Question
Find the angles between the lines whose direction cosines $l, m, n$ satisfy the equations $5l + m + 3n = 0$ and $5mn − 2nl + 6lm = 0$
✓
Answer
$ 51+m+3 n=0\ldots[Given]$
$\therefore m=-(5 I+3 n)\ldots(i)$
$5 m n-2 n l+6 l m=0\ldots[Given] \$
$therefore-5(5 l+3 n) n-2 n l-6 l(5 l+3 n)=0\ldots[[From (i)]$
$\therefore-25\left|n-15 n^2-2 n l-30\right|^2-18 n l=0$
$\therefore-\left.30\right|^2-45 \ln -15 n^2=0$
$\therefore 21^2+3 \ln +n^2=0$
$\therefore(2 l+n)(1+n)=0$
$\therefore n =-21 \text { or } n =-1$
If $n=-2 I$, then from (i), we get
$ m=-[5 \mid+3(-2 l)]$
$\therefore m=1$
$\therefore m=1, n=-21 $
$\therefore$ Direction ratios of the first line are proportional to $I , I ,-2 l$
i.e., $1,1,-2$
If $n=-1$, then from (i), we get
$ m=-[5 \mid+3(-1)]$
$\therefore m=-21$
$\therefore m=-21, n=-1 $
$\therefore$ Direction ratios of the second line are proportional to $I ,-2 l ,- I$
i.e., $1,-2,-1$
Let $\theta$ be the angle between the two lines.
$ \therefore \cos \theta=\left|\frac{1(1)+1(2)+(-2)(-1)}{\sqrt{1^2+1^2+(-2)^2} \cdot \sqrt{1^2+(-2)^2+(-1)^2}}\right|$
$=\left|\frac{1-2+2}{\sqrt{1+1+4} \cdot \sqrt{1+4+1}}\right|$
$=\left|\frac{1}{\sqrt{6} \cdot \sqrt{6}}\right|$
$=\frac{1}{6}$
$\therefore \theta=\cos ^{-1}\left(\frac{1}{6}\right) $
$\therefore m=-(5 I+3 n)\ldots(i)$
$5 m n-2 n l+6 l m=0\ldots[Given] \$
$therefore-5(5 l+3 n) n-2 n l-6 l(5 l+3 n)=0\ldots[[From (i)]$
$\therefore-25\left|n-15 n^2-2 n l-30\right|^2-18 n l=0$
$\therefore-\left.30\right|^2-45 \ln -15 n^2=0$
$\therefore 21^2+3 \ln +n^2=0$
$\therefore(2 l+n)(1+n)=0$
$\therefore n =-21 \text { or } n =-1$
If $n=-2 I$, then from (i), we get
$ m=-[5 \mid+3(-2 l)]$
$\therefore m=1$
$\therefore m=1, n=-21 $
$\therefore$ Direction ratios of the first line are proportional to $I , I ,-2 l$
i.e., $1,1,-2$
If $n=-1$, then from (i), we get
$ m=-[5 \mid+3(-1)]$
$\therefore m=-21$
$\therefore m=-21, n=-1 $
$\therefore$ Direction ratios of the second line are proportional to $I ,-2 l ,- I$
i.e., $1,-2,-1$
Let $\theta$ be the angle between the two lines.
$ \therefore \cos \theta=\left|\frac{1(1)+1(2)+(-2)(-1)}{\sqrt{1^2+1^2+(-2)^2} \cdot \sqrt{1^2+(-2)^2+(-1)^2}}\right|$
$=\left|\frac{1-2+2}{\sqrt{1+1+4} \cdot \sqrt{1+4+1}}\right|$
$=\left|\frac{1}{\sqrt{6} \cdot \sqrt{6}}\right|$
$=\frac{1}{6}$
$\therefore \theta=\cos ^{-1}\left(\frac{1}{6}\right) $
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