What is the equation of the plane which passes through the z-axis and is perpendicular to the line `(x-a)/(costheta)=(y+2)/(sintheta)=(z-3)/0?`

Options

A. `x+ytantheta=0`
B. `y+xtantheta=0`
C. `xcostheta-ysintheta=0`
D. `xsintheta-ycostheta=0`

Answer

a

Explanation

The plane is perpendicular to the line
`" "(x-a)/(costheta)=(y+2)/(sintheta)=(z-3)/(0)`
Hence, the direction ratios of the normal of the plane are `costheta, sintheta and 0`. `" "` (i)
Now, the required plane passes through the z-axis.
Hence, the point `(0, 0, 0)` lies on the plane.
From Eqs. (i) and (ii) , we get equation of the plane as
`" "costheta (x-0)+ sin theta(y-0) + 0(z-0) =0`
`" "costhetax + sin theta y =0`
`" "x+ytantheta=0`