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Find vec a . vec b, when vec a = hat i ...

Find `vec a . vec b`, when `vec a = hat i +3 hat j + hat k` and `vec b = 2 hat i - hat j - hat k`

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The correct Answer is:
a, c
Plane `P_(1)` contains the line
`vecr=hati+hatj+hatk+lamda(hati-hatj-hatk)`, hence contains the point `hati+hatj+hatk` and is normal to vector `(hati+hatj)`.
Hence, equation of plane is
`" "(vecr-(hati+hatj+hatk))*(hati+hatj)=0`
or `" "x+y=2`
Plane `P_(2)` contains the line
`" "vecr=hati+hatj=hatk+lamda(hati-hatj-hatk)` and point `hatj`
Hence, equation of plane is
`|{:(x-0,,y-1,,z-0),(1-0,,1-1,,1-0),(1,,-1,,-1):}|=0`
or `" "x+2y-z=2`
If `theta` is the acute angle between `P_(1) and P_(2)`, then
`" "costheta= (vec(n_1)*vec(n_2))/(|vec(n_1)||vec(n_2)|)=|((hati+hatj)*(hati+2hatj-hatk))/(sqrt2*sqrt6)|`
`" "= (3)/(sqrt2*sqrt6)= (sqrt(3))/(2)`
`" "theta=cos^(-1)""(sqrt(3))/(2)= (pi)/(6)`
As `L` is containec in `P_(2) rArr theta=0`
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