Question

Find a vector ü normal to the plane 2x + 2y + 2 = 3

Answer 1

Answer:

ü=2i+2j+0k

Step-by-step explanation:

The given plane 2x + 2y + 2 = 3 can also be written as:

2x+2y=3-2

2x+2y=1

The general equation for a plane is Ax+By+Cz=D and by definition the normal vector of that plane is n=Ai+Bj+Ck

Where i,j,k are the unit vectors

In order to demostrate that the vector n is normal to the plane, let R1=(a1,b1,c1) and R2=(a2,b2,c2) be two vectors that are in the plane.

If R1 ∈ Ax+By+Cz=D then Aa1+Bb1+Cc1=D

If R2 ∈ Ax+By+Cz=D then Aa2+Bb2+Cc2=D

Therefore, the vector R1R2=R2-R1=(a2-a1)i+(b2-b1)j+(c2-c1)k

You can apply the dot product. If the dot product of the two vectors is zero then the vectors are normal.

$R1R2_{o}n= [(a2-a1)i+(b2-b1)j+(c2-c1)k]_{o}(Ai+Bj+Ck)\\R1R2_{o}n = A(a2-a1) + B(b2-b1) + C(c2-c1)\\R1R2_{o}n = Aa2 + Bb2 +Cc2 - (Aa1+Bb1+Cc1)\\R1R2_{o}n = D - D\\R1R2_{o}n = 0$

So, the vector which components are A,B,C is normal to the plane becase it is normal to any vector contained in the plane.

In this case:

A=2, B=2, C=0

ü=2i+2j+0k