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QJEEAdv08Ma358c: JEE Advanced 2008 Mathematics - Class 11 JEEAdv08Ma358c Solution: A particle P stats from the point = 1 +2i, where . It moves…

A particle P stats from the point z0{z_0} = 1 +2i, where i=−1i = \sqrt { - 1} . It moves horizontally away from origin by 5 unit and then vertically away from origin by 3 units to reach a point z1{z_1} . From z1{z_1} the particle moves 2\sqrt 2 units in the direction of the vector i^+j^\hat i + \hat j and then it moves through an angle π2{\pi \over 2} in anticlockwise direction on a circle with centre at origin, to reach a point z2{z_2} . The point z2{z_2} is given by

Multiple Choice Options

  • A 6 + 7i
  • B -7 + 6i
  • C 7 + 6i
  • D - 6 + 7i

Step-by-Step Verified Solution

given −- α\alpha particle

In the direction of, or, Now rotation about origin through angle of means multiply by

z0=1+2i=(1,2)=(x0,y0){z_0} = 1 + 2i = (1,2) = ({x_0},{y_0})

z1=(x0+5,y0+3){z_1} = ({x_0} + 5,{y_0} + 3)

=(6,5)=6+5i= (6,5) = 6 + 5i

⇒2\Rightarrow 2 in the direction of i=ji = j

x1=2cos⁡45∘{x_1} = 2\cos 45^\circ

y1=2sin⁡45∘{y_1} = 2\sin 45^\circ

z2(7+6i){z_2}(7 + 6i)

Now, rotation about origin through an angle of 2 π\pi means multiply z 2_2 by →eiπ/2\to {e^{i\pi /2}}

cos⁡90∘+isin⁡90∘=i\cos 90^\circ + i\sin 90^\circ = i

z3=i(7+6i){z_3} = i(7 + 6i)

=−6+7i= - 6 + 7i

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