Find the distance of the point (1,2,5) from the plane `vecr.(hati+hatj+hatk)+17=0`

Options

A. `25/sqrt(2)` units
B. `25/sqrt(3)` units
C. `25sqrt(2)` units
D. `25sqrt(3)` units

Answer

b

Explanation

We know that the perpendicular distance of a point have position vector `veca` from the plane `vecr. Vecn=q` is given by ,
`p=|veca.vecn-q|/(|vecen|)`
Here, `veca=(hati+2hatj+5hatk), vecn=(-hati-hatj-hatk)` and q=17.
`therefore p=|(hati+2hatj+5hatk).(-hatk-hatj-hatk)-17|/|-hati-hatj-hatk|`
`=|(-1)+(-2)+(-5)+(-7)|/sqrt((-1)^(2)+(-1)^(2)+(-1)^(2))=25/sqrt(3)` units