Question

(3.1) Find an expression for 12 || ~ u + ~ v || 2 + 12 || ~ u − ~ v || 2 in terms of || ~ u || 2 + || ~ v || 2

(3.2) Find an expression for || ~ u + ~ v || 2 − || ~ u − ~ v || 2 in terms of ~ u · ~ v

3.3) Use the result of (3.2) to deduce an expression for || ~ u + ~ v || 2 whenever ~ u and ~ v are orthogonal

to each other.

Answer 1

(3.1). Given arbitrary vectors $u$ and $v.$ We get

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

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

Then

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

Therefore

$$\dfrac{1}{2}||u+v||^2+\dfrac{1}{2}||u-v||^2=||u||^2+||v||^2$$

(3.2).

$$||u+v||^2-||u-v||^2$$ $$=||u||^2+2(u,v)+||v||^2-||u||^2+2(u,v)-||v||^2$$ $$=4(u,v)$$

Therefore

$$||u+v||^2-||u-v||^2=4u\cdot v$$

(3.3). If u and v are orthogonal to each other, then $u\cdot v=0.$

Hence

$$||u+v||^2-||u-v||^2=0$$ $$=>||u+v||^2=|u-v||^2$$

$$\dfrac{1}{2}||u+v||^2+\dfrac{1}{2}||u-v||^2$$ $$=\dfrac{1}{2}||u+v||^2+\dfrac{1}{2}||u+v||^2$$ $$=||u+v||^2$$ $$=\dfrac{1}{2}||u-v||^2+\dfrac{1}{2}||u-v||^2$$ $$=||u-v||^2$$ $$=||u||^2+||v||^2$$ $$||u-v||^2=||u+v||^2=||u||^2+||v||^2$$