The angle between the lines,whose direction c | vedclass

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(B) Given equations are $l+m+n=0$ and $2l^2+2m^2-n^2=0$.From the first equation,$n = -(l+m)$.Substituting this into the second equation:$2l^2 + 2m^2 -

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$180$

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(B) Given equations are $l+m+n=0$ and $2l^2+2m^2-n^2=0$.From the first equation,$n = -(l+m)$.Substituting this into the second equation:$2l^2 + 2m^2 - (-(l+m))^2 = 0$$2l^2 + 2m^2 - (l^2 + m^2 + 2lm) = 0$$l^2 + m^2 - 2lm = 0$$(l-m)^2 = 0 \Rightarrow l=m$.If $l=m$,then $n = -(l+l) = -2l$.The direction ratios of the lines are proportional to $(l, l, -2l)$,which simplifies to $(1, 1, -2)$.Since the direction ratios are the same for both lines,the lines are parallel.The angle $	heta$ between parallel lines is $0^{\circ}$ or $180^{\circ}$.Given the options,$180^{\circ}$ is the correct choice.

The angle between the lines,whose direction cosines $l, m, n$ satisfy the equations $l+m+n=0$ and $2l^2+2m^2-n^2=0$,is (in $^{\circ}$)

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