Question

Find the equation of a plane which is at a distance `3sqrt(3)` units from origin and the normal to which is equally inclined to coordinate axis.

Answer 1

Since, the normal to the plane is equally inclined to the axes

∴ cos α = cos β = cos ϒ

⇒ cos²α + cos²α + cos²α = 1

⇒ 3 cos²α = 1

⇒ cos α = `1/sqrt(3)`

⇒ cos α = cos β = cos ϒ = `1/sqrt(3)`

So, the normal is `vec"N" = 1/sqrt(3)hat"i" + 1/sqrt(3)hat"j" + 1/sqrt(3)hat"k"`

∴ Equation of the plane is `vec"r" . vec"N"` = d

⇒ `vec"r" . vec"N"/|vec"N"|` = d

⇒ `(vec"r"*(1/sqrt(3)hat"i" + 1/sqrt(3)hat"j" + 1/sqrt(3)hat"k"))/1 = 3sqrt(3)`

⇒ `vec"r"*(1/sqrt(3)hat"i" + 1/sqrt(3)hat"j" + 1/sqrt(3)hat"k") = 3sqrt(3)`

⇒ `(xhat"i" + yhat"j" + zhat"k") * 1/sqrt(3) (hat"i" + hat"j" + hat"k") = 3sqrt(3)`

⇒ x + y + z = `3sqrt(3) * sqrt(3)`

⇒ x + y + z = 9

Hence, the required equation of plane is x + y + z = 9.