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.

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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