Problem Analysis:
The task involves understanding the subtraction of vectors and deriving the resultant vector when two vectors are subtracted. The provided text explains the mathematical formulation for vector subtraction and includes a diagram to illustrate the process. Let's break it down step by step.
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Key Concepts:
1.
Vector Subtraction:
- Vector subtraction is defined as:
\[
\vec{A} - \vec{B} = \vec{A} + (-\vec{B})
\]
- Here, \(-\vec{B}\) is the vector \(\vec{B}\) with its direction reversed.
2.
Magnitude of the Resultant Vector:
- The magnitude of the resultant vector \(\vec{R}\) when two vectors \(\vec{A}\) and \(\vec{B}\) are subtracted is given by:
\[
|\vec{R}| = |\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos\theta}
\]
- This formula is derived from the law of cosines, where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).
3.
Direction of the Resultant Vector:
- The angle \(\alpha_2\) between the resultant vector \(\vec{R}\) and vector \(\vec{A}\) is given by:
\[
\tan\alpha_2 = \frac{B \sin\theta}{A - B \cos\theta}
\]
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Derivation of the Formulae:
#### 1. Magnitude of the Resultant Vector:
The magnitude of the resultant vector \(\vec{R} = \vec{A} - \vec{B}\) can be derived using the law of cosines. Consider the triangle formed by vectors \(\vec{A}\), \(-\vec{B}\), and \(\vec{R}\):
- The sides of the triangle are:
- \(|\vec{A}|\) (magnitude of \(\vec{A}\))
- \(|-\vec{B}| = |\vec{B}|\) (magnitude of \(\vec{B}\))
- \(|\vec{R}|\) (magnitude of the resultant vector)
- The angle between \(\vec{A}\) and \(-\vec{B}\) is \(180^\circ - \theta\), where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).
Using the law of cosines:
\[
|\vec{R}|^2 = A^2 + B^2 - 2AB \cos(180^\circ - \theta)
\]
Since \(\cos(180^\circ - \theta) = -\cos\theta\), we have:
\[
|\vec{R}|^2 = A^2 + B^2 - 2AB (-\cos\theta) = A^2 + B^2 + 2AB \cos\theta
\]
Thus, the magnitude of the resultant vector is:
\[
|\vec{R}| = \sqrt{A^2 + B^2 - 2AB \cos\theta}
\]
#### 2. Direction of the Resultant Vector:
To find the angle \(\alpha_2\) between the resultant vector \(\vec{R}\) and vector \(\vec{A}\), we use trigonometry. From the diagram, consider the right triangle formed by projecting \(\vec{B}\) onto \(\vec{A}\):
- The horizontal component of \(\vec{B}\) along \(\vec{A}\) is \(B \cos\theta\).
- The vertical component of \(\vec{B}\) perpendicular to \(\vec{A}\) is \(B \sin\theta\).
The effective horizontal component of the resultant vector is:
\[
A - B \cos\theta
\]
The vertical component of the resultant vector is:
\[
B \sin\theta
\]
Using the tangent function:
\[
\tan\alpha_2 = \frac{\text{vertical component}}{\text{horizontal component}} = \frac{B \sin\theta}{A - B \cos\theta}
\]
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Final Answer:
The magnitude of the resultant vector \(\vec{R}\) is:
\[
\boxed{|\vec{R}| = \sqrt{A^2 + B^2 - 2AB \cos\theta}}
\]
The angle \(\alpha_2\) between the resultant vector \(\vec{R}\) and vector \(\vec{A}\) is:
\[
\boxed{\tan\alpha_2 = \frac{B \sin\theta}{A - B \cos\theta}}
\]
Parent Tip: Review the logic above to help your child master the concept of adding and subtracting vectors graphically.
