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Question
Let the angle between two nonzero vectors \[\vec{A}\] and \[\vec{B}\] be 120° and its resultant be \[\vec{C}\].
Options
C must be equal to \[\left| A - B \right|\]
C must be less than \[\left| A - B \right|\]
C must be greater than \[\left| A - B \right|\]
C may be equal to \[\left| A - B \right|\]
Solution
C must be less than \[\left| A - B \right|\]
Here, we have three vector A, B and C.
\[\left| \vec{A} + \vec{B} \right|^2 = \left| \vec{A} \right|^2 + \left| \vec{B} \right|^2 + 2 \vec{A} . \vec{B} . . . (i)\]
\[ \left| \vec{A} - \vec{B} \right|^2 = \left| \vec{A} \right|^2 + \left| \vec{B} \right|^2 - 2 \vec{A} . \vec{B} . . . (ii)\]
Subtracting (i) from (ii), we get:
\[\left| \vec{A} + \vec{B} \right|^2 - \left| \vec{A} - \vec{B} \right|^2 = 4 \vec{A} . \vec{B}\]
Using the resultant property \[\vec{C} = \vec{A} + \vec{B}\],we get:
\[\left| \vec{C} \right|^2 - \left| \vec{A} - \vec{B} \right|^2 = 4 \vec{A} . \vec{B} \]
\[ \Rightarrow \left| \vec{C} \right|^2 = \left| \vec{A} - \vec{B} \right|^2 + 4 \vec{A} . \vec{B} \]
\[ \Rightarrow \left| \vec{C} \right|^2 = \left| \vec{A} - \vec{B} \right|^2 + 4\left| \vec{A} \right|\left| \vec{B} \right|\cos120^\circ\]
Since cosine is negative in the second quadrant, C must be less than \[\left| A - B \right|\] .
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