ABCDEF is a regular hexagon and forces represented in magnitude and direction by AB, AC,A

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3-1.Vectors

medium

$ABCDEF$ is a regular hexagon and forces represented in magnitude and direction by $AB, AC,AD, AE$ and $AF$ act at $A$. Their resultant is :

$6\, AD$

$4\, AD$

$3\, AD$

$2\, AD$

Solution

$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{AE}}+\overrightarrow{\mathrm{AF}}$

$=(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AE}})+(\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{AF}})+\overrightarrow{\mathrm{AD}}$

$=\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{AD}}=3 \overrightarrow{\mathrm{AD}}$

Standard 11

Physics

For the given vector $\vec A =3\hat i -4\hat j+10\hat k$ , the ratio of magnitude of its component on the $x-y$ plane and the component on $z-$ axis is

medium

Two forces $P + Q$ and $P -Q$ make angle $2 \alpha$ with each other and their resultant make $\theta$ angle with bisector of angle between them. Then :

hard

The magnitude of pairs of displacement vectors are given. Which pair of displacement vectors cannot be added to give a resultant vector of magnitude $13\, cm$?

easy

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