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 find the magnitude of \( \mathbf{c} \) using the formula for the magnitude of a vector sum:

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

Substituting the known magnitudes:

\[ 1^2 = 1^2 + 1^2 + 2(\mathbf{a} \cdot \mathbf{b}) \]

This simplifies to:

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

Rearranging gives:

\[ 1 = 2 + 2(\mathbf{a} \cdot \mathbf{b}) \implies 2(\mathbf{a} \cdot \mathbf{b}) = -1 \implies \mathbf{a} \cdot \mathbf{b} = -\frac{1}{2} \]

Finding the Dot Products

Next, we need to calculate \( \mathbf{b} \cdot \mathbf{c} \) and \( \mathbf{c} \cdot \mathbf{a} \). We can express these dot products in terms of \( \mathbf{a} \) and \( \mathbf{b} \):

Calculating \( \mathbf{b} \cdot \mathbf{c} \)

Using \( \mathbf{c} = \mathbf{a} + \mathbf{b} \):

\[ \mathbf{b} \cdot \mathbf{c} = \mathbf{b} \cdot (\mathbf{a} + \mathbf{b}) = \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} \]

Since \( \mathbf{b} \cdot \mathbf{b} = |\mathbf{b}|^2 = 1 \), we have:

\[ \mathbf{b} \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{b} + 1 = -\frac{1}{2} + 1 = \frac{1}{2} \]

Calculating \( \mathbf{c} \cdot \mathbf{a} \)

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

\[ \mathbf{c} \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{c} = \mathbf{a} \cdot (\mathbf{a} + \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot \mathbf{b} \]

Again, since \( \mathbf{a} \cdot \mathbf{a} = 1 \), we have:

\[ \mathbf{c} \cdot \mathbf{a} = 1 + \mathbf{a} \cdot \mathbf{b} = 1 - \frac{1}{2} = \frac{1}{2} \]

Summing the Dot Products

Now we can sum all the dot products:

\[ \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a} = -\frac{1}{2} + \frac{1}{2} + \frac{1}{2} \]

Calculating this gives:

\[ -\frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{1}{2} \]

Final Result

The value of \( \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a} \) is \( \frac{1}{2} \).