Q. 85.0( 1 Vote )

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

Answer :

We are given that,

A plane is at a distance of 3√3 units from the origin.


Also, the normal is equally inclined to coordinate axis.


We need to find the equation of the plane.


We know that, vector equation of a plane at a distance d from the origin is given by




lx + my + nz = d …(i)


where,


l, m and n are direction cosines of the normal of the plane.


And since, normal is equally inclined to coordinate axis, then


l = m = n


cos α = cos β = cos γ …(ii)


Also, we know that,


cos2 α + cos2 β + cos2 γ = 1


cos2 α + cos2 α + cos2 α = 1 [from (ii)]


3cos2 α = 1




This means,



Substituting values of l, m and n in equation (i), we get



Here, d = 3√3


So,




x + y + z = 3√3 × √3


x + y + z = 3 × 3


x + y + z = 9


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


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