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

ANS:$\sqrt{24}$

I found the orthogonal vector as $\begin{bmatrix}1& -2& 1\end{bmatrix}$ but I don't see the connection between $H$ and $H$ orthogonal.

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You already found a basis for the orthogonal space, let us call $v_1=(2,1,0)$, $v_2=(0,1,2)$, $v_3=(1,-2,1)$, you know that these three vectors form a basis of $\mathbb{R}^3$ because the direct sum of a vector subspace and its orthogonal is the space itself.

You already know that $u=\alpha v_1+\beta v_2$ where $\alpha$ and $\beta$ are scalars because $u\in H$, and you know also that $v=\gamma v_3$ whew $\gamma$ is a scalar because $v \in H \perp$. Can you see how to go on from here?

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  • $\begingroup$ no I'm still stuck @chris $\endgroup$
    – Andrew
    Apr 21, 2017 at 18:17
  • $\begingroup$ well you know what I said and that $(4,12,8)=u+v$, just solve the system of equation $(4,12,8)=\alpha v_1+\beta v_2 + \gamma v_3$, then having $\gamma$ you can directly calculate $|v|=|\gamma v_3|$ $\endgroup$
    – user438666
    Apr 21, 2017 at 18:21

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