frac{y}{-3}=\frac{z+5}{-6}$ and $\frac{x-1}{10}=\frac{y+1}{-2}=\frac{z-3}{11}$. The direction ratios of the first line are $\vec{b_1} = (2, -3, -6)$ and 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 $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_1^2}}$. 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)$.

The angle between two lines $\frac{x+3}{2}=\frac{-y}{3}=\frac{z+5}{-6}$ and $\frac{x-1}{10}=\frac{y+1}{-2}=\frac{z-3}{11}$ is . . . . . . .

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