1.

Find the angle between the lines `(x+1)/(1)=(2y-3)/(3) =(z-6)/(2) " and "(x-4)/(3)=(y+3)/(-2) ,z=5`

Answer» The given equations may be written as
`(x+1)/(1) =(y-(3)/(2))/((3)/(2)) =(z-6)/(2) rArr (x+1)/(2)=(y-(3)/(2))/(3)=(z-6)/(4)....(i)`
and `(x-4)/(3) =(y+3)/(-2) =(z-5)/(0)`
Here `a_(1) =2,b_(1) =3 , c_(1)=4 " and " a_(2) =3 ,b_(2) =-2 ,c_(2) =0`
Let `theta` be the angle between the given lines. Then
`" cos " theta = (|a_(1)a_(2) +b_(1)b_(2)+c_(1)c_(2)|)/((sqrt(a_(1)^(2) +b_(1)^(2) +c_(1)^(2)))(sqrt(a_(2)^(2) +b_(2)^(2)+c_(2)^(2))))`
`=(|(2xx3)+3xx(-2)+4xx0|)/((sqrt(4+9+16))(sqrt(19+4+0)))=(0)/((sqrt(29)xxsqrt(13)))=0`
`:. theta =(pi)/(2)`
Hence the angle between the given lines is `(pi)/(2)`


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