Q13:

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, find the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.

Step 1: Establishing Given Parameters and Definitions

Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, their magnitudes are defined as:

$|\vec{a}| = 1, |\vec{b}| = 1, |\vec{c}| = 1$

The square of the magnitude of any vector $\vec{v}$ is given by the dot product of the vector with itself [Per the properties of the Scalar Product]:

$|\vec{a}|^2 = \vec{a} \cdot \vec{a} = 1^2 = 1$

$|\vec{b}|^2 = \vec{b} \cdot \vec{b} = 1^2 = 1$

$|\vec{c}|^2 = \vec{c} \cdot \vec{c} = 1^2 = 1$

Step 2: Utilizing the Given Vector Equation

It is given that:

$\vec{a} + \vec{b} + \vec{c} = \vec{0}$

To evaluate the expression $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$, we perform the dot product of the sum vector with itself:

$(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = \vec{0} \cdot \vec{0}$

Step 3: Algebraic Expansion using the Distributive Property

Applying the distributive property of the scalar product [By the Distributive Law of Dot Product over Addition]:

$\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c}) + \vec{b} \cdot (\vec{a} + \vec{b} + \vec{c}) + \vec{c} \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$

$\vec{a}\cdot\vec{a} + \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} + \vec{c}\cdot\vec{b} + \vec{c}\cdot\vec{c} = 0$

Step 4: Simplification using Commutative Property

Since the dot product is commutative, $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}$, $\vec{a}\cdot\vec{c} = \vec{c}\cdot\vec{a}$, and $\vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{b}$ [Per Commutative Law of Dot Product]:

$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$

Step 5: Substitution and Final Calculation

Substitute the known magnitudes calculated in Step 1:

$1 + 1 + 1 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$

$3 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$

$2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = -3$

$\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = -\frac{3}{2}$

Visual Representation of Vectors in Equilibrium

Final Solution: $-\frac{3}{2}$