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

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Aicharya · Accepted Answer

Given lines are $\begin{aligned} & \frac{x-0}{1}=\frac{y-0}{-1}=\frac{z-0}{0} 	ext { and } \frac{x-0}{0}=\frac{y-0}{1}=\frac{z-0}{0} \ & 	ext { Now, the angle }=\cos ^{-1}\left(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \cdot \sqrt{a_2^2+b_2^2+c_2^2}}ight) \ & =\cos ^{-1}\left(\frac{1 	imes 0+(-1) 	imes 1+0 	imes 0}{\sqrt{1^2+(-1)^2+0^2} \cdot \sqrt{0^2+1^2+0^2}}ight) \ & =\cos ^{-1}\left|\frac{-1}{\sqrt{2}}ight|[	ext { for acute angle }] \ & =\cos ^{-1}\left(\frac{1}{\sqrt{2}}ight)=\frac{\pi}{4} \end{aligned}$

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

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