Find a vector ü normal to the plane 2x + 2y + 2 = 3
1 answer:
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
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Answer and explanation:
Given : Strands Copper wire from a manufacturer are analyzed forstrenghth and conductivity. The Results from 100 strands are as follows :
Strength Strength
High Low
High conductivity 74 8
Low conductivity 15 3
To find :
a) If a stand is randomly selected, the probability that is conductivity is high and its strength is high
The favorable outcome is 74
The probability is given by,
b) If a stand is randomly selected, the probability that its conductivity is low or strength is low
Conductivity is low A= 15+3=18
Strength is low B= 8+3=11
Conductivity is low and strength is low
Probability is given by,
c) Consider the event that a strand low conductivity and the event that the strand has a low strength. Are these tow events mutually exclusive?
Since the events the stand has low conductivity and the stand has low strength are not mutually exclusive, since there exists some cases in which both the events coincide. i.e. Intersection of both the events exists with probability 0.03.