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The scalar product of vectors \(\vec{A}\) and \(\vec{B}\) is \(+48.0 \mathrm{m}^{2} .\) Vector \(\vec{A}\) has magnitude 9.00 \(\mathrm{m}\) and direction \(28.0^{\circ}\) west of south.If vector \(\vec{B}\) has direction \(39.0^{\circ}\) south of east, what is the magnitude of \(\vec{B} ?\)

Short Answer

Expert verified
The magnitude of \( \vec{B} \) is approximately 13.68 m.

Step by step solution

01

Understand Scalar Product Formula

The scalar product (dot product) of two vectors \( \vec{A} \) and \( \vec{B} \) is given by the formula \( \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \), where \( \theta \) is the angle between the directions of the two vectors.
02

Determine Angle Between Vectors

The direction of \( \vec{A} \) is 28° west of south, and \( \vec{B} \) is 39° south of east. When representing these directions in the standard position (relative to the north), \( \vec{A} \) points 208° (180° + 28°) clockwise from north, and \( \vec{B} \) points 321° (360° - 39°) clockwise from north. Thus, the angle \( \theta \) between them is \( |208° - 321°| = 113° \).
03

Apply Scalar Product Formula to Find Magnitude of \( \vec{B} \)

Rewrite the scalar product equation \( 48.0 = 9.00 \times |\vec{B}| \times \cos(113^{\circ}) \). Use cosine to find the magnitude: solve for \( |\vec{B}| \), \( |\vec{B}| = \frac{48.0}{9.00 \times \cos(113^{\circ})} \).
04

Calculate Cosine and Solve for \(|\vec{B}|\)

Calculate \( \cos(113^{\circ}) \), which is approximately \(-0.3907\). Substitute back into the equation: \( |\vec{B}| = \frac{48.0}{9.00 \times (-0.3907)} \). This results in \( |\vec{B}| \approx 13.68 \mathrm{m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Vector Magnitude
Vectors are quantities that have both a magnitude and a direction. The magnitude of a vector is a scalar number that measures the size or length of the vector. To find the magnitude, visualize it as the distance from the starting point to the endpoint of the vector. If you're given the vector in terms of its components, you can calculate the magnitude using the Pythagorean theorem.For example, if a vector \( \vec{A} \) has components \( A_x \) and \( A_y \), its magnitude \( |\vec{A}| \) can be calculated as:
  • \( |\vec{A}| = \sqrt{A_x^2 + A_y^2} \)
In the original problem, vector \( \vec{A} \) is given with a magnitude of 9.00 meters. Knowing this gives us vital information required for further computations, such as using the dot product to solve for another vector's magnitude.
Finding the Angle Between Vectors
When dealing with two vectors, the angle between them provides insight into their relational direction. This angle is crucial for calculating the dot product because it affects the scalar product directly through the cosine function.
To determine this angle, you'll convert each vector's direction into a standard form relative to the north.In vector problems, angles are often given with directional references, such as "west of south," which can initially seem complex to analyze. To simplify, imagine rotating a vector's direction into a standard position from the north, measured clockwise.
  • For instance, a vector described as 28° west of south translates into an angle of 208° measured from north (180° + 28°).
  • Similarly, a vector labeled as 39° south of east can be seen as 321° from north (360° - 39°).
From these positions, calculate the absolute difference between their angles to find the angle between the vectors, in this case, \( 113° \). This angle directly influences the result of their scalar product.
Applying the Dot Product Formula
The dot product, also known as the scalar product, is a mathematical operation that takes two vectors and returns a single scalar. The formula is:
  • \( \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \)
Here, \( \theta \) is the angle between the vectors, and |\vec{A}| and |\vec{B}| are their magnitudes.
This operation is especially useful in physics and engineering because it combines both magnitude and directional information of vectors into a single value.In our specific problem, we're given the dot product value of 48.0 m², the magnitude of \( \vec{A} \) as 9.00 m, and an angle of \( 113° \). By rearranging the scalar product equation, we can solve for \( |\vec{B}| \):
  • \( |\vec{B}| = \frac{48.0}{9.00 \times \cos(113°)} \)
After calculating, \( \cos(113°) \approx -0.3907 \), substituting this value yields:
  • \( |\vec{B}| = \frac{48.0}{9.00 \times (-0.3907)} \approx 13.68 \mathrm{m} \)
Understanding this formula allows you to interrelate different vectors' properties, making it a powerful tool in vector analysis.

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