If two straight lines whose direction cosines | vedclass

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(A) Given the relations for direction cosines $l, m, n$ are $l+m-n=0$ and $3l^{2}+m^{2}+cnl=0$. From the first equation,we have $n = l+m$. Substitutin

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$6$

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(A) Given the relations for direction cosines $l, m, n$ are $l+m-n=0$ and $3l^{2}+m^{2}+cnl=0$. From the first equation,we have $n = l+m$. Substituting this into the second equation: $3l^{2}+m^{2}+cl(l+m)=0$. Expanding this,we get $3l^{2}+m^{2}+cl^{2}+clm=0$. Grouping the terms,we have $(3+c)l^{2}+clm+m^{2}=0$. Dividing by $m^{2}$ (assuming $m 
eq 0$),we get $(3+c)(\frac{l}{m})^{2}+c(\frac{l}{m})+1=0$. Since the two lines are parallel,the roots of this quadratic equation in $(\frac{l}{m})$ must be equal. Therefore,the discriminant $D = b^{2}-4ac = 0$. $c^{2}-4(3+c)(1) = 0$. $c^{2}-4c-12=0$. Factoring the quadratic,we get $(c-6)(c+2)=0$. This gives $c=6$ or $c=-2$. Since we need the positive value of $c$,we have $c=6$.

If two straight lines whose direction cosines are given by the relations $l+m-n=0$ and $3l^{2}+m^{2}+cnl=0$ are parallel,then the positive value of $c$ is

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