The angle between the lines with direction ratios `(1,0,pmcosalpha)" is "60^@`. What is the value of `alpha`?

Options

A. `cos^(-1)(1//sqrt2)`
B. `cos^(-1)(1//sqrt3)`
C. `cos^(-1)(1//3)`
D. `cos^(-1)(1//2)`

Answer

B

Explanation

Let the angle between two lines whose direction ratios are `a_1,b_1,c_1 and a_2,b_2,c_2` respectively is 0.
Then , `costheta=(a_1a_2+b_1b_2+c_1c_2)/(sqrt(a_1^2+b_1^2+c_1^2)sqrt(a_2^2+b_2^2+c_2^2))`
`:.cos60^@=(1xx1+0xx0+(cosalpha)(-cosalpha))/(sqrt(1^2+(0)^2+cos^2alpha)sqrt(1^2+(0)^2+(-cosalpha)^2))`
`rArr 1/2(1-cos^2alpha)/(sqrt(1+cos^2alpha)sqrt(1+cos^2alpha))`
`1/2=(1-cos^2alpha)/(1+cos^2alpha)rArr(1+2)/(1-2)=2/(-2cos^2alpha)`
`rArr 3/(-1)=1/(-cos^2alpha)`
`rArr cosalpha=1/sqrt3rArralpha=cos^(-1)(1/sqrt3)`