The angle between the lines x=y, z=0 and y=0, | vedclass

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(B) The first line is given by $x=y$ and $z=0$. This line lies in the $xy$-plane. Its direction vector $\vec{v_1}$ can be written as $(1, 1, 0)$.The s

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$45$

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(B) The first line is given by $x=y$ and $z=0$. This line lies in the $xy$-plane. Its direction vector $\vec{v_1}$ can be written as $(1, 1, 0)$.The second line is given by $y=0$ and $z=0$. This is the $x$-axis. Its direction vector $\vec{v_2}$ is $(1, 0, 0)$.The angle $	heta$ between two lines with direction vectors $\vec{v_1}$ and $\vec{v_2}$ is given by $\cos 	heta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.Calculating the dot product: $\vec{v_1} \cdot \vec{v_2} = (1)(1) + (1)(0) + (0)(0) = 1$.Calculating the magnitudes: $|\vec{v_1}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$ and $|\vec{v_2}| = \sqrt{1^2 + 0^2 + 0^2} = 1$.Substituting these values: $\cos 	heta = \frac{1}{\sqrt{2} 	imes 1} = \frac{1}{\sqrt{2}}$.Therefore,$	heta = 45^{\circ}$.

The angle between the lines $x=y, z=0$ and $y=0, z=0$ is (in $^{\circ}$)

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