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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 $H=spanleft{(2,1,0),(0,1,2)right}$ be a subspace of $mathbb{R}^3$. If $(4,12,8)=u+v$, with $u$ in H and $vepsilon$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:

$v=spanleft{(1,-2,1)right}$

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

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

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 $v$.

You found already a basis $w = (1,-2,1)$ for $H^{perp}$.

Now,

• $(4,12,8) = u + underbrace{sw}_{=v} Rightarrow (4,12,8)cdot w stackrel{u perp w}{=} s||w||^2 Rightarrow -12 = 6s Rightarrow boxed{s=-2}$


• $||v|| = ||sw|| = |s|cdot||w|| = 2 sqrt{6} = boxed{sqrt{24}}$

$bf v$ 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$$|mathbf v|=begin{vmatrix}displaystylefrac{langle(1,-2,1),(4,12,8)rangle}{|(1,-2,1)|}end{vmatrix}=frac{12}{sqrt6}=sqrt{24}$$

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 $|u|=sqrt{24}$.