4
$\begingroup$

Well, this is question from a test that I had, I didn't know how to answer it so I am forwarding this to you:

Consider $v\:=\:\begin{pmatrix}\frac{1}{3} \\\frac{2}{3}\: \\\frac{2}{3}\end{pmatrix}$. Let $U\:=(\:span\left(v\right))^⊥$ and let $v_2\:=\:\begin{pmatrix}9 \\0 \\0\end{pmatrix}$

How to find the distance between $v_2$ and $U$? I don't know the right method, tnx!

$\endgroup$
1
$\begingroup$

The projection matrix is given as $P=A(A^TA)^{-1}A$. In this case $A=v$, so the projection matrix reduces to $$ P = \frac19\begin{bmatrix}1&2&2\\2&4&4\\2&4&4\end{bmatrix}.$$ Thus evaluating $Pv_2$, we get $$ Pv_2 = \frac19\begin{bmatrix}9\\18\\18\end{bmatrix} = \begin{bmatrix}1\\2\\2\end{bmatrix}. $$ The question is reduced to finding the distance between $v_2$ and this vector. This is easy; the answer is $$\sqrt{(9-1)^2+2^2+2^2} = \sqrt{72} = 6\sqrt{2}.$$

$\endgroup$
1
$\begingroup$

Using Lagrange multipliers: minimum of $$(x - 9)^2 + y^2 + z^2$$ (distance from $(x,y,z)$ to $(9,0,0)$)

with restriction $$x + 2y + 2z = 0.$$ ($(x,y,z)\in U$)

The Lagrange system: $$2(x-9) = \lambda,$$ $$2y = 2\lambda,$$ $$2z = 2\lambda,$$ $$x + 2y + 2z = 0.$$ Solution:

$y = \lambda = z\implies -8z = 2x = 18 + \lambda = 18 + z\implies y = z = -2\implies x = 9 + z/2 = 8.$

$\endgroup$
0
$\begingroup$

The projection of $v_2$ on $U^\perp=\operatorname{span}(v)$ is: $$\pi_{U^\perp}(v_2) = \frac{(v_2 \cdot v)}{(v\cdot v)}v$$

This is the connecting vector between $U$ and $v_2$, which is perpendicular to U.

Therefore the requested distance is its length: $$d(v_2,U)=\|\pi_{U^\perp}(v_2)\| = \left\|\frac{(v_2 \cdot v)}{(v\cdot v)}v\right\|$$

Fill in the numbers...

$\endgroup$
6
  • 1
    $\begingroup$ What is $v_1$?? $\endgroup$
    – Bernard
    Feb 14 '16 at 10:41
  • $\begingroup$ Apparently that should be $v$ now. It seems the problem statement was edited after I had answered. Anyway, I've update my answer. $\endgroup$ Feb 15 '16 at 13:52
  • 2
    $\begingroup$ How come $\pi_{U^\perp}(v_2)$ is in $U = \operatorname{span}(v)$ when it should be orthogonal to $\operatorname{span}(v)$? $\endgroup$ Jul 12 '16 at 8:36
  • $\begingroup$ @IlikeSerena Note that $U$ is a span of the orthogonal. Therefore the distance between $v,U$ should be the norm of the projection onto the span of $v$. Since this question recieved many attention, I think you should fix it. $\endgroup$
    – Mickey
    Jul 6 '17 at 5:32
  • $\begingroup$ @IlikeSerena, that is not the projection of $v_2$ onto $U$. $\endgroup$
    – Laz
    Jul 26 '18 at 0:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.