A vector `vecn` is inclined to `x`-axis at `45^(@)`, to y-axis at `60^(@)` and at an angle to z-axis. If `vecn` is a normal to the plane passing through the point `(sqrt(2),-1,1)`, then the equation of plane is

Options

A. `3sqrt(2)x-4y-3z=7`
B. `4sqrt(2)x+7y+z=2`
C. `sqrt(2)x+y+z=2`
D. `sqrt(2)x-y-z=2`

Answer

C

Explanation

Let plane make angle `theta` with z-axis.
`therefore cos^(2)45^(@) + cos^(2)60^(@)+cos^(2)theta=1`
`therefore 1/2+1/4+cos^(2)theta=1`
`costheta=1/2`
`rArr theta=60^(@)`
So, equation of plane is
`(x-sqrt(2))/sqrt(2) +(y+1)/2+(z-1)/2=0`
`rArr x/sqrt(2)+y/2+z/2=1`
`rArr sqrt(2)x+y+z=2`