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`r_(2)` are normals to the planes
`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, `cos theta = |(n_(1).n_(2))/(|n_(1)||n_(2)|)|`
`:. cos 'pi/3 ((mhati-hatj+2hatk).(2hati-3hatj-hatk))/((sqrt(m^(2)+(-1)^(2)+2^(2))))`
`(sqrt(2^(2)+(-m)^(2)+(-1)^(2)))`
`rArr 1/2 =|(2m+m-2)/(sqrt(m^(2)+1+4)sqrt(4+m^(2) +1))|`
`[ :' cos'pi/3=1/2]`
`= |(3m-2)/(sqrt((m^(2)+5)^(2)))|`
`rArr +-1/2=(3m-2)/(m^(2)+5)`
`rArr+-(m^(2)+5)=6m-4`
`rArrm^(2)+5=6m-4`
or `-m^(2)-5=6m-4`
`m^(2)-6m+9=0` or `m^(2)+6m+1=0`
But `m^(2)+6m+1=0` not give real values.
`:. (m-3)^(2)=0`
`rArr m = 3`