# If a^2 + b^2 = c^2 + d^2 = 1 and ac + bd = 0 , which is the value for bc - ad = ?

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a^2 + b^2 = c^2 + d^2= 1....(1)

ac + bd = 0..........(2)

we need:

bc-ad = ?

From (1) let us multiply (by c^2 + d^2) both sides:

(a^2 + b^2)*(c^2 + d^2) = c^2 + d^2

(ac)^2 + (ad)^2 + (bc)^2 + (bd)^2 = 1

(ac + bd)^2 - 2acbd + (bc-ad)^2 +2abcd = 1

From (2) we know that ac + bd= 0

==> 0 -2abcd + (bc-ad)^2 + 2abcd= 1

Reduce similar:

==> (bc-ad)^2 = 1

==> bc - ad = 1

or:

==> bc - ad = -1

From identity (a^2 + b^2)*(c^2 + d^2)= a^2* c^2 + a^2* d^2+ b^2*c^2 + b^2*d^2=1

We’ll try to form in the relation obtained, square binomials.

a^2* c^2+ b^2*d^2=( a* c+ b*d)^2 – 2a*c*b*d

b^2*c^2+ a^2* d^2 = (b*c-a*d)^2 +2 a*c*b*d

(a^2 + b^2)*(c^2 + d^2)= ( a* c+ b*d)^2 – 2a*c*b*d + (b*c-a*d)^2 +2 a*c*b*d

We'll eliminate like terms and we'll get:

(a^2 + b^2)*(c^2 + d^2)= ( a* c+ b*d)^2+ (b*c-a*d)^2=1

But from enunciation a* c+ b*d=0, so ( a* c+ b*d)^2=0, too.

Substituting the value for ( a* c+ b*d)^2,

(a^2 + b^2)*(c^2 + d^2)=0+(b*c-a*d)^2=1, so (b*c-a*d)^2=1

From (b*c-a*d)^2=1, the posibilites are:

**(b*c-a*d)=-1 or (b*c-a*d)=1**

We use a = sinx and b=cosx, c = cosy and d = cosy

Then a^2+b^2= 1 and c^2+d^2= 1 are satisfied,

Also by data ac+bd =0 Or sinx siny+cosx cos y = 0.

cos (x-y) = 0

Or (x-y )/2 = + or - pi/2. So x = pi+y , x = -pi+y .....(1)

Now to find bc-ad = cos xsiny - sinx cosy

= sin (y-x) = sin (-pi) = -1 or sin (y-x) = sin(pi) = 1.

So ac-bd = -1 or +1.