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The problem involves finding the scalar (dot) product of two vectors \(\vec{m}\) and \(\vec{n}\), where:

\[
\vec{m} = \langle 2, -5, 0 \rangle
\]
\[
\vec{n} = \langle 3, 0, -4 \rangle
\]

The scalar (dot) product of two vectors \(\vec{m}\) and \(\vec{n}\) is given by the formula:

\[
\vec{m} \cdot \vec{n} = m_1 \cdot n_1 + m_2 \cdot n_2 + m_3 \cdot n_3
\]

Now, substitute the components of the vectors \(\vec{m} = \langle 2, -5, 0 \rangle\) and \(\vec{n} = \langle 3, 0, -4 \rangle\):

\[
\vec{m} \cdot \vec{n} = (2)(3) + (-5)(0) + (0)(-4)
\]
\[
\vec{m} \cdot \vec{n} = 6 + 0 + 0 = 6
\]

So, the scalar (dot) product of the vectors \(\vec{m}\) and \(\vec{n}\) is \(6\).

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. How do you calculate the magnitude of a vector?
2. What is the geometric interpretation of the dot product?
3. How does the dot product help determine if two vectors are perpendicular?
4. What is the difference between the scalar product and the vector product?
5. Can the scalar product ever be negative, and what does that signify?

**Tip:** The dot product is useful for determining the angle between two vectors. If the result is zero, the vectors are perpendicular.

The vectors \(\vec{m} = \langle 2, -5, 0 \rangle\) and \(\vec{n} = \langle 3, 0, -4 \rangle\) are set. Find the scalar value of the vectors \(\vec{m}\) and \(\vec{n}\).

This problem involves calculating the scalar (dot) product of two vectors, \(\vec{m}\) and \(\vec{n}\). The step-by-step solution covers key vector algebra concepts and the dot product formula. Ideal for students in Grades 10-12.

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