To solve the expression [tex]\(\frac{1}{2} \vec{u} - 3 \vec{v}\)[/tex] given the vectors [tex]\(\vec{u} = (1, -3, 2)\)[/tex] and [tex]\(\vec{v} = (-1, 4, 3)\)[/tex], follow these steps:
1. Scale the vector [tex]\(\vec{u}\)[/tex] by [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[
\frac{1}{2} \vec{u} = \frac{1}{2} \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \times 1 \\ \frac{1}{2} \times -3 \\ \frac{1}{2} \times 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \\ -\frac{3}{2} \\ 1 \end{pmatrix}
\][/tex]
2. Scale the vector [tex]\(\vec{v}\)[/tex] by 3:
[tex]\[
3 \vec{v} = 3 \begin{pmatrix} -1 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \times -1 \\ 3 \times 4 \\ 3 \times 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 12 \\ 9 \end{pmatrix}
\][/tex]
3. Subtract the scaled vectors:
[tex]\[
\frac{1}{2} \vec{u} - 3 \vec{v} = \begin{pmatrix} \frac{1}{2} \\ -\frac{3}{2} \\ 1 \end{pmatrix} - \begin{pmatrix} -3 \\ 12 \\ 9 \end{pmatrix}
\][/tex]
4. Perform the subtraction component-wise:
[tex]\[
\begin{pmatrix} \frac{1}{2} - (-3) \\ -\frac{3}{2} - 12 \\ 1 - 9 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} + 3 \\ -\frac{3}{2} - 12 \\ 1 - 9 \end{pmatrix} = \begin{pmatrix} 3.5 \\ -13.5 \\ -8 \end{pmatrix}
\][/tex]
Therefore, the result of the expression [tex]\(\frac{1}{2} \vec{u} - 3 \vec{v}\)[/tex] is:
[tex]\[
\boxed{\begin{pmatrix} 3.5 \\ -13.5 \\ -8 \end{pmatrix}}
\][/tex]
