Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

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

MCQ

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 $\left| A - B \right|$ .

Concept: What is Physics?
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 2 Physics and Mathematics
MCQ | Q 3 | Page 28
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