If \(\vec{a}\:and\:\vec{b}\) are unit vectors and θ is the angle between them, then what is \({{\sin }^{2}}\left( \frac{\theta }{2} \right)\) equal to?
Finding \( {{\sin }^{2}}\left( \frac{\theta }{2} \right) \) for Unit Vectors
The problem asks us to find the value of \( {{\sin }^{2}}\left( \frac{\theta }{2} \right) \) given that \( \vec{a} \) and \( \vec{b} \) are unit vectors and \( \theta \) is the angle between them. Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, their magnitudes are equal to 1.
- Magnitude of \( \vec{a} \): \( |\vec{a}| = 1 \)
- Magnitude of \( \vec{b} \): \( |\vec{b}| = 1 \)
Using the Magnitude of the Difference of Vectors
Let's consider the magnitude squared of the difference between the two unit vectors, \( |\vec{a} - \vec{b}|^2 \). We can expand this using the dot product property \( |\vec{v}|^2 = \vec{v} \cdot \vec{v} \):
\( |\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) \)
Expanding the dot product:
\( |\vec{a} - \vec{b}|^2 = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} \)
Using the properties \( \vec{v} \cdot \vec{v} = |\vec{v}|^2 \) and \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \):
\( |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \)
Substitute the magnitudes since \( \vec{a} \) and \( \vec{b} \) are unit vectors:
\( |\vec{a} - \vec{b}|^2 = (1)^2 - 2(\vec{a} \cdot \vec{b}) + (1)^2 \)
\( |\vec{a} - \vec{b}|^2 = 1 - 2(\vec{a} \cdot \vec{b}) + 1 \)
\( |\vec{a} - \vec{b}|^2 = 2 - 2(\vec{a} \cdot \vec{b}) \)
Applying the Dot Product Definition
The dot product of two vectors \( \vec{a} \) and \( \vec{b} \) is defined as \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \), where \( \theta \) is the angle between them. Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, \( |\vec{a}|=1 \) and \( |\vec{b}|=1 \).
So, \( \vec{a} \cdot \vec{b} = (1)(1) \cos \theta = \cos \theta \).
Substitute this back into the equation for \( |\vec{a} - \vec{b}|^2 \):
\( |\vec{a} - \vec{b}|^2 = 2 - 2 \cos \theta \)
\( |\vec{a} - \vec{b}|^2 = 2(1 - \cos \theta) \)
Using the Half-Angle Identity
Recall the trigonometric half-angle identity for cosine: \( \cos \theta = 1 - 2 {{\sin }^{2}}\left( \frac{\theta }{2} \right) \). Rearranging this identity to solve for \( 1 - \cos \theta \):
\( 1 - \cos \theta = 2 {{\sin }^{2}}\left( \frac{\theta }{2} \right) \)
Substitute this into the equation for \( |\vec{a} - \vec{b}|^2 \):
\( |\vec{a} - \vec{b}|^2 = 2 \left( 2 {{\sin }^{2}}\left( \frac{\theta }{2} \right) \right) \)
\( |\vec{a} - \vec{b}|^2 = 4 {{\sin }^{2}}\left( \frac{\theta }{2} \right) \)
Solving for \( {{\sin }^{2}}\left( \frac{\theta }{2} \right) \)
Now, we can isolate \( {{\sin }^{2}}\left( \frac{\theta }{2} \right) \) by dividing both sides by 4:
\( {{\sin }^{2}}\left( \frac{\theta }{2} \right) = \frac{{{\left| \vec{a}-\vec{b} \right|}^{2}}}{4} \)
This result matches one of the given options.
Revision Table: Key Formulas Used
| Concept | Formula |
|---|---|
| Unit Vector Magnitude | \( |\vec{v}| = 1 \) |
| Magnitude Squared | \( |\vec{v}|^2 = \vec{v} \cdot \vec{v} \) |
| Dot Product Definition | \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \) |
| Half-Angle Identity | \( 1 - \cos \theta = 2 {{\sin }^{2}}\left( \frac{\theta }{2} \right) \) |
Additional Information: Magnitude of Sum of Vectors
We can also consider the magnitude squared of the sum of the two unit vectors, \( |\vec{a} + \vec{b}|^2 \). Following a similar process:
\( |\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) \)
\( |\vec{a} + \vec{b}|^2 = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} \)
\( |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \)
For unit vectors \( |\vec{a}|=1, |\vec{b}|=1 \):
\( |\vec{a} + \vec{b}|^2 = 1^2 + 2(\cos \theta) + 1^2 \)
\( |\vec{a} + \vec{b}|^2 = 2 + 2 \cos \theta \)
\( |\vec{a} + \vec{b}|^2 = 2(1 + \cos \theta) \)
Using the identity \( 1 + \cos \theta = 2 {{\cos }^{2}}\left( \frac{\theta }{2} \right) \):
\( |\vec{a} + \vec{b}|^2 = 2 \left( 2 {{\cos }^{2}}\left( \frac{\theta }{2} \right) \right) \)
\( |\vec{a} + \vec{b}|^2 = 4 {{\cos }^{2}}\left( \frac{\theta }{2} \right) \)
So, \( {{\cos }^{2}}\left( \frac{\theta }{2} \right) = \frac{{{\left| \vec{a}+\vec{b} \right|}^{2}}}{4} \). This shows the relationship between \( |\vec{a}+\vec{b}|^2 \) and \( {{\cos }^{2}}\left( \frac{\theta }{2} \right) \), which is useful for other problems involving unit vectors and angles.
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