First slide

Scalars and Vectors

Question

Let $\vec{C} = \vec{A} + \vec{B}$ then

Easy

A

$|\vec{C}| is always greater than | \vec{A} |$
B

$It is possible to have | \vec{C} | < | \vec{A} | and | \vec{C} | < | \vec{B} |$
C

$C i s a l w a y s e q u a l t o A + B$
D

$C i s n e v e r e q u a l t o A + B$

Solution

$\vec{C} = \vec{A} + \vec{B}$
The value of C lies between $|A - B|$ and $|A + B|$
Let say $\vec{A} = 10 \hat{i}$ and $\vec{B} = - 8 \hat{i}$
then $\vec{C} = 2 \hat{i}$
$∴ | \vec{C} | < | \vec{A} | or | \vec{C} | < | \vec{B} |$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App