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

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.

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

i think it's ф or x=6,-6

step-by-step explanation:

-x^2=-36

x^2=-36

ф

or

if you mean by : -(x^2)=-36

it'll be: x^2=36

x=6,x=-6

hope it'll ! have a good day!

you for asking!