Find the angel between the lines `2x=3y=-za n d6x=-y=-4zdot`

Explanation

We have to find the angle between the lines `2x=3y=-z and6x=-y=-4z`

These lines can be written as

`x/(1/2)=y/(1/3)=z/(-1) and x/(1/6)=y/(-1)=z/(-1/4)`

Now the direction ratios of the given lines are proportional to `1/2,1/3,-1` and `1/6,-1,-1/4`

And the lines are parallel to vectors

`vec(b_1)=1/2hat(i)+1/3hat(j)-1hat(k)` and `vec(b_2)=1/6hat(i)-1hat(j)-1/4hat(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)|)`

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