1.

Find the angle between the lines whose direction cosines are given by the equations `3l + m + 5n = 0` and `6mn - 2nl + 5lm = 0`

Answer» Direction Cosines of the two lines are given by the equations `3l+m+5n=0` and `6mn-2l+5lm=0`
From first equation we get, `m=-5n-3l`
Put it in second equation we get,
`-30n^2-45nl-15l^2=0`
`2n^2+3nl+l^2=0`
`2n^2+2nl+nl+l^2=0`
`(2n+l)(n+l)=0`
If `l=-2n`, then
`m=n` (by putting it in equation 1)
and if `l=-n`
we get `m=-2n`
`costheta=(a_1a_2+b_1b_2+c_1c_2)/(sqrt(a_1^2+b_1^2+c_1^2)sqrt(a_2^2+b_2^2+c_2^2)`
We get,
`theta=cos^(-1)(-1/6)`


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