If the angle between the planes `r.(mhati-hatj+2hatk)+3=0 and r*(2hati-mhatj-hatk)-5=0` is `(pi)/(3)`, then m=

Options

A. 2
B. `+-3`
C. `3`
D. `-2`

Answer

C

Explanation

Let `n_(1) and n_(2)` are normals to the planes ,brgt `F*(mhati-hatj+2hatk)+3=0` and
`r*(2hati-mhatj-hatk)-5=0`, respectively.
Also, `theta=(pi)/(3)` be the angle between the given planes.
we know that, `costheta=|(n_(1)*n_(2))/(|n_(1)|n_(2)|)|`
`therefore"cos"(pi)/(3)=((mhati-hatj+2hatk)*(2hati-mhatj-hatk))/((sqrt(m^(2)+(-1)^(2)+2^(2)))(sqrt(2^(2)+(-m)^(2)+(-1)^(2))))`
`implies(1)/(2)=|(2m+m-2)/(sqrt(m^(2)+1+4)sqrt(4+m^(2)+-1))|" "[because"cos"(pi)/(3)=(1)/(2)]`
`=|(3m-2)/(sqrt((m^(2)+5)^(2)))|`
`implies+-(1)/(2)=(3m-2)/(m^(2)+5)`
`implies+-(m^(2)+5)=6m-4`
`impliesm^(2)+5=6m-4` or `-m^(2)-5=-6m-4`
`m^(2)-6m+9=0` or `m^(2)=6m+1=0`
`(m-3)^(2)=0` or `m^(2)+6m+1=0`
But `m^(2)+6m+1=0` not give real values.
`therefore(m-3)^(2)=0`
`impliesm=3`