A vector ` vec n` f magnitude 8 units is inclined to x-axis at `45^0` , y-axis at `60^0` and an acute angle with z-axis. If a plane passes through a point `(sqrt(2),-1,1)` and is normal to ` vec n` , find its equation in vector form.

Answer

`vecr.(sqrt(2)hati+hatj+hatk)=2`

Explanation

We know that `vecn=hatn/(|nvecn|)=(lhati+mhatj+nhatk)`.
Here, `l=cos45^(@)=1/sqrt(2),m=cos 60^(@)=1/2` and let `n=cosgamma)`. Then,
`l^(2)+m^(2)+n^(2)=1 rArr 1/2+1/4 +cos^(2)gamma=1`
`rArr cos^(2)gamma=1/4`
`rArr cosgamma=1/2`.
`therefore vecn=|vecn|(lhati+mhatj+nhatk) rArr vecn=8(1/sqrt(2)hati+1/2hatj+1/2hatk)=(4sqrt(2)hati+4hatj+4hatk)`.
So, the equation of the plane is
`vecr.vecn=veca.vecn rArr vecr.(4sqrt(2)hati+4hatj+4hatk)=(sqrt(2)hati+4hatj+4hatk)`.
So, the equation of the plane is
`vecr.vecn=veca.vecn rArr vecr.(4sqrt(2)hati+4hatj+4hatk)=(sqrt(2)hati-hatj+hatk).(4sqrt(2)hati+4hatj+4hatk)`
`rArr vecr.(4sqrt(2)hati+4hatj+4hatk)=(8-4+4)=8`.
`rArr vecr.(sqrt(2)hati+hatj+hatk)=2`.