Find the magnitude of the vector  A2→
  represented by the pair of components:  Ax2=
  -6.70  m
 ,  Ay2=
  -3.00  m
 .
Express your answer in meters.

To find the magnitude of the vector \( \mathbf{A}_2 \) represented by the components \( A_{x2} = -6.70 \, \text{m} \) and \( A_{y2} = -3.00 \, \text{m} \), you can use the following formula for the magnitude of a vector in two dimensions:

\[
|\mathbf{A}_2| = \sqrt{A_{x2}^2 + A_{y2}^2}
\]

Substituting the given values:

\[
|\mathbf{A}_2| = \sqrt{(-6.70)^2 + (-3.00)^2}
\]

Let's calculate that:

\[
|\mathbf{A}_2| = \sqrt{44.89 + 9.00} = \sqrt{53.89}
\]

\[
|\mathbf{A}_2| \approx 7.34 \, \text{m}
\]

So, the magnitude of the vector \( \mathbf{A}_2 \) is approximately **7.34 meters**.

Would you like to know more details or have any questions?

Here are 5 related questions for further exploration:
1. How would the magnitude change if \( A_{y2} \) were positive?
2. What is the direction (angle) of the vector \( \mathbf{A}_2 \) relative to the positive x-axis?
3. How would you find the magnitude of a 3-dimensional vector?
4. Can a vector's magnitude be negative? Why or why not?
5. How would the magnitude change if both components were doubled?

**Tip:** Always square the components before summing when calculating the magnitude of a vector.

Learn how to find the magnitude of vector A2 represented by components Ax2 = -6.70 m and Ay2 = -3.00 m. Use the Pythagorean Theorem to calculate |A2| = sqrt((-6.70)^2 + (-3.00)^2) = 7.34 m. Understand the concept of vector magnitude in two dimensions. Suitable for Grades 9-12.

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