Chapter 3: Problem 61
Find the direction and magnitude of the vector \(2 \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\), where \(\overrightarrow{\mathbf{A}}=(12.1 \mathrm{m}) \hat{\mathrm{x}}\) and \(\overrightarrow{\mathrm{B}}=(-32.2 \mathrm{m}) \hat{\mathrm{y}}\)
Short Answer
Expert verified
Magnitude: 40.27 m, Direction: -53.13° from positive x-axis.
Step by step solution
01
Multiply Vector A by Scalar
First, we need to find the vector \( 2 \overrightarrow{\mathbf{A}} \). Since \( \overrightarrow{\mathbf{A}} = (12.1 \text{ m}) \hat{\mathbf{x}} \), multiplying by 2 gives us: \( 2 \overrightarrow{\mathbf{A}} = 2 \times (12.1 \hat{\mathbf{x}}) = (24.2 \text{ m}) \hat{\mathbf{x}} \).
02
Add Vectors
Now add the resulting vector \( 2 \overrightarrow{\mathbf{A}} = (24.2 \text{ m}) \hat{\mathbf{x}} \) to \( \overrightarrow{\mathbf{B}} = (-32.2 \text{ m}) \hat{\mathbf{y}} \). The sum of these vectors is: \((24.2 \hat{\mathbf{x}}) + (-32.2 \hat{\mathbf{y}}) = (24.2 \hat{\mathbf{x}} - 32.2 \hat{\mathbf{y}})\).
03
Calculate Magnitude
The magnitude of a vector \( \overrightarrow{\mathbf{C}} = a \hat{\mathbf{x}} + b \hat{\mathbf{y}} \) is given by the formula \( \sqrt{a^2 + b^2} \). Here, \( a = 24.2 \) and \( b = -32.2 \). Therefore, the magnitude is: \( \sqrt{(24.2)^2 + (-32.2)^2} \). Calculate this to find \( \sqrt{585.64 + 1036.84} \approx \sqrt{1622.48} \approx 40.27 \text{ m} \).
04
Determine Direction
The direction \( \theta \) of the vector is given by \( \tan \theta = \frac{b}{a} \). Here, \( \tan \theta = \frac{-32.2}{24.2} \). Solve this to find \( \theta = \tan^{-1}\left(\frac{-32.2}{24.2}\right) \approx \tan^{-1}(-1.33) \approx -53.13^\circ \). The negative sign indicates the vector is measured clockwise from the positive x-axis.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnitude of a Vector
The magnitude of a vector is a measure of its length in space, reflecting how far the vector extends from its initial point to its endpoint.
For a vector expressed in terms of components, such as \( \overrightarrow{\mathbf{C}} = a \hat{\mathbf{x}} + b \hat{\mathbf{y}} \), the magnitude is calculated using the Pythagorean theorem:
\[ \text{Magnitude} = \sqrt{a^2 + b^2} \] This formula is derived from considering the vector as the hypotenuse of a right triangle formed by its components on the x and y axes.
For a vector expressed in terms of components, such as \( \overrightarrow{\mathbf{C}} = a \hat{\mathbf{x}} + b \hat{\mathbf{y}} \), the magnitude is calculated using the Pythagorean theorem:
\[ \text{Magnitude} = \sqrt{a^2 + b^2} \] This formula is derived from considering the vector as the hypotenuse of a right triangle formed by its components on the x and y axes.
- Let’s say a vector has components \( a = 24.2 \) and \( b = -32.2 \).
- The magnitude calculation will be \( \sqrt{24.2^2 + (-32.2)^2} \).
- After carrying out the computation, you'll find a magnitude of approximately 40.27 meters.
Direction of a Vector
Determining the direction of a vector involves finding the angle that the vector makes with a reference direction, commonly the positive x-axis.
This direction is usually represented as an angle \( \theta \), and can be calculated using the tangent trigonometric function:
\[ \tan \theta = \frac{b}{a} \] where \( b \) and \( a \) are the y and x components, respectively.
This direction is usually represented as an angle \( \theta \), and can be calculated using the tangent trigonometric function:
\[ \tan \theta = \frac{b}{a} \] where \( b \) and \( a \) are the y and x components, respectively.
- For our example, we use \( b = -32.2 \) and \( a = 24.2 \).
- The angle is calculated as \( \theta = \tan^{-1}\left(\frac{-32.2}{24.2}\right) \), which results in an angle of approximately \(-53.13^\circ\).
- The negative sign here indicates that the vector is measured in a clockwise direction from the positive x-axis.
Scalar Multiplication of a Vector
Scalar multiplication is a simple yet fundamental operation involving a vector and a number, known as a scalar.
When you multiply a vector by a scalar, you are essentially stretching or shrinking the vector.
\[ k \overrightarrow{\mathbf{A}} = (k \cdot a) \hat{\mathbf{x}} + (k \cdot b) \hat{\mathbf{y}} \] In our exercise, we had \( \overrightarrow{\mathbf{A}} = (12.1 \text{ m}) \hat{\mathbf{x}} \) and a scalar of 2, which results in:
When you multiply a vector by a scalar, you are essentially stretching or shrinking the vector.
- If the scalar is greater than one, the vector becomes longer.
- If the scalar is between zero and one, the vector becomes shorter.
- If the scalar is zero, the vector reduces to a point.
\[ k \overrightarrow{\mathbf{A}} = (k \cdot a) \hat{\mathbf{x}} + (k \cdot b) \hat{\mathbf{y}} \] In our exercise, we had \( \overrightarrow{\mathbf{A}} = (12.1 \text{ m}) \hat{\mathbf{x}} \) and a scalar of 2, which results in:
- \( 2 \overrightarrow{\mathbf{A}} = 2 \times (12.1 \text{ m}) \hat{\mathbf{x}} = (24.2 \text{ m}) \hat{\mathbf{x}} \).