Correct answer: A. It is possible to have | C | 90^ ^ ^ ), the term 2| A || B | θ becomes negative, which can make | C | smaller than both | A | and | B | . For example, if A = 5 i and B = -6 i , then C = - i . Here, | A | = 5 , | B | = 6 , and | C | = 1 . Thus, | C | < | A | and | C | < | B | is possible.

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Physics and Mathematics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Physics and Mathematics

Suppose C = A + B .

Suppose

\vec{C} = \vec{A} + \vec{B}

Options

AIt is possible to have $|\vec{C}| < |\vec{A}|$ as well as $|\vec{C}| < |\vec{B}|$
B $|\vec{C}|$ always equals $|\vec{A}| + |\vec{B}|$
C $|\vec{C}|$ never equals $|\vec{A}| + |\vec{B}|$
D $|\vec{C}|$ is always larger than $|\vec{A}|$

Correct answer

A. It is possible to have $|\vec{C}| < |\vec{A}|$ as well as $|\vec{C}| < |\vec{B}|$

Step-by-step solution

Given

\vec{C} = \vec{A} + \vec{B}

.

The magnitude of

\vec{C}

is given by

|\vec{C}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta}

, where

\theta

is the angle between

\vec{A}

and

\vec{B}

.

If the angle between the vectors is obtuse (

\theta > 90^{{}^{{}^{\circ}}}

), the term

2|\vec{A}||\vec{B}|\cos\theta

becomes negative, which can make

|\vec{C}|

smaller than both

|\vec{A}|

and

|\vec{B}|

.

For example, if

\vec{A} = 5\hat{i}

and

\vec{B} = -6\hat{i}

, then

\vec{C} = -\hat{i}

.

Here,

|\vec{A}| = 5

|\vec{B}| = 6

, and

|\vec{C}| = 1

.

Thus,

|\vec{C}| < |\vec{A}|

and

|\vec{C}| < |\vec{B}|

is possible.

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