Find the angle between the following pair of lines: `(x-2)/2=(y-1)/5=(z+3)/(-3)`and `(x+2)/(-1)=(y-4)/8=(z-5)/4`

Explanation

(i) Equation of given lines are
`(x-2)/(2)=(y-1)/(5)=(z+3)/(-3)`
and `(x+2)/(-1)=(y-4)/(8) = (z-5)/(4)`
`:.` The direction ratios of the two lines are `(2,5,-3)` and `(-1,8,4)`.
Let `theta` be the angle between two 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)))`
`rArr cos theta = (2xx(-1)+5xx8+(-3)xx4)/(sqrt(2^(2)+5^(2)+(-3)^(2))sqrt((-1)^(2)+8^(2)+4^(2)))`
`=((-2)40-12)/(sqrt(4+25+9)sqrt(1+64+16))`
`= (26)/(sqrt(38)sqrt(81)) = (26)/(9sqrt(38))`
`rArr theta cos^(-1) = ((26)/(9sqrt(38)))`
(ii) Equation of given lines are
`x/2 = y/2 = z/1` and `(x-5)/(4)=(y-2)/(1)=(z-3)/(8)`
`:.` The direction ratios of the lines are (2,2,1) and (4,1,8) respectively.
Let `theta` be the acute angle between these 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)))`

`rArr cos theta = (2xx4+2xx1+1xx8)/(sqrt(2^(2)+2^(2)+1^(2))sqrt(4^(2)+1^(2)+8^(2)))`
`=(8+2+8)/(sqrt(4+4+1)sqrt(16+1+64))`
`=(18)/(sqrt(9)sqrt(81)) = (18)/(3xx9) = 2/3`.
`rArr theta cos^(-1) = (2/3)`