If vector c = vector a + vector b and |c| = |a| = |b| =1, then the value of a.b + b.c + c.a is

To solve the problem where vector \( \mathbf{c} = \mathbf{a} + \mathbf{b} \) and the magnitudes of all three vectors are equal to 1, we need to analyze the relationships between these vectors. Given that \( |\mathbf{c}| = |\mathbf{a}| = |\mathbf{b}| = 1 \), we can derive the value of \( \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a} \).

Understanding the Vector Relationships

Since \( \mathbf{c} = \mathbf{a} + \mathbf{b} \), we can use the property of magnitudes to express \( |\mathbf{c}|^2 \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \).

Calculating the Magnitude of Vector c

The magnitude squared of vector \( \mathbf{c} \) can be expressed as:

• \( |\mathbf{c}|^2 = |\mathbf{a} + \mathbf{b}|^2 \)
• \( = \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + 2 \mathbf{a} \cdot \mathbf{b} \)

Since \( |\mathbf{a}| = 1 \) and \( |\mathbf{b}| = 1 \), we have:

• \( |\mathbf{c}|^2 = 1 + 1 + 2 \mathbf{a} \cdot \mathbf{b} \)
• \( = 2 + 2 \mathbf{a} \cdot \mathbf{b} \)

Given that \( |\mathbf{c}| = 1 \), we can set up the equation:

• \( 1 = 2 + 2 \mathbf{a} \cdot \mathbf{b} \)

Solving for the Dot Product

Rearranging the equation gives us:

• \( 2 \mathbf{a} \cdot \mathbf{b} = 1 - 2 \)
• \( 2 \mathbf{a} \cdot \mathbf{b} = -1 \)
• \( \mathbf{a} \cdot \mathbf{b} = -\frac{1}{2} \)

Finding the Other Dot Products

Next, we need to find \( \mathbf{b} \cdot \mathbf{c} \) and \( \mathbf{c} \cdot \mathbf{a} \). We can express these using the definition of \( \mathbf{c} \):

• \( \mathbf{b} \cdot \mathbf{c} = \mathbf{b} \cdot (\mathbf{a} + \mathbf{b}) = \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} \)
• \( = \mathbf{b} \cdot \mathbf{a} + 1 \)
• \( = -\frac{1}{2} + 1 = \frac{1}{2} \)

Similarly, for \( \mathbf{c} \cdot \mathbf{a} \):

• \( \mathbf{c} \cdot \mathbf{a} = \mathbf{c} \cdot \mathbf{a} = (\mathbf{a} + \mathbf{b}) \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{a} \)
• \( = 1 - \frac{1}{2} = \frac{1}{2} \)

Summing the Dot Products

Now we can combine all the dot products we calculated:

• \( \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a} \)
• \( = -\frac{1}{2} + \frac{1}{2} + \frac{1}{2} \)
• \( = -\frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{1}{2} \)

Thus, the final value of \( \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a} \) is \( \frac{1}{2} \).