Let’s clarify the vectors first: \[ \vec{z} = \begin{bmatrix} 4 \\ 4 \\ -4 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \quad \vec{y} = \begin{bmatrix} -8 \\ -2 \\ -4 \\ 1 \\ 2 \\ 1 \end{bmatrix} \] We will use the Gram-Schmidt process to find an orthonormal basis for the subspace of \(\mathbb{R}^6\) spanned by \(\vec{z}\) and \(\vec{y}\). --- **Step 1: Normalize \(\vec{z}\) to get the first orthonormal vector \(\vec{u}_1\)** First, compute the norm of \(\vec{z}\): \[ \|\vec{z}\| = \sqrt{4^2 + 4^2 + (-4)^2 + 0^2 + 0^2 + 0^2} = \sqrt{16 + 16 + 16} = \sqrt{48} = 4\sqrt{3} \] So, \[ \vec{u}_1 = \frac{1}{4\sqrt{3}} \begin{bmatrix} 4 \\ 4 \\ -4 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \\ 0 \\ 0 \\ 0 \end{bmatrix} \] **Explanation:** We normalized \(\vec{z}\) by dividing it by its length to get the first orthonormal vector. --- **Step 2: Make \(\vec{y}\) orthogonal to \(\vec{u}_1\), then normalize to get \(\vec{u}_2\)** First, compute the projection of \(\vec{y}\) onto \(\vec{u}_1\): \[ \text{proj}_{\vec{u}_1}(\vec{y}) = (\vec{y} \cdot \vec{u}_1) \vec{u}_1 \] Calculate \(\vec{y} \cdot \vec{u}_1\): \[ \vec{y} \cdot \vec{u}_1 = (-8)\left(\frac{1}{\sqrt{3}}\right) + (-2)\left(\frac{1}{\sqrt{3}}\right) + (-4)\left(-\frac{1}{\sqrt{3}}\right) + 1(0) + 2(0) + 1(0) \] \[ = \frac{-8 - 2 + 4}{\sqrt{3}} = \frac{-6}{\sqrt{3}} = -2\sqrt{3} \] So, \[ \text{proj}_{\vec{u}_1}(\vec{y}) = -2\sqrt{3} \begin{bmatrix} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} -2 \\ -2 \\ 2 \\ 0 \\ 0 \\ 0 \end{bmatrix} \] Now, subtract this projection from \(\vec{y}\): \[ \vec{y}_\perp = \vec{y} - \text{proj}_{\vec{u}_1}(\vec{y}) = \begin{bmatrix} -8 \\ -2 \\ -4 \\ 1 \\ 2 \\ 1 \end{bmatrix} - \begin{bmatrix} -2 \\ -2 \\ 2 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} -6 \\ 0 \\ -6 \\ 1 \\ 2 \\ 1 \end{bmatrix} \] Now, compute the norm of \(\vec{y}_\perp\): \[ \|\vec{y}_\perp\| = \sqrt{(-6)^2 + 0^2 + (-6)^2 + 1^2 + 2^2 + 1^2} = \sqrt{36 + 0 + 36 + 1 + 4 + 1} = \sqrt{78} \] Therefore, the second orthonormal vector is: \[ \vec{u}_2 = \frac{1}{\sqrt{78}} \begin{bmatrix} -6 \\ 0 \\ -6 \\ 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} -\frac{6}{\sqrt{78}} \\ 0 \\ -\frac{6}{\sqrt{78}} \\ \frac{1}{\sqrt{78}} \\ \frac{2}{\sqrt{78}} \\ \frac{1}{\sqrt{78}} \end{bmatrix} \] **Explanation:** We made \(\vec{y}\) orthogonal to \(\vec{u}_1\) by subtracting its projection, then normalized it to get the second orthonormal vector. --- ### Final Answer: An orthonormal basis for the subspace of \(\mathbb{R}^6\) spanned by \(\vec{z}\) and \(\vec{y}\) is: \[ \boxed{ \left\{ \begin{bmatrix} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \\ 0 \\ 0 \\ 0 \end{bmatrix} , \begin{bmatrix} -\frac{6}{\sqrt{78}} \\ 0 \\ -\frac{6}{\sqrt{78}} \\ \frac{1}{\sqrt{78}} \\ \frac{2}{\sqrt{78}} \\ \frac{1}{\sqrt{78}} \end{bmatrix} \right\} } \]