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 π_{1} whose equation is r^{->}. nˆ=d.

Consider a plane π_{2} through P parallel to the plane π_{1}.

The unit vector normal to π_{2} = nˆ.

Therefore it equation will be (r^{->} - a^{->}). nˆ=0. i.e. r^{->}. nˆ=a ⃗. nˆ

Therefore, the distance PQ from the plane π_{1} 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 π_{2} 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(x_{1}, y_{1}, z_{1}) is the given point with position vector a ⃗.

The Cartesian equation is given by Ax+By+Cz+D=0.

Then a ⃗ = x_{1}î +y_{1}ĵ +z_{1}k̂ and N=A î +B_{1} 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

** Answer: -** Here, a ⃗=2î +5ĵ -3k̂, N

Therefore, the distance of the point (2, 5, – 3) from the given plane is

= 13/7

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