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To find the vector parallel to \(\vec{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\) with a magnitude of 7, follow these steps:

### Step 1: Find the magnitude of the given vector \(\vec{u}\).
The magnitude of \(\vec{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\) is calculated using the formula:

\[
|\vec{u}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}
\]

So, the magnitude of \(\vec{u}\) is \(\sqrt{14}\).

### Step 2: Normalize the vector \(\vec{u}\).
To get a unit vector in the direction of \(\vec{u}\), divide each component of \(\vec{u}\) by its magnitude:

\[
\hat{u} = \frac{1}{\sqrt{14}} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{14}} \\ \frac{2}{\sqrt{14}} \\ \frac{3}{\sqrt{14}} \end{bmatrix}
\]

### Step 3: Scale the unit vector to the desired magnitude.
Now, multiply the unit vector by 7 to obtain a vector with a magnitude of 7:

\[
\vec{v} = 7 \times \hat{u} = 7 \times \begin{bmatrix} \frac{1}{\sqrt{14}} \\ \frac{2}{\sqrt{14}} \\ \frac{3}{\sqrt{14}} \end{bmatrix} = \begin{bmatrix} \frac{7}{\sqrt{14}} \\ \frac{14}{\sqrt{14}} \\ \frac{21}{\sqrt{14}} \end{bmatrix}
\]

Simplifying:

\[
\vec{v} = \begin{bmatrix} \frac{7}{\sqrt{14}} \\ \frac{14}{\sqrt{14}} \\ \frac{21}{\sqrt{14}} \end{bmatrix} = \sqrt{\frac{7}{2}} \cdot expansion 
making"



Find the vector parallel to \(\vec{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\) that has a magnitude/length equal to 7.

Learn how to find a vector parallel to \(\vec{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\) with a magnitude of 7. This solution includes vector magnitude, unit vector, and scalar multiplication for high school level students.

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