QJEEM18Ma7027: JEE Main 2018 Mathematics - Class 12 JEEM18Ma7027 Solution: An angle between the lines whose direction cosines are gien by…

Given

l + 3m + 5n = 0

and 5 $l$ m $-$ 2mn + 6n $l$ = 0

From eq. (1) we have

$l$ = $-$ 3m $-$ 5n

Put the value of $l$ in eq. (2), we get ;

5 ( $-$ 3m $-$ 5n) m $-$ 2mn + 6n ( $-$ 3m $-$ 5n) = 0

$\Rightarrow$   15m² + 45mn + 30n² = 0

$\Rightarrow$  m² + 3mn + 2n² = 0

$\Rightarrow$   m² + 2mn + mn + 2n² = 0

$\Rightarrow$ $\,\,\,$ (m + n) (m + 2n) = 0

$\therefore$   m = $-$ n   or   m = $-$ 2n

For m = $-n,$ $l$ = $-$ 2n

And for m = $-$ 2n, $l$ = n

$\therefore$    ( $l$ , m, n) = ( $-$ 2n, $-$ n, n) Or ( $l$ , m, n) = (n, $-$ 2n, n)

$\Rightarrow$   ( $l$ , m, n) = ( $-$ 2, $-$ 1, 1) Or ( $l$ , m, n) = (1, $-$ 2, 1)

Therefore, angle between the lines is given as :

cos ( $\theta$ ) = ${{\left( { - 2} \right)\left( 1 \right) + \left( { - 1} \right).\left( { - 2} \right) + \left( 1 \right)\left( 1 \right)} \over {\sqrt 6 .\sqrt 6 }}$

$\Rightarrow$ cos ( $\theta$ ) = ${1 \over 6}$ $\Rightarrow$   $\theta$ =cos ^{$-$ 1} $\left( {{1 \over 6}} \right)$