Question

The condition for the vector w to be equidistant from u and v is (w−u)(w−u)=(w−v)(w−v) Show that this is equivalent to

(a) ∣u∣ 2 −2wu=∣v∣ 2−2wv
(b) w− 2u+v (u−v)=0

Answer 1

To prove that the condition (w−u)·(w−u) = (w−v)·(w−v) is equivalent to the given statements, we use the properties of dot product to expand and simplify the initial condition resulting in two alternative forms that match the given expressions (a) and (b).

Step-by-step explanation:

The condition for vector w to be equidistant from u and v is given by the equation (w−u)·(w−u) = (w−v)·(w−v). This condition can be rewritten in two equivalent forms. To show this, we consider the scalar dot product properties and manipulate the equations accordingly.

Firstly, let's expand both sides using the distributive property of dot product:


(w−u)·(w−u) = w·w - 2w·u + u·u,


and


(w−v)·(w−v) = w·w - 2w·v + v·v.

Setting these two equal gives us:


w·w - 2w·u + u·u = w·w - 2w·v + v·v,


which simplifies to

|u|·|u| - 2w·u = |v|·|v| - 2w·v


This is the first alternate form (a).

For form (b), we start with the same equation and rearrange terms to obtain:


2w·u - 2w·v = u·u - v·v,


then we factor out the terms with w which gives us:


w(2u - 2v) = u·u - v·v,


and adding 2v - u to both sides yields:

w = 2u - v + (u - v)


Finally, we can simplify to:

w - 2u + v(u - v)= 0, which is the desired form (b).