CBSE-XI-Physics
02: Physics and Mathematics
- #3Let the angle between two nonzero vectors
A→and B→be 120° and its resultant be
C→.
-(a) C must be equal to
A-B
(b) C must be less than
A-B
(c)
C must be greater than
A-B
(d) C may be equal to
A-BdigAnsr: cAns : (b) C must be less than `` \left|A-B\right|``
Here, we have three vector A, B and C.
`` {\left|\stackrel{\to }{A}+\stackrel{\to }{B}\right|}^{2}={\left|\stackrel{\to }{A}\right|}^{2}+{\left|\stackrel{\to }{B}\right|}^{2}+2\stackrel{\to }{A}.\stackrel{\to }{B}...\left(i\right)
{\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}={\left|\stackrel{\to }{A}\right|}^{2}+{\left|\stackrel{\to }{B}\right|}^{2}-2\stackrel{\to }{A}.\stackrel{\to }{B}...\left(ii\right)``
Subtracting (i) from (ii), we get:
`` {\left|\stackrel{\to }{A}+\stackrel{\to }{B}\right|}^{2}-{\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}=4\stackrel{\to }{A}.\stackrel{\to }{B}``
Using the resultant property `` \stackrel{\to }{C}=\stackrel{\to }{A}+\stackrel{\to }{B}``, we get:
`` {\left|\stackrel{\to }{C}\right|}^{2}-{\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}=4\stackrel{\to }{A}.\stackrel{\to }{B}
\Rightarrow {\left|\stackrel{\to }{C}\right|}^{2}={\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}+4\stackrel{\to }{A}.\stackrel{\to }{B}
\Rightarrow {\left|\stackrel{\to }{C}\right|}^{2}={\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}+4\left|\stackrel{\to }{A}\right|\left|\stackrel{\to }{B}\right|\mathrm{cos}120°``
Since cosine is negative in the second quadrant, `` C``must be less than `` \left|A-B\right|``.
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