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)
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
Substituting values of l, m and n in equation (i), we get
Here, d = 3√3
⇒ 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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