Question

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 Hi := {x E Rn·Xi Xi = 0 (b) the distance sk from 20 to nil Hi, 1 〈 k < n (c) the distance mk from ro to k+1Hi,1 S k< n (d) calculate skmk

Answer 1

a) Note that , and $x_i$ is orthogonal to $H_i$ . Thus, a unit normal to $H_i$ is

"a) Note that 0 element H_i, and x_i is orthogonal to H_i. Thus, a unit normal to H_i is eta_i: = x_i/

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

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

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

c) Note that implies and $x_i$ are orthogonal; in particular, $x\in\cap_{i=k+1}^n H_i$ iff 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,

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