18 Show that the set of all points in R3 lying in a planeis

18. Show that the set of all points in R3 lying in a planeis a vector space with respect to the standard operations of vector addi- tion and scalar multiplication if and only if the plane passes through the origin.

Solution

If the set of points in R³ lying in a plane i.e. W which passes through the point P₀(x₀,y₀,z₀) and W is perpendicular to a non-zero vector v = <v₁,v₂,v₃>

If P = (x,y,z) belongs to W

Then, the equation of plane, W is given by

(P-P₀).v = 0

On expanding above equation, we get

xv₁+yv₂+zv₃ = x₀v₁ + y₀v₂+z₀v₃

Suppose P₀(x₀,y₀,z₀) is not an origin

For M=(a,b,c) and N=(p,q,r) belongs to W

Now, (a+p)v₁+(b+q)v₂+(c+r)v₃ = av₁+bv₂+cv₃+pv₁+qv₂+rv₃

                                                     = 2(x₀v₁+y₀v₂+z₀v₃) is not equal to (x₀v₁+y₀v₂+z₀v₃)

That means, M+N doesnot belong to W.

Hence, W can\'t be vector space in R³ if P₀ is not an origin.

Suppose P₀ (x₀,y₀,z₀) is an origin then the equation of plane W becomes

xv₁+yv₂+zv₃ = 0 i.e.(x,y,z).(v₁,v₂,v₃) = 0 where (x,y,z) belongs to W

To show, W is a subspace in R³

Suppose M(x,y,z) and N(p,q,r) belongs to W and a,b belongs to R

(aM + bN).(v₁,v₂,v₃)

= (a(x,y,z) + b(p,q,r)).(v₁,v₂,v₃)

= (ax+bp,ay+bq,az+br).(v₁,v₂,v₃)

= (ax+bp)v₁ + (ay+bq)v₂ + (az+br)v₃

= a(xv₁+yv₂+zv₃) + b(pv₁+qv₂+rv₃)

= a*0 + b*0

= 0

Hence, (aM + bN) belongs to W

Therefore, W is a subspace of R³

That means W is a vector space since R³ is also a vector space.