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The problem states that we are given the vectors:

\[
\vec{a} = (1; 2; 0), \quad \vec{b} = (m; 5; 6)
\]
and the condition that \(\vec{a} \perp \vec{b}\), which means the vectors \(\vec{a}\) and \(\vec{b}\) are orthogonal. The task is to find the value of \(m\) such that \(\vec{a} \cdot \vec{b} = 0\).

The dot product of two vectors \(\vec{a} = (a_1, a_2, a_3)\) and \(\vec{b} = (b_1, b_2, b_3)\) is given by:

\[
\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3
\]

For the given vectors, we compute:

\[
\vec{a} \cdot \vec{b} = 1 \cdot m + 2 \cdot 5 + 0 \cdot 6
\]
\[
\vec{a} \cdot \vec{b} = m + 10
\]

Since \(\vec{a} \perp \vec{b}\), the dot product must be zero:

\[
m + 10 = 0
\]

Solving for \(m\), we get:

\[
m = -10
\]

Thus, the value of \(m\) is \(-10\).

Would you like more details or have any questions?

Here are 5 related questions to expand this topic:
1. What is the geometric interpretation of two vectors being orthogonal?
2. How do we calculate the magnitude of a vector?
3. What happens if vectors are not orthogonal but have a non-zero dot product?
4. Can two non-zero vectors be orthogonal in a higher-dimensional space?
5. How can the cross product help in finding a vector perpendicular to two given vectors?

**Tip:** When calculating the dot product of two vectors, always check if each component contributes correctly to avoid mistakes in simplifications.

Given vectors a = (1, 2, 0) and b = (m, 5, 6) with a ⊥ b, find the value of m.

Learn how to find the value of m such that the vectors a = (1, 2, 0) and b = (m, 5, 6) are orthogonal. This problem covers key vector concepts including the dot product and orthogonality. Suitable for students in Grades 9-11.

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