What is the square root of 1200?
2 answers:
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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. <em>If the dot product of the two vectors is zero then the vectors are normal.</em>
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