The angle between the lines whose direction c | vedclass

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(D) Given the equations for direction cosines $(l, m, n)$:$l + m + n = 0$  $(i)$$l^2 + m^2 - n^2 = 0$  (ii)From $(i)$,$n = -(l + m)$. Substituting thi

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$\frac{\pi}{3}$

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(D) Given the equations for direction cosines $(l, m, n)$:$l + m + n = 0$  $(i)$$l^2 + m^2 - n^2 = 0$  (ii)From $(i)$,$n = -(l + m)$. Substituting this into (ii):$l^2 + m^2 - (-(l + m))^2 = 0$$l^2 + m^2 - (l^2 + m^2 + 2lm) = 0$$-2lm = 0$,which implies $l = 0$ or $m = 0$.Case $1$: If $l = 0$,then $m + n = 0 \implies n = -m$. Since $l^2 + m^2 + n^2 = 1$,we have $0^2 + m^2 + (-m)^2 = 1 \implies 2m^2 = 1 \implies m = \pm \frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ and $(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.Case $2$: If $m = 0$,then $l + n = 0 \implies n = -l$. Similarly,$l^2 + 0^2 + (-l)^2 = 1 \implies 2l^2 = 1 \implies l = \pm \frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$ and $(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.Let the direction vectors be $\vec{a} = (0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ and $\vec{b} = (\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.The cosine of the angle $	heta$ is given by $\cos 	heta = |\vec{a} \cdot \vec{b}| = |(0)(\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}})(0) + (-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})| = |0 + 0 + \frac{1}{2}| = \frac{1}{2}$.Therefore,$	heta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

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

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