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I need guidance in solving this problem: Find an orthogonal basis for $\mathbb{R}^3$ containing the set $S = \{v_1, v_2\}$ where $v_1 = \begin{bmatrix} -4 \\ 1 \\ 2 \end{bmatrix}$ and $v_2 = \begin{bmatrix} 0 \\ -1 \\ 2 \end{bmatrix}, \langle v_1, v_2\rangle := a_1b_1 + 4a_2b_2 + a_3b_3$

My approach: Applying the Gram-Schmidt Method, \begin{align*} x_1 &= v_1\\ x_2 &= v_2 - \dfrac{{v_2}^Tx_1}{{x_1}^Tx_1}x_1 = \begin{bmatrix} 0 \\ -1 \\ 2 \end{bmatrix} - \dfrac{\begin{bmatrix} 0 & -1 & 2 \end{bmatrix}\begin{bmatrix} -4 \\ 1 \\ 2 \end{bmatrix}}{\begin{bmatrix} -4 & 1 & 2 \end{bmatrix}\begin{bmatrix} -4 \\ 1 \\ 2 \end{bmatrix}}\begin{bmatrix} -4 \\ 1 \\ 2 \end{bmatrix}\\ &= \begin{bmatrix} 0 \\ -1 \\ 2 \end{bmatrix} - \dfrac{3}{21}\begin{bmatrix} -4 \\ 1 \\ 2 \end{bmatrix}\\ &= \begin{bmatrix} 0 \\ -1 \\ 2 \end{bmatrix} - \begin{bmatrix} -4/7 \\ 1/7 \\ 2/7 \end{bmatrix} = \begin{bmatrix} 4/7 \\ -8/7 \\ 12/7 \end{bmatrix}\\\\ \text{Orthogonal basis } &= \text{ span}\left\{x_1, x_2\right\}\\ &= \text{ span}\left\{\begin{bmatrix} -4 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 4/7 \\ -8/7 \\ 12/7 \end{bmatrix}\right\} \end{align*} Is my solution and conclusion correct?

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  • $\begingroup$ can 2 vectors span $\mathbb{R}^3$? $\endgroup$
    – mathma
    May 19, 2021 at 16:55
  • $\begingroup$ This was given as a seatwork for us which makes me wonder, could this possibly be a trick question? $\endgroup$
    – muw
    May 19, 2021 at 16:59
  • $\begingroup$ Are you sure there is no accidental swapping of $y$ and $z$ coordinate for one of the two given vectors? $\endgroup$ May 19, 2021 at 17:01
  • $\begingroup$ Do you know what the cross product (or: vector product) of vectors is ? $\endgroup$
    – KonKan
    May 19, 2021 at 17:03
  • $\begingroup$ @HagenvonEitzen those were the given. There was also an inner product endowed but I am not exactly sure where to use it. Updated my post. $\endgroup$
    – muw
    May 19, 2021 at 17:07

1 Answer 1

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With respect to the given inner product, you have $\langle v_1,v_2\rangle=0$; in other words, they're orthogonal.

So, find a vector$$u=\begin{bmatrix}a\\b\\c\end{bmatrix}$$which is orthogonal to both and which os not the null vector. That is, solve the system$$\left\{\begin{array}{l}\langle v_1,u\rangle=0\\\langle v_2,u\rangle=0.\end{array}\right.$$Every solution is of the form$$u=\begin{bmatrix}2c\\c\\2c\end{bmatrix}$$and you can take, say,$$u=\begin{bmatrix}2\\1\\2\end{bmatrix}.$$

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  • $\begingroup$ There is an endowed inner product, please see my updated post. $\endgroup$
    – muw
    May 19, 2021 at 17:10
  • $\begingroup$ They are orthogonal w.r.t. given dot product. $\endgroup$ May 19, 2021 at 17:12
  • $\begingroup$ @muw I've edited my answer. $\endgroup$ May 19, 2021 at 17:18
  • $\begingroup$ @Yalikesifulei The “given dot product” was not given in the original version of the question. Actually, no mention was made to the fact there was some non-standard dot product involved. $\endgroup$ May 19, 2021 at 17:20
  • $\begingroup$ @JoséCarlosSantos ok, sorry, I've seen the question with dot product already given :) $\endgroup$ May 19, 2021 at 17:22

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