Understanding the Problem: Unit Vectors and Angle Between Them

The question asks us to find the magnitude of a specific vector expression involving two unit vectors. We are given two vectors, \(\bar a\) and \(\bar b\). The key information is that they are unit vectors, which means their magnitudes are equal to 1. In mathematical terms, this is written as \(|\bar a| = 1\) and \(|\bar b| = 1\).

We are also told that \(\theta\) is the angle between these two unit vectors. We need to calculate the value of \(\left| {\frac{{\bar a - \bar b}}{2}} \right|\). This involves vector subtraction and finding the magnitude of the resulting vector scaled by a factor of one-half. This is a fundamental calculation in Vector Angle Calculation.

Calculating the Magnitude of the Vector Difference

To find \(\left| {\frac{{\bar a - \bar b}}{2}} \right|\), let's first focus on finding the magnitude of the vector difference, \(|\bar a - \bar b|\). The magnitude squared of any vector \(\bar v\) can be found using the dot product: \(|\bar v|^2 = \bar v \cdot \bar v\). Applying this to \(\bar a - \bar b\), we get:

\(|\bar a - \bar b|^2 = (\bar a - \bar b) \cdot (\bar a - \bar b)\)

We can expand this dot product similar to how we expand algebraic expressions:

\(|\bar a - \bar b|^2 = \bar a \cdot \bar a - \bar a \cdot \bar b - \bar b \cdot \bar a + \bar b \cdot \bar b\)

Since the dot product is commutative (\(\bar a \cdot \bar b = \bar b \cdot \bar a\)), this simplifies to:

\(|\bar a - \bar b|^2 = \bar a \cdot \bar a - 2(\bar a \cdot \bar b) + \bar b \cdot \bar b\)

Now, we use the properties of the dot product and the fact that \(\bar a\) and \(\bar b\) are unit vectors. As mentioned earlier, \(\bar a \cdot \bar a = |\bar a|^2\) and \(\bar b \cdot \bar b = |\bar b|^2\). Since \(|\bar a| = 1\) and \(|\bar b| = 1\), we have:

\(\bar a \cdot \bar a = 1^2 = 1\)

\(\bar b \cdot \bar b = 1^2 = 1\)

The dot product of two vectors can also be expressed in terms of their magnitudes and the angle between them: \(\bar a \cdot \bar b = |\bar a| |\bar b| \cos \theta\). Substituting the magnitudes of the unit vectors, we get:

\(\bar a \cdot \bar b = (1)(1) \cos \theta = \cos \theta\)

Substitute these values back into the expression for \(|\bar a - \bar b|^2\):

\(|\bar a - \bar b|^2 = 1 - 2(\cos \theta) + 1\)

\(|\bar a - \bar b|^2 = 2 - 2 \cos \theta\)

\(|\bar a - \bar b|^2 = 2(1 - \cos \theta)\)

This calculation shows the relationship between the vector magnitude of the difference and the angle \(\theta\).

Applying Trigonometry to the Vector Angle Calculation

To simplify further, we use a common trigonometric identity related to \(1 - \cos \theta\). The identity is \(1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}\). Using this identity allows us to directly incorporate half the angle into our expression.

Substitute this identity into the equation for \(|\bar a - \bar b|^2\):

\(|\bar a - \bar b|^2 = 2 \left( 2 \sin^2 \frac{\theta}{2} \right)\)

\(|\bar a - \bar b|^2 = 4 \sin^2 \frac{\theta}{2}\)

Now, to find \(|\bar a - \bar b|\), we take the square root of both sides:

\(|\bar a - \bar b| = \sqrt{4 \sin^2 \frac{\theta}{2}}\)

\(|\bar a - \bar b| = \sqrt{4} \sqrt{\sin^2 \frac{\theta}{2}}\)

\(|\bar a - \bar b| = 2 \left| \sin \frac{\theta}{2} \right|\)

The angle \(\theta\) between two vectors is usually taken in the range \(0 \le \theta \le \pi\). For this range, \(\frac{\theta}{2}\) lies in the range \(0 \le \frac{\theta}{2} \le \frac{\pi}{2}\). In this interval, \(\sin x \ge 0\). Therefore, \(\left| \sin \frac{\theta}{2} \right| = \sin \frac{\theta}{2}\).

So, the vector magnitude of the difference is:

\(|\bar a - \bar b| = 2 \sin \frac{\theta}{2}\)

Completing the Vector Angle Calculation

The problem asks for the magnitude of \(\left| {\frac{{\bar a - \bar b}}{2}} \right|\). We can use the property that for any scalar \(c\) and vector \(\bar v\), \(|c \bar v| = |c| |\bar v|\). Here, \(c = \frac{1}{2}\) and \(\bar v = \bar a - \bar b\).

\(\left| {\frac{{\bar a - \bar b}}{2}} \right| = \left| \frac{1}{2} (\bar a - \bar b) \right| = \left| \frac{1}{2} \right| |\bar a - \bar b|\)

Since \(\left| \frac{1}{2} \right| = \frac{1}{2}\), we have:

\(\left| {\frac{{\bar a - \bar b}}{2}} \right| = \frac{1}{2} |\bar a - \bar b|\)

Now, substitute the expression we found for \(|\bar a - \bar b|\):

\(\left| {\frac{{\bar a - \bar b}}{2}} \right| = \frac{1}{2} \left( 2 \sin \frac{\theta}{2} \right)\)

Multiplying by \(\frac{1}{2}\), we get:

\(\left| {\frac{{\bar a - \bar b}}{2}} \right| = \sin \frac{\theta}{2}\)

This step-by-step Vector Angle Calculation, using properties of unit vectors and vector magnitude, leads to the final answer.

In summary, understanding unit vectors, vector magnitude, dot products, and trigonometric identities is key to solving this type of Vector Angle Calculation problem.