The acute angle between the lines x=-y, z=0 a | vedclass

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(C) The given lines are $x = -y, z = 0$ and $x = 0, z = 0$. We can write these lines in symmetric form: Line $1$: $\frac{x}{1} = \frac{y}{-1} = \frac{

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$\frac{\pi}{4}$

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(C) The given lines are $x = -y, z = 0$ and $x = 0, z = 0$. We can write these lines in symmetric form: Line $1$: $\frac{x}{1} = \frac{y}{-1} = \frac{z}{0}$,so the direction vector is $\vec{v_1} = (1, -1, 0)$. Line $2$: $\frac{x}{0} = \frac{y}{1} = \frac{z}{0}$,so the direction vector is $\vec{v_2} = (0, 1, 0)$. The angle $	heta$ between the lines is given by $\cos 	heta = \left| \frac{\vec{v_1} \cdot \vec{v_2}}{|\vec{v_1}| |\vec{v_2}|} ight|$. Calculating the dot product: $\vec{v_1} \cdot \vec{v_2} = (1)(0) + (-1)(1) + (0)(0) = -1$. Calculating the magnitudes: $|\vec{v_1}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}$ and $|\vec{v_2}| = \sqrt{0^2 + 1^2 + 0^2} = 1$. Thus,$\cos 	heta = \left| \frac{-1}{\sqrt{2} 	imes 1} ight| = \frac{1}{\sqrt{2}}$. Therefore,$	heta = \cos^{-1}\left(\frac{1}{\sqrt{2}}ight) = \frac{\pi}{4}$.

The acute angle between the lines $x=-y, z=0$ and $x=0, z=0$ is

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