Dot Product

Step 1: Establish Basic Geometric Relations

We are given that $$\vec{a}$$, $$\vec{b}$$, and $$\vec{c}$$ are unit vectors:

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

Since the angle between any pair of these vectors is $$\frac{\pi}{3}$$, their mutual dot products are:

$$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 1 \times 1 \times \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$

Step 2: Set Up the System of Linear Equations

We take the dot product of the vector expansion $$\vec{v} = x\vec{a} + y\vec{b} + z\vec{c}$$ with each of the basis vectors $$\vec{a}$$, $$\vec{b}$$, and $$\vec{c}$$:

1. Dot product with $$\vec{a}$$:$$\vec{v} \cdot \vec{a} = x(\vec{a} \cdot \vec{a}) + y(\vec{b} \cdot \vec{a}) + z(\vec{c} \cdot \vec{a})$$  $$1 = x(1) + y\left(\frac{1}{2}\right) + z\left(\frac{1}{2}\right) \implies 2x + y + z = 2 \quad \text{--- (Equation 1)}$$
2. Dot product with $$\vec{b}$$:$$\vec{v} \cdot \vec{b} = x(\vec{a} \cdot \vec{b}) + y(\vec{b} \cdot \vec{b}) + z(\vec{c} \cdot \vec{b})$$  $$2 = x\left(\frac{1}{2}\right) + y(1) + z\left(\frac{1}{2}\right) \implies x + 2y + z = 4 \quad \text{--- (Equation 2)}$$
3. Dot product with $$\vec{c}$$:$$\vec{v} \cdot \vec{c} = x(\vec{a} \cdot \vec{c}) + y(\vec{b} \cdot \vec{c}) + z(\vec{c} \cdot \vec{c})$$  $$3 = x\left(\frac{1}{2}\right) + y\left(\frac{1}{2}\right) + z(1) \implies x + y + 2z = 6 \quad \text{--- (Equation 3)}$$

Step 3: Solve for Components $$x$$, $$y$$, and $$z$$

Add Equation 1, Equation 2, and Equation 3 together:

$$(2x + y + z) + (x + 2y + z) + (x + y + 2z) = 2 + 4 + 6$$  $$4x + 4y + 4z = 12 \implies x + y + z = 3 \quad \text{--- (Equation 4)}$$

Now, subtract Equation 4 individually from each of the first three equations to isolate the variables:

• Finding $$x$$: $$\text{(Equation 1)} - \text{(Equation 4)} \implies x = 2 - 3 = -1$$
• Finding $$y$$: $$\text{(Equation 2)} - \text{(Equation 4)} \implies y = 4 - 3 = 1$$
• Finding $$z$$: $$\text{(Equation 3)} - \text{(Equation 4)} \implies z = 6 - 3 = 3$$

Step 4: Compute the Squared Magnitude $$|\vec{v}|^2$$

Instead of fully expanding $$|\vec{v}|^2 = (x\vec{a} + y\vec{b} + z\vec{c}) \cdot (x\vec{a} + y\vec{b} + z\vec{c})$$, we can use the linear breakdown trick via dot products:

$$|\vec{v}|^2 = \vec{v} \cdot \vec{v} = \vec{v} \cdot (x\vec{a} + y\vec{b} + z\vec{c})$$  $$|\vec{v}|^2 = x(\vec{v} \cdot \vec{a}) + y(\vec{v} \cdot \vec{b}) + z(\vec{v} \cdot \vec{c})$$

Substitute our computed values of $$x, y, z$$ along with the given scalar dot product parameters:

$$|\vec{v}|^2 = (-1)(1) + (1)(2) + (3)(3)$$  $$|\vec{v}|^2 = -1 + 2 + 9 = 10$$

Correct Option: B