Magnitude of Vector Cross Product in a Triangle

In a triangle $PQR$, the vectors representing the sides taken in order satisfy the relation: $\vec{a} + \vec{b} + \vec{c} = \vec{QR} + \vec{RP} + \vec{PQ} = \vec{0}$

From this relation, we can express $\vec{c}$ in terms of $\vec{a}$ and $\vec{b}$: $\vec{c} = -(\vec{a} + \vec{b})$

We are given: $|\vec{a}| = 3 \quad \text{and} \quad |\vec{b}| = 4$

Now, let us simplify the given equation: $\frac{\vec{a} \cdot (\vec{c} - \vec{b})}{\vec{c} \cdot (\vec{a} - \vec{b})} = \frac{|\vec{a}|}{|\vec{a}| + |\vec{b}|}$

First, simplify the numerator of the left-hand side: $\vec{a} \cdot (\vec{c} - \vec{b}) = \vec{a} \cdot (-\vec{a} - \vec{b} - \vec{b}) = \vec{a} \cdot (-\vec{a} - 2\vec{b}) = -|\vec{a}|^2 - 2(\vec{a} \cdot \vec{b})$ Substituting $|\vec{a}| = 3$: $\vec{a} \cdot (\vec{c} - \vec{b}) = -9 - 2(\vec{a} \cdot \vec{b})$

Next, simplify the denominator of the left-hand side: $\vec{c} \cdot (\vec{a} - \vec{b}) = -(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = - (|\vec{a}|^2 - |\vec{b}|^2)$ Substituting $|\vec{a}| = 3$ and $|\vec{b}| = 4$: $\vec{c} \cdot (\vec{a} - \vec{b}) = -(3^2 - 4^2) = -(9 - 16) = 7$

Now, evaluate the right-hand side of the given equation: $\frac{|\vec{a}|}{|\vec{a}| + |\vec{b}|} = \frac{3}{3 + 4} = \frac{3}{7}$

Equating the left-hand side and right-hand side: $\frac{-9 - 2(\vec{a} \cdot \vec{b})}{7} = \frac{3}{7}$

Solving for $\vec{a} \cdot \vec{b}$: $-9 - 2(\vec{a} \cdot \vec{b}) = 3$ $-2(\vec{a} \cdot \vec{b}) = 12 \implies \vec{a} \cdot \vec{b} = -6$

We are required to find the value of $|\vec{a} \times \vec{b}|^2$. Using Lagrange's identity: $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$

Substituting the known values into the identity: $|\vec{a} \times \vec{b}|^2 = (3)^2 \cdot (4)^2 - (-6)^2$ $|\vec{a} \times \vec{b}|^2 = 9 \times 16 - 36$ $|\vec{a} \times \vec{b}|^2 = 144 - 36 = 108$