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Problem 3: Let r be given n mutually orthogonal vectors in Rn, and ro E R be also given. Find: (a) the distance di from ao to

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Answer #1

a) Note that 0\in H_i, and x_i is orthogonal to H_i. Thus, a unit normal to H_i is

\eta_i:={\frac{x_i}{||x_i||}}

Therefore, the distance d_i is the length of the projection of x_0 on \eta_i, showing that

d_i=\eta_i\cdot x_0={\frac{x_i\cdot x_0}{||x_i||}}

b) Note that TEH implies x and x_i are orthogonal; in particular, x\in\cap_{i=1}^k H_i iff x and x_1,\cdots,x_k are orthogonal. The space n1Hi has unit normal vectors

7t 7l

Therefore, the distance s_k is the length of the projection of x_0 on any of these vectors, showing that

7t Sk = Tk+1 7t

c) Note that TEH implies x and x_i are orthogonal; in particular, x\in\cap_{i=k+1}^n H_i iff x and x_{k+1},\cdots,x_n are orthogonal. The space \cap_{i=k+1}^n H_i has unit normal vectors

{\frac{x_1}{||x_1||}},\cdots,{\frac{x_{k}}{||x_{k}||}}

Therefore, the distance m_k is the length of the projection of x_0 on any of these vectors, showing that

m_k={\frac{x_1\cdot x_0}{||x_1||}}=\cdots={\frac{x_{k}\cdot x_0}{||x_{k}||}}

d) From the previous parts, we get

s_k={\frac{x_{k+1}\cdot x_0}{||x_{k+1}||}}=\cdots={\frac{x_n\cdot x_0}{||x_n||}}={\frac{(x_{k+1}+\cdots+x_n)\cdot x_0}{||x_{k+1}||+\cdots+||x_n||}}

and

m_k={\frac{x_1\cdot x_0}{||x_1||}}=\cdots={\frac{x_k\cdot x_0}{||x_k||}}={\frac{(x_1+\cdots+x_k)\cdot x_0}{||x_1||+\cdots+||x_k||}}

Therefore,

m_k+s_k={\frac{(x_1+\cdots+x_k)\cdot x_0}{||x_1||+\cdots+||x_k||}}+{\frac{(x_{k+1}+\cdots+x_n)\cdot x_0}{||x_{k+1}||+\cdots+||x_n||}}

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