Question

if a particle is displaced from (0,0,0) to a point in xy plane which is at a distance of 4 units in a direction making an angle clock wise 60° with negative x axis. what is final position of vector​

Answer 1

Explanation:

component along x-axis

ax=4cos60°=2

ay=4sin60°=2root3

clockwise direction

with negative x axis

a = ax(-i^) +ay(j^)

=-2i^+2root3j^

Answer 2

Answer:

The final position of vector is

$- 2i + 2 \sqrt{3} j$

Explanation:

Given that,

The particle is initially at (0,0,0)

Later it is displaced in XY Plane at a distance=4units

The angle made with negative x-axis is 60°

According to resolution of vectors,

The horizontal component of vector is

$- 4cos60 = - 2 \: ( - ve \: x - axis)$

The vertical component of vector is

$4sin60 = 2 \sqrt{3}$

Therefore,the final position of Vector is written as

$- 2i + 2 \sqrt{3} j$

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