|Class 12 Maths Three Dimensional Geometry||Distance of a Point from a Plane: vector|
Distance of a Point from a Plane: vector
Consider a point P with position vector a ⃗ and a plane π1 whose equation is r->. nˆ=d.
Consider a plane π2 through P parallel to the plane π1.
The unit vector normal to π2 = nˆ.
Therefore it equation will be (r-> - a->). nˆ=0. i.e. r->. nˆ=a ⃗. nˆ
Therefore, the distance PQ from the plane π1 is (Fig. (a)), i.e.
ON –ON’= | d - (a ⃗. nˆ) |
This gives the length of the perpendicular from a point to the given plane.
If the equation of the plane π2 is in the form r->. N =d, where N= normal to the plane, then the perpendicular distance is given as | ((a ⃗.N)-d)/N|.
The length of the perpendicular from the origin O to the plane
r->. N=d =| d |/| N| (since a ⃗=0).
Let P(x1, y1, z1) is the given point with position vector a ⃗.
The Cartesian equation is given by Ax+By+Cz+D=0.
Then a ⃗ = x1î +y1ĵ +z1k̂ and N=A î +B1 ĵ +C k̂
Therefore by using the result | ((a ⃗.N)-d)/N|, the perpendicular from P to the plane is
Problem: -Find the distance of a point (2, 5, – 3) from the plane r->. (6î -3ĵ +2k̂)=4?
Answer: - Here, a ⃗=2î +5ĵ -3k̂, N =6î -3ĵ +2k̂ and d=4.
Therefore, the distance of the point (2, 5, – 3) from the given plane is