First slide

Motion in a plane

Question

It is found that | $\vec A\$ + $\vec B\$ | = | $\vec A\$ | . This necessarily implies.

Difficult

A

$\vec B$ must be zero
B

$\vec A$ , $\vec B$ are antiparallel
C

$\vec A$ and $\vec B$ are perpendicular
D

$\vec A$ . $\vec B$ < 0

Solution

If α be the angle between $\vec{A} and \vec{B}$ , then R =
$\large \sqrt {{A^2} + {B^2} + 2AB\cos \alpha }$
'R' may be equal to A only when α is obtuse.
In that Case $\vec{A} . \vec{B}$ = AB cosα < 0
So option -(4) is correct.

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