The distance of a plane `Ax+By+Cz=D` from the point `(x_(1),y_(1),z_(1))` is

Options

A. `|{:Ax_(1)+By_(1)+Cz_(1)-D:}|`
B. `|{:(Ax_(1)+By_(1)+Cz_(1)-D)/(A^(2)+B^(2)+C^(2)):}|`
C. `|{:(Ax_(1)+By_(1)+Cz_(1)-D)/(sqrt(A^(2)+B^(2)+C^(2))):}|`
D. `|{:(Ax_(1)+By_(1)+Cz_(1))/(sqrt(A^(2)+B^(2)+C^(2))):}|`

Answer

C

Explanation

Let `P(x_(1),y_(1),z_(1))` be the given points with position vector a and `Ax+By+Cz=D` be the cartesian equation of the given plane. Then,
`a=x_(1)hat(i)+y_(1)hat(j)+z_(1)hat(k)`
`n=Ahat(i)+Bhat(j)+Chat(k)`
The perpendicular from P to the plane is `(|a*n-d|)/(|n|)`
`=|{:((x_(1)hat(i)+y_(1)hat(j)+z_(1)hat(k))*(Ahat(i)+Bhat(j)Chat(k))-D)/(sqrt(A^(2)+B^(2)+C^(2))):}|`
`=|{:(Ax_(1)+By_(1)+Cz_(1)-D)/(sqrt(A^(2)+B^(2)+C^(2))):}|`