Let G be the plane defined by 2x 3y 5z 0 Find the point p

Let G be the plane defined by 2x + 3y + 5z = 0. Find the point p on G that is closest to the point q = (1, 1, 1). Show that the vector going from p - q is orthogonal to every vector lying in G. Find the equation of the plane passing through (1, 1, 1) that is parallel to G. Find 2 vectors in R^3 whose span is equal to G.

Solution

Line parallel to normal to plane passing through given point is

(1,1,1)+t(2,3,5)

(1+2t,1+3t,1+5t)

We need to find intersection of this line with given plane Substitute this in the plane and we get

2(1+2t)+3(1+3t)+5(1+5t)=0

10+4t+9t+25t=0

10+38t=0

t=-5/19

So point on plane closest to the q is

p=(1+2t,1+3t,1+5t)=(9/19,4/19,-6/19)

b)

Vector from p to q is

(2t,3t,5t) which is orthogonal to the plane as it is parallel to the normal vector of the plane

c)

The required equation is

2x+3y+5z=c

Substituting (1,1,1) we get

2+3+5=c=10

So the equation of plane is

2x+3y+5z=10

d)

On G we have

x=-3y/2-5z/2

(x,y,z)=(-3y/2-5z/2,y,z)=y(-3/2,1,0)+z(-5/2,0,1)

So two vectors that span G are

{(-3/2,1,0),(-5/2,0,1)}