The angle between the lines `(x-1)/1=(y-1)/1=(z-1)/2a n d(x-1)/(-sqrt(3)-1)=(y-1)/(sqrt(3)-1)=(z-1)/4` is `cos^(-1)(1/(65))` b. `pi/6` c. `pi/3` d. `pi/4`

Explanation

We have to find the angle between the lines `(x-1)/1=(y-1)/1=(z-1)/2a n d(x-1)/(-sqrt(3)-1)=(y-1)/(sqrt(3)-1)=(z-1)/4`

Now the direction ratios of the given lines are proportional to 1,1,2 and -sqrt3 -1, sqrt3 -1, 4

And the lines are parallel to vectors

`vec(b_1)=hat(i)+hat(j)+2hat(k)` and `vec(b_2)=(-sqrt3 -1)hat(i)+(sqrt3 -1)hat(j)+4hat(k)`

Then the angle between the given lines is given by

`cos theta=(vec(b_1).vec(b_2))/(|vec(b_1)||vec(b_2)|)`

`cos theta=((hat(i)+hat(j)+hat(k)).((-sqrt3 -1)hat(i)+(sqrt3 -1)hat(j)+4hat(k)))/(sqrt(1^2+1^2+1^2)sqrt((-sqrt3 -1)^2+(sqrt3-1)^2+4^2))`

`cos theta=(-sqrt3 -1+sqrt3-1+8)/(sqrt3 sqrt24)`

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