Find the angle between the following pairs of lines.
(i) `hatr = 2hati-5hatj+hatk+lambda(3hati+2hatj+6hatk)` and `vecr = 7 hati-6hatk+mu(hati+2hatj+2hatk)`
(ii) `vecr = 3hati+hatj-2hatk+lambda(hati-hatj-2hatk)` and
`vecr= 2hati-hatj-56hatk+mu(3hati-5hatj-4hatk)`

Explanation

Let `theta` be the acute angle between the given lines than
`cos theta = |(vecb_(1).vecb_(2))/(|vecb_(1)||vecb_(2)|)|`
Given lines are parallel to the vectors `vecb_(1) = 3hati+2hatj+6hatk` and `vecb_(2) = hati+2hatj+2hatk` respectively.
`:.|vecb_(1)| = sqrt(3^(2)+2^(2)+6^(2)) = sqrt(9+4+36)=sqrt(49) = 7`
`|vecb_(2)| = sqrt(1^(2)+2^(2)+2^(2))=sqrt(1+4+4)=sqrt(9) = 3`
and ` vecb_(1).vecb_(2)=(3hati+2hatj+6hatk).(hati+2hatj+2hatk)`
`=3xx1+2xx2+6xx2`
`=3+4+12=19`
`:. cos theta = |(vecb_(1).vecb_(2))/(|vecb_(1)||vecb_(2)|)| = (19)/(7xx3)=19/21`
(ii) Let `theta` be the angle between given lines.
`cos theta = |(vecb_(1).vecb_(2))/(|vecb_(1)||vecb_(2)|)|`
Given linea are parallel to the vectors `vecb_(1) = hati-hatj-2hatk` and `vecb_(2) = 3hati-5hatj-4hatk`.
`:.|vecb_(1)|=sqrt((1)^(2)+(-1)^(2)+(-2)^(2))=sqrt(1+1+4)=sqrt(6)`
`|vecb_(2)| = (hati-hatj-2hatk).(3hati-5hatj-4hatk)`
`= 1xx3-1xx(-5)-2xx(-4)`
`=3+5+8=16`
`cos theta = |(vecb_(1).vecb_(2))/(|vecb_(1)||vecb_(2)|)| = (16)/(sqrt(6)xx5sqrt(2))`
`=(16)/(10sqrt(3)) = (8)/(5sqrt(3))`
`:. theta = cos^(-1)((8)/(5sqrt(3)))`