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Let the Angle Between Two Nonzero Vectors → a and → B Be 120° and Its Resultant Be → C . - Physics

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प्रश्न

Let the angle between two nonzero vectors \[\vec{A}\] and \[\vec{B}\] be 120° and its resultant be \[\vec{C}\].

पर्याय

  • 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 \[\left| A - B \right|\]

MCQ

उत्तर

 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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पाठ 2: Physics and Mathematics - MCQ [पृष्ठ २८]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 2 Physics and Mathematics
MCQ | Q 3 | पृष्ठ २८

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