If the direction ratios of two lines are given by `a+b+c=0 and 2ab+2ac-bc=0`, then the angle between the lines is

Options

A. `pi`
B. `(2pi)/3`
C. `(pi)/2`
D. `(pi)/3`

Answer

B

Explanation

We have
`a+b+c=0` and `2ab+2ac-bc=0`
`impliesa=-(b+c)` and `2a(b+c)-bc=0`
`implies-2(b+c)^(2)-bc=0`
`implies2b^(2)+5bc+2c^(2)=0`
`implies(2bc+c)(b+2c)=0`
`implies2b+c=0` or `b+2c=0`
If `2b+c=0` then `a=-(b+c)impliesa=b`
`:.a=b` and `c=-2bimpliesa/1=b/1=c/(-2)`
If `b+2c=0` then `a=-(b+c)impliesa=c`
`:.a=c` and `b=-2cimpliesa/1=b/(-2)=c/1`
Thus, the direction ratios of two lines are proportional to 1,1,-2 and 1,-2,1 respectively. So, the angle `theta` between them is given by
`cos theta=(1-2-2)/(sqrt(1+1+4)sqrt(1+4+1))=(-1)/2impliestheta=(2pi)/3`