Find the angle between the following pair of lines :
i. `vecr=2hati-5hatj+hatk+lamda(3hati+2hatj+6hatk) and vecr=7hati-6hatk+mu(hati+2hatj+2hatk)`
ii. `(x)/(2)=(y)/(2)=(z)/(1) and (x-5)/(4)=(y-2)/(1)=(z-3)/(8)`
i. `vecr=2hati-5hatj+hatk+lamda(3hati+2hatj+6hatk) and vecr=7hati-6hatk+mu(hati+2hatj+2hatk)`
ii. `(x)/(2)=(y)/(2)=(z)/(1) and (x-5)/(4)=(y-2)/(1)=(z-3)/(8)`
i. The given lines are parallel to the vectors `vec(b_(1))=3hati+2hatj+6hatk and vec(b_(2))=hati+2hatj+2hatk`, respectively. If `theta` is the angle between the given pair of lines, then
`” “costheta=(vec(b_(1))*vec(b_(2)))/(|vec(b_(1))||vec(b_(2))|)=((3)(1)+(2)(2)+(6)(2))/(sqrt(3^(2)+2^(2)+6^(2))sqrt(1^(2)+2^(2)+2^(2)))=(19)/(7xx3)`
`therefore” “theta=cos^(-1)((19)/(21))`
ii. The given lines are parallel to the vectors `vec(b_(1))=2hati+2hatj+hatk and vec(b_(2))=4hati+hatj+8hatk`, respectively. If `theta` is the angle between the given pair of lines, then
`” “costheta=(vec(b_(1))*vec(b_(2)))/(|vec(b_(1))||vec(b_(2))|)=((2)(4)+(2)(1)+(1)(8))/(sqrt(2^(2)+2^(2)+1^(2))sqrt(4^(2)+1^(2)+8^(2)))=(18)/(3xx9)=(2)/(3)`
`therefore ” “theta=cos^(-1) ((2)/(3))`