 ## Question

junior college 2 | H2 Maths

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##### Alina

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Date Posted: 1 week ago
Views: 6
J
1 week ago
Equation of plane : r.n = a.n , where

r is the position vector of any point on the plane.
a is the position vector of a fixed point on the plane.
n is a vector perpendicular to the plane.

There are many planes which R can lie on. Let's find the equation of a plane where OR lies in.

Let AB be perpendicular to this plane. Since AR = BR, let R be the midpoint of AB. So the plane bisects AB, and AB is perpendicular to OR and bisects OR.

The result is an isosceles triangle OAB, where length of OA is equal to OB. It is part of a plane that is orthogonal/perpendicular to the plane π.

vector BA can be used as the normal to the plane.

vector BA = vector OA - vector OB
= a - b

(If we compare the equation to be shown with the general equation of the plane, we can see intuitively that a - b is the normal vector to be used)

We use vector OR as the position vector of the point that lies in the plane.

vector OR = vector OB + vector BR

= vector OB + ½BA
(BR is part of BA. It has the same direction but half the magnitude)

= b + ½(a - b)
= ½(a + b)

So equation of the plane is :

r.(a - b) = ½(a + b).(a - b)

2r.(a - b) = (a + b).(a - b)

(Shown)