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.

Note:-

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).

Cartesian form:-

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

= 13/7

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