A direction perpendicular to the plane 2x-y+z=9

Question

The question asks.. A direction perpendicular to the plane 2x-y+z=9.

It gives me four answers to chose from;

a)  (2x, -y, z)     b)  (4,0,1)     c) (1,1,-1)     and d)  (2,-1,1)

 

How do i find out the answer?

Answer 1

I think it's d.

Consider a point in the plane P(x₀,y₀,z₀) then 2x₀-y₀+z₀=9.

The point becomes a vector v=<x₀,y₀,z₀> in relation to an origin O.

Let Q(x,y,z) be another point on the plane. Its vector is u=<x,y,z>.

OP+PQ=OR, so v+PQ=u, PQ=u-v=<x-x₀,y-y₀,z-z₀>.

The normal n to PQ is the normal to the plane and the dot product n.PQ=0:

n.<x-x₀,y-y₀,z-z₀>=0. Let n=<a,b,c>, then a(x-x₀)+b(y-y₀)+c(z-z₀)=0.

That is, ax+by+cz=ax₀+by₀+cz₀. The RHS is a constant and we have the equation of the plane, so:

ax₀+by₀+cz₀=9 and  ax+by+cz=2x-y+z, making n=<2,-1,1>.