An angle between the lines whose direction cosines are gien by the equations, l + 3m + 5n …

Question

An angle between the lines whose direction cosines are gien by the equations,

$l$ + 3m + 5n = 0 and 5$l$m - 2mn + 6n$l$ = 0, is :

Accepted Answer

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 $$\left( {{1 \over 6}} \right)$$

Solutions · 1