The angle between the lines whose direction r | vedclass

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(B) Given equations for direction ratios $(l, m, n)$ are $l+m+n=0$ and $l^2=m^2+n^2$.From $l+m+n=0$,we have $l=-(m+n)$.Substituting this into $l^2=m^2

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$\frac{\pi}{3}$

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(B) Given equations for direction ratios $(l, m, n)$ are $l+m+n=0$ and $l^2=m^2+n^2$.From $l+m+n=0$,we have $l=-(m+n)$.Substituting this into $l^2=m^2+n^2$,we get $(-(m+n))^2 = m^2+n^2$.$m^2+n^2+2mn = m^2+n^2$,which implies $2mn=0$,so $mn=0$.This gives two cases:Case $I$: $m=0$. Then $l=-n$. Let the direction ratios be $(k, 0, -k)$. The unit vector is $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.Case $II$: $n=0$. Then $l=-m$. Let the direction ratios be $(k, -k, 0)$. The unit vector is $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)$.Let $\vec{a} = (\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$ and $\vec{b} = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)$.The cosine of the angle $	heta$ between them is given by $\cos 	heta = |\vec{a} \cdot \vec{b}|$.$\cos 	heta = |(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (0)(-\frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}})(0)| = |\frac{1}{2} + 0 + 0| = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,we have $	heta = \frac{\pi}{3}$.

The angle between the lines whose direction ratios satisfy the equations $l+m+n=0$ and $l^2=m^2+n^2$ is

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