A particle P starts from the point z_{0}=1+2 i, where i=\sqrt{-1}. It moves first horizontally away from the origin by 5 units and then vertically away from origin by 3 units to reach a point z_{1}. From z_{1} the particle moves \sqrt{2} units in the direction of the vector \hat{1}+\hat{j} to reach z_{2}, and then it moves through an angle \frac{\pi}{2} in an anticlock-wise direction on a circle with center at origin, to reach a point z_{3}. The point z_{3} is given by:

NIMCET 2019 Question Paper with solutions

Question : 3 of 120

Marks: +1, -0

A particle

P

starts from the point

z_{0}=1+2 i

, where

i=√{-1}

. It moves first horizontally away from the origin by 5 units and then vertically away from origin by 3 units to reach a point

z_{1}

. From

z_{1}

the particle moves

√{2}

units in the direction of the vector

{1}↖{∧}+{j}↖{∧}

to reach

z_{2}

, and then it moves through an angle

\;{π}/{2}

in an anticlock-wise direction on a circle with center at origin, to reach a point

z_{3}

. The point

z_{3}

is given by:

Solution:

Given

A=Z_0=1+2i

Moving horizontally 5 units w< reach at /?(IS f 2l), then moving vertically 3 units,we reach

C=Z _1(6+5i)

,then moves

√2

unit in the direction of

i+j

we reach at

D(7+6i)

and then moves through an angle

π/2

with anti-clockwise direction,we reach at ∊

Z_2(-6+7i)

∴ The point

Z_2=-6+7i

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