If the direction cosines l, m, n of two lines | vedclass

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(A) Given relations are $l-5m+3n=0$ and $7l^2+5m^2-3n^2=0$. From the first equation,$l = 5m - 3n$. Substituting this into the second equation: $7(5m-3

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$\frac{2}{\sqrt{6}}$ or $\frac{6}{\sqrt{14}}$

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(A) Given relations are $l-5m+3n=0$ and $7l^2+5m^2-3n^2=0$. From the first equation,$l = 5m - 3n$. Substituting this into the second equation: $7(5m-3n)^2 + 5m^2 - 3n^2 = 0$. $7(25m^2 - 30mn + 9n^2) + 5m^2 - 3n^2 = 0$. $175m^2 - 210mn + 63n^2 + 5m^2 - 3n^2 = 0$. $180m^2 - 210mn + 60n^2 = 0$. Dividing by $30$,we get $6m^2 - 7mn + 2n^2 = 0$. Factoring the quadratic: $(3m - 2n)(2m - n) = 0$. Case $1$: $3m = 2n \Rightarrow m = \frac{2n}{3}$. Then $l = 5(\frac{2n}{3}) - 3n = \frac{10n-9n}{3} = \frac{n}{3}$. Using $l^2 + m^2 + n^2 = 1$: $(\frac{n}{3})^2 + (\frac{2n}{3})^2 + n^2 = 1 \Rightarrow \frac{n^2}{9} + \frac{4n^2}{9} + n^2 = 1 \Rightarrow \frac{14n^2}{9} = 1 \Rightarrow n = \frac{3}{\sqrt{14}}$. Then $l = \frac{1}{\sqrt{14}}$ and $m = \frac{2}{\sqrt{14}}$. $l+m+n = \frac{1+2+3}{\sqrt{14}} = \frac{6}{\sqrt{14}}$. Case $2$: $2m = n \Rightarrow m = \frac{n}{2}$. Then $l = 5(\frac{n}{2}) - 3n = \frac{5n-6n}{2} = -\frac{n}{2}$. Using $l^2 + m^2 + n^2 = 1$: $(-\frac{n}{2})^2 + (\frac{n}{2})^2 + n^2 = 1 \Rightarrow \frac{n^2}{4} + \frac{n^2}{4} + n^2 = 1 \Rightarrow \frac{6n^2}{4} = 1 \Rightarrow n^2 = \frac{2}{3} \Rightarrow n = \sqrt{\frac{2}{3}}$. Then $l = -\frac{1}{\sqrt{6}}$ and $m = \frac{1}{\sqrt{6}}$. $l+m+n = \frac{-1+1+\sqrt{4}}{\sqrt{6}} = \frac{2}{\sqrt{6}}$. Thus,the possible values are $\frac{2}{\sqrt{6}}$ or $\frac{6}{\sqrt{14}}$.

If the direction cosines $l, m, n$ of two lines are connected by relations $l-5m+3n=0$ and $7l^2+5m^2-3n^2=0$,then the value of $l+m+n$ is

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