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Let $\vec{x}=\left[\begin{array}{c}2 \\ 1 \\ -2\end{array}\right]$ and $\vec{y}=\left[\begin{array}{c}1 \\ -1 \\ -4\end{array}\right]$. Use the Gram-Schmidt process to find an orthonormal basis for the subspace of $\mathbb{R}^{3}$ spanned by $\vec{x}$ and $\vec{y}$.

Answer:

Hint: Find an orthogonal basis first, then normalize the vectors.

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Step 1 ：Define the vectors \(\vec{x} = \left[\begin{array}{c}2 \ 1 \ -2\end{array}\right]\) and \(\vec{y} = \left[\begin{array}{c}1 \ -1 \ -4\end{array}\right]\).

Step 2 ：Calculate the first vector of the orthogonal basis, \(\vec{u}_1\), which is just \(\vec{x}\). So, \(\vec{u}_1 = \vec{x} = \left[\begin{array}{c}2 \ 1 \ -2\end{array}\right]\).

Step 3 ：Calculate the second vector, \(\vec{u}_2\), by subtracting the projection of \(\vec{y}\) onto \(\vec{u}_1\) from \(\vec{y}\). This gives \(\vec{u}_2 = \vec{y} - \frac{\vec{y} \cdot \vec{u}_1}{\vec{u}_1 \cdot \vec{u}_1} \vec{u}_1 = \left[\begin{array}{c}-1 \ -2 \ -2\end{array}\right]\).

Step 4 ：Normalize \(\vec{u}_1\) and \(\vec{u}_2\) to get the orthonormal basis. This gives \(\vec{u}_{1_{norm}} = \frac{\vec{u}_1}{||\vec{u}_1||} = \left[\begin{array}{c}0.66666667 \ 0.33333333 \ -0.66666667\end{array}\right]\) and \(\vec{u}_{2_{norm}} = \frac{\vec{u}_2}{||\vec{u}_2||} = \left[\begin{array}{c}-0.33333333 \ -0.66666667 \ -0.66666667\end{array}\right]\).

Step 5 ：\(\boxed{\text{Final Answer: The orthonormal basis for the subspace of } \mathbb{R}^{3} \text{ spanned by } \vec{x} \text{ and } \vec{y} \text{ is } \left\{\left[\begin{array}{c}0.66666667 \ 0.33333333 \ -0.66666667\end{array}\right], \left[\begin{array}{c}-0.33333333 \ -0.66666667 \ -0.66666667\end{array}\right]\right\}}\)