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+A3ˆk→B=B1ˆi+B2ˆj+B3ˆk→C=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=1a′and, a′b=a′c=b′a=b′c=c′a=c′b=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×c≠0c′=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+3ˆj−ˆk→b=ˆi−ˆj−2ˆk→c=−ˆi+2ˆj+2ˆk Now,(→b×→c=|ˆiˆjˆk1−1−2−122)|=2ˆi+ˆk →a.(→b×→c)=4−1=3 (→c×→a)=|ˆiˆjˆk−12223−1|=−8ˆi+3ˆj−7ˆk (→a×→b)=|ˆiˆjˆk23−11−1−2|=−7ˆi+3ˆj−5ˆk ∴a′=b×ca.b×c=23ˆi+13ˆkb′=c×aa.b×c=−83ˆi+ˆj−73ˆkc′=a×ba.b×c=−73ˆi+ˆj−53ˆk
• Direction Cosine: cosα=x|→r|=x√x2+y2+z2 cosβ=y|→r|=y√x2+y2+z2 cosγ=z|→r|=z√x2+y2+z2
• Direction Ratio:
→r=xˆi+yˆj+zˆkx:y:zˆl=x→rˆm=y→rˆn=z→r l2+m2+n2=1
♦ Vector linear dependency and linear independence:
Vectors
→A1,→A2,→A3,……→An
are linearly dependent if there exist scalers
a1,a2,a3……an
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+ˆj−ˆk, →b=ˆi−ˆj+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,
x→a+y→b+z→c=x(ˆi+ˆj−ˆk)+y(ˆi−ˆj+ˆk)+z(−ˆi+ˆj+ˆk)=ˆi(x+y+z)+ˆj(x−y+z)+ˆk(−x+y+z)
This will be linearly dependent if,
ˆi(x+y+z)+ˆj(x−y+z)+ˆk(−x+y+z)=0
∴x+y−z=0z−y+z=0−x+yz=0 ∴x=y=z=0
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