Question 1203666
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(a) Show that 4u · v = ||u + v||^2 − ||u − v||^2. 
(b) Show that u and v {{{highlight(cross(is))}}} {{{highlight(highlight(are))}}} orthogonal if and only if ||u + v|| = ||u − v||.
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<pre>
             <U>Part (a)</U>


    ||u + v||^2 = (u,u) + 2(u,v) + (v,v)      (1)

    ||u - v||^2 = (u,u) - 2(u,v) + (v,v)      (2)


Here (u,v) is the scalar product of vectors u and v, or, in your designations, (u,v) = u · v.


Subtract equation (2) from equation (1) .  You will get 

    ||u + v||^2 − ||u − v||^2 = 4u · v.


It is exactly what should be proven in part (a).



             <U>Part (b)</U>


Vectors u and v are orthogonal if and only if  u · v = 0.


In part (a), we show that  4u · v = ||u + v||^2 − ||u − v||^2.


So, u · v = 0  if and only if  ||u + v||^2 − ||u − v||^2 = 0,  or  ||u + v||^2 = ||u − v||^2.


It is exactly what should be proven in part (b).
</pre>

Solved, proved and completed.