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To solve the given problem, we need to address two parts:

### 1. Find a vector of magnitude 14 units orthogonal to vectors \( \mathbf{a} = -\mathbf{i} + 3\mathbf{j} \) and \( \mathbf{b} = \mathbf{i} + 2\mathbf{k} \).
An orthogonal vector to both \( \mathbf{a} \) and \( \mathbf{b} \) can be found by calculating the **cross product** of \( \mathbf{a} \) and \( \mathbf{b} \).

Given the vectors:
\[
\mathbf{a} = -\mathbf{i} + 3\mathbf{j} + 0\mathbf{k} = (-1, 3, 0)
\]
\[
\mathbf{b} = \mathbf{i} + 0\mathbf{j} + 2\mathbf{k} = (1, 0, 2)
\]

The cross product \( \mathbf{a} \times \mathbf{b} \) is calculated as:
\[
\mathbf{a} \times \mathbf{b} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-1 & 3 & 0 \\
1 & 0 & 2
\end{vmatrix}
\]

Expanding this determinant:
\[
\mathbf{a} \times \mathbf{b} = \mathbf{i} \left( 3 \times 2 - 0 \times 0 \right) - \mathbf{j} \left( -1 \times 2 - 0 \times 1 \right) + \mathbf{k} \left( -1 \times 0 - 3 \times 1 \right)
\]
\[
\mathbf{a} \times \mathbf{b} = \mathbf{i}(6) - \mathbf{j}(-2) + \mathbf{k}(-3)
\]
\[
\mathbf{a} \times \mathbf{b} = 6\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}
\]

Thus, the vector orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \) is \( \mathbf{v} = 6\mathbf{i} + 2\mathbf{j} - 3\mathbf{k} \).

Next, we scale this vector to have a magnitude of 14 units.

The magnitude of \( \mathbf{v} \) is:
\[
|\mathbf{v}| = \sqrt{6^2 + 2^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7
\]

To get a vector with magnitude 14, we multiply \( \mathbf{v} \) by 2 (since \( \frac{14}{7} = 2 \)):

\[
\mathbf{v}_{14} = 2(6\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) = 12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}
\]

Thus, the vector of magnitude 14 units orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \) is:
\[
\boxed{12\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}}
\]

### 2. Find the angle between the vectors \( \mathbf{a} \) and \( \mathbf{b} \).

The formula to find the angle \( \theta \) between two vectors is:
\[
\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}
\]

First, compute the dot product \( \mathbf{a} \cdot \mathbf{b} \):
\[
\mathbf{a} \cdot \mathbf{b} = (-1)(1) + (3)(0) + (0)(2) = -1
\]

Now, find the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \):
\[
|\mathbf{a}| = \sqrt{(-1)^2 + 3^2 + 0^2} = \sqrt{1 + 9} = \sqrt{10}
\]
\[
|\mathbf{b}| = \sqrt{1^2 + 0^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}
\]

Substitute into the cosine formula:
\[
\cos \theta = \frac{-1}{\sqrt{10} \times \sqrt{5}} = \frac{-1}{\sqrt{50}} = \frac{-1}{5\sqrt{2}}
\]

Thus, the angle \( \theta \) is:
\[
\theta = \cos^{-1} \left( \frac{-1}{5\sqrt{2}} \right)
\]

You can further simplify or approximate using a calculator, but this is the exact form for the angle.

Would you like to know more details about any step, or would you like me to further simplify this?

### Follow-up Questions:
1. How do we verify the orthogonality of the resulting vector to the original ones?
2. What is the geometric interpretation of the cross product in this context?
3. Can you explain why the cross product results in a vector perpendicular to both input vectors?
4. How do you find the angle between vectors without a calculator?
5. How does changing the magnitude of a vector affect its direction?

### Tip:
Remember, the cross product of two vectors is always perpendicular to both, but the direction depends on the right-hand rule.

Find a vector of magnitude 14 units orthogonal to vectors a = -i + 3j and b = i + 2k both. Also find angle between the vectors a and b.

Learn how to find a vector of magnitude 14 units that is orthogonal to vectors a = -i + 3j and b = i + 2k, and also calculate the angle between these vectors. This problem covers vector algebra, cross product, dot product, and vector magnitudes. Suitable for students in Grades 10-12.

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