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 \[\left| A - B \right|\]

Answer 1

 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 \[\left| A - B \right|\] .