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Can someone point out what I am fundamentally doing wrong in this question?

Consider the vector space $\mathbb{R}^3$ with the standard inner product (dot product) and let $H=span\left\{(2,1,0),(0,1,2)\right\}$ be a subspace of $\mathbb{R}^3$. If $(4,12,8)=u+v$, with $u$ in H and $v\epsilon$H$^\perp$, then $||v||$ equals:

$(A)$ $\sqrt6$

$(B)$ $1$

$(C)$ $3$

$(D)$ $\sqrt{24}$

$(E)$ $\sqrt3$

So my reasoning was that I needed what $||v||$ was. $v$ is the orthogonal complement of $H$. Since the vectors in $H$ are written as rows, we can find the Null space of $H$ in order to determine the orthogonal complement and hence $v$.

Thus:

$ \left[ \begin{array}{ccc|c} 2 & 1 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ \end{array} \right] $

$...$

$ \left[ \begin{array}{ccc|c} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0 \\ \end{array} \right] $

Therefore, $x_1=t,x_2=-2t,x_3=t$ and then:

$v=span\left\{(1,-2,1)\right\}$

Thus, $||v||=\sqrt{1^2+(-2)^2+1^2}$ $||v||=\sqrt6$

However, the answer says it should be $(D)$, $\sqrt{24}$. Why is that?

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  • $\begingroup$ Your $v=span\left\{(1,-2,1)\right\}$ is a subspace, not a vector! You are searching a vector in this subspace. $\endgroup$ – Emilio Novati Dec 21 '18 at 9:44
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You don't get $\|v\|$ by just finding some vector orthogonal to the two given vectors. The formula for $v$ is $v=\frac {|\langle (4,12,8), (1,-2,1) \rangle |} {\ {\|(1,-2,1\|}}$ which works out to $\sqrt {24}$.

When you write the given vector $(4,12,8)$ in terms of the span of $(2,1,0), (0,1,2)$ and its orthogonal complement you will get $(4,12,8)=a(2,1,0)+b(0,1,2)+c(1,-2,1)$ and $v =c((1,-2,1)$. You are missing the coefficient $c$ in your calculation. To find the value of $c$ all you have to do is take the inner product of both sides with $(1,2,1)$. That gives you the formula for $v$.

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  • $\begingroup$ Where is this formula from? $\endgroup$ – Future Math person Dec 21 '18 at 9:44
  • $\begingroup$ @FutureMathperson I have added some more steps to my answer. $\endgroup$ – Kavi Rama Murthy Dec 21 '18 at 9:49
  • $\begingroup$ Can't I just solve the system of equations instead to find the value of $a,b,c$ simultaneously? I get the same answer doing it that way. $\endgroup$ – Future Math person Dec 21 '18 at 9:57
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You found already a basis $w = (1,-2,1)$ for $H^{\perp}$.

Now,

  • $(4,12,8) = u + \underbrace{sw}_{=v} \Rightarrow (4,12,8)\cdot w \stackrel{u \perp w}{=} s||w||^2 \Rightarrow -12 = 6s \Rightarrow \boxed{s=-2}$
  • $||v|| = ||sw|| = |s|\cdot||w|| = 2 \sqrt{6} = \boxed{\sqrt{24}}$
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  • $\begingroup$ I like this too! I ended up doing a system of equations to solve for $s$ but this is a good way too. I think that's what the others were trying to tell me as well. $\endgroup$ – Future Math person Dec 21 '18 at 10:08
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$\bf v$ is not the span of $(1,-2,1); H^\perp$ is the span of $(1,-2,1).\bf v$ is the component of $(4,12,8)$ in $H^\perp$ and $\|\bf v\|$ is the norm of $\bf v$, that is, the absolute value of the projection of $(4,12,8)$ along $(1,-2,1)$. You may recall this is given by$$\|\mathbf v\|=\begin{vmatrix}\displaystyle\frac{\langle(1,-2,1),(4,12,8)\rangle}{\|(1,-2,1)\|}\end{vmatrix}=\frac{12}{\sqrt6}=\sqrt{24}$$

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Just to suggest another way.

We have the vectors $\vec a=(2,1,0)$ and $\vec b=(0,1,2)$, so the vector $\vec c=\vec a \times \vec b=(2,-4,2)$ spans the orthogonal complement of the subspace spanned by $\vec a $ and $\vec b$.

Normalizing this vector we have $$\hat c=\frac{\vec c}{|\vec c|}=\frac{1}{\sqrt{24}}(2,-4,2) $$

so the component of $\vec d=(4,12,8)$ in the subspace spanned by $\hat c$ is: $$ u=(\vec d \cdot \hat c)=\frac{8-48+16}{\sqrt{24}}=\frac{-24}{\sqrt{24}} $$ and $|u|=\sqrt{24}$.

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