Question

An ant is moving on a plane, x+y+z=9. find its location on plane such that, it is closest from the origin​

Answer 1

Given: A plane, X+y+z= 9

To find: point on the plane closest to the origin

Solution: The given equation of plane can be written in the form ax+by+cz=d where, a= 1, b= 1, c= 1, d= 9

for a given equation of plane ( ax+by+cz=d) , points nearest to the origin is

(da/√(a2+b2+c2), db/√(a2+b2+c2), dc/√(a2+b2+c2))

putting the value of a, b, c, d from the given equation of plane we will get

(9/√3, 9/√3, 9/√3,)

That is (3√3, 3√3, 3√3)

and it's distance from the plane can be get by the following formula.

distance= (aPx+ bPy+ cPz- d) / √(a2+b2+c2)

where (Px, Py, Pz ) are the point closest to origin

By substituting the values,

distance= (1×3√3 + 1×3√3 + 1×3√3 - 9 )/ √3

distance= 3×3√3-9/√3 = 3√3(√3-1)

Therefore, the point on the plane x+y+z= 9 closest to the origin is (3√3, 3√3, 3√3) having it's distance from origin of 3√3(√3-1).