IIT JEE , AIEEE Forum - Mathematics , Vectors

rajat

1. z(bar) + i w(bar) = 0

and arg(zw) = pie

then find arg z

2. a vector of magnitude 1/4 is resolved into OA and OB such that it makes an angle of 30 with OA and 45 with OB . find magnitude of vector along OB

3. a line has direction cosiines proportional to 2,1,2 meets the lines

x = y+a = z and x+a = 2y = 2z .find the coordiantes of point of intersection for both the lines

prakhar.in

ans 2

for any vector R resolved in to part A and B such that it makes

a with A and

b with B then

|A| = Rsinb/(sin(a+b))

so OB = (1/4)sin30/sin75

Do rate me for my answer

aysh

1)Let z=(a+ib) & w=(c+id).
z(bar)+iw(bar)=0.
(a-ib)+i(c-id)=0.
(a+d)+i(b-c)=0.
Thus,
a=-d.
b=c.
Also, arg(zw)=pi.
Therefore, img. part of zw is 0.
(a+ib).(c+id)=(ac-bd)+i(bc+ad)
Clearly,
bc+ad=0.
Using above eq.s we get,
-(b^2)+(a^2)=0.
a^2=b^2
a=b or, a=-b
tan[arg(z)]=b/a.
=+1 or, -1
arg(z)=pi/4 or, -pi/4.

PLEAZZZZ.. RATE IF YOU LIKE MY EFFORT...

nigitha_17

i do accept with aysh but plz see this

ifwe consider z=(x,y) w=(a,b)

we get x=-b,a=y (as mentioned above )

consi,

arg(zw)=arg(z)+arg(w) (standared result)

implies tan^-1(-y/x)+tan^-1(b/a)

implies tan^-1(-y/x)+tan^-1(-x/y)=pi (given)

implies tan^-1(inf)=pi

implies pi/2=pi

which is not possible

i think there is some mistake in the ques

N IF THERE IS ANY MIS IN MY METHOD PLZ INFORM ME

rajat

even i got pie/4

but the anwer is given is 3pie/4

and nigitha ,theres nothing wrong wid the question

ROHIIT

!} z<bar> = i w<bar.

-argz=argiw=pie/2-argw.......1

argz+argw=pie/2.................2

solve this get u r ans

ROHIIT

used formula as argac= arga+ argc

arga<bar> = -arga

please rate me good as method is absoulately correct

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