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प्रश्न
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.
उत्तर
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.
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