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Lecture 5 MATH 127 (Maj Sultana)

 

MATH 127 

Vector Analysis


    1) Prove that,  |A.B|2+|A×B|2=|A|2.|B|2  
Solution|A.B|2=|A|2+|B|2cos2θ|A×B|2=|A|2.|B|2sin2θ|ˆn|2=|A|2.|B|2sin2θ L.S=|A|2.|B|2(sin2θ+cos2θ)=|A|2|B|2= R.S  
    2) Prove that,  A.(B×C)=B.(C×A)=C.(A×B)  
Solution: [Hint:   Let,  A=A1ˆi+A2ˆj+A3ˆkB=B1ˆi+B2ˆj+B3ˆkC=C1ˆi+C2ˆj+C3ˆk  


   ♦ Reciprocal sets of vectors: 
the set a,b,c and a,b,c are called reciprocal sets of reciprocal system of vectors if  a.a=b.b=c.c=1           a=1aand, ab=ac=ba=bc=ca=cb=0                                                       
    That is, each vector is orthogonal to the reciprocal of the other two vectors in the system.  a=1a  
    

   ♦ Theorem: The sets a,b,c and a,b,c are reciprocal sets of vectors if and only if  a=b×ca.b×cb=c×aa.b×c         |where,a.b×c0c=a×ba.b×c a.b×c=b.c×a=c.a×b  

    Example: Find a set of vectors reciprocal to the set vector. 
Solution: Let,  a=2ˆi+3ˆjˆkb=ˆiˆj2ˆkc=ˆi+2ˆj+2ˆk   Now,(b×c=|ˆiˆjˆk112122)|=2ˆi+ˆk    a.(b×c)=41=3    (c×a)=|ˆiˆjˆk122231|=8ˆi+3ˆj7ˆk  (a×b)=|ˆiˆjˆk231112|=7ˆi+3ˆj5ˆk  a=b×ca.b×c=23ˆi+13ˆkb=c×aa.b×c=83ˆi+ˆj73ˆkc=a×ba.b×c=73ˆi+ˆj53ˆk  


    • Direction Cosine:  cosα=x|r|=xx2+y2+z2 cosβ=y|r|=yx2+y2+z2 cosγ=z|r|=zx2+y2+z2  

    • Direction Ratio:  r=xˆi+yˆj+zˆkx:y:zˆl=xrˆm=yrˆn=zr l2+m2+n2=1  

    ♦ Vector linear dependency and linear independence: 
Vectors  A1,A2,A3,An are linearly dependent if there exist scalers  a1,a2,a3an not all zero, such that,
  A1a1+A2a2+A3a3+Anan=0   Linear Combination  
Otherwise, the vectors are linearly independent.

    Example: Find the linear combination of the vectors  a=ˆi+ˆjˆk, b=ˆiˆj+ˆk and c=ˆi+ˆj+ˆk . Test whether they are linearly dependent or not.
Solution: Let,  x,y,z are scalers.
The linear combination is, 
  xa+yb+zc=x(ˆi+ˆjˆk)+y(ˆiˆj+ˆk)+z(ˆi+ˆj+ˆk)=ˆi(x+y+z)+ˆj(xy+z)+ˆk(x+y+z)  
This will be linearly dependent if,  ˆi(x+y+z)+ˆj(xy+z)+ˆk(x+y+z)=0  
  x+yz=0zy+z=0x+yz=0 x=y=z=0  



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