Question

Let \( \vec{a} \) and \( \vec{b} \) be two co-initial vectors forming adjacent sides of a parallelogram such that:
\[ |\vec{a}| = 10, \quad |\vec{b}| = 2, \quad \vec{a} \cdot \vec{b} = 12 \] Find the area of the parallelogram.

Answer 1

The area of a parallelogram formed by vectors \( \vec{a} \) and \( \vec{b} \) is given by: \[ \text{Area} = |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \] We are given: \[ |\vec{a}| = 10, \quad |\vec{b}| = 2, \quad \vec{a} \cdot \vec{b} = 12 \] Recall that: \[ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \Rightarrow 12 = 10 \cdot 2 \cdot \cos\theta = 20 \cos\theta \Rightarrow \cos\theta = \frac{12}{20} = \frac{3}{5} \] Now, compute \( \sin\theta \) using the identity: \[ \sin^2\theta = 1 - \cos^2\theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \Rightarrow \sin\theta = \frac{4}{5} \] Now compute the area: \[ \text{Area} = 10 \cdot 2 \cdot \frac{4}{5} = 20 \cdot \frac{4}{5} = \boxed{16} \]

Final Answer:

\[ \boxed{16 \text{ square units}} \]

Answer 2

Given: \( |\vec a| = 10,\; |\vec b| = 2,\; \vec a \cdot \vec b = 12 \).

Method 1 (using angle):
\[ \vec a \cdot \vec b = |\vec a||\vec b|\cos\theta \;\Rightarrow\; \cos\theta = \frac{12}{10\cdot 2} = \frac{3}{5}. \] \[ \sin\theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\frac{9}{25}} = \frac{4}{5}. \] Area of parallelogram \(= |\vec a||\vec b|\sin\theta = 10\cdot 2 \cdot \frac{4}{5} = 16. \]

Method 2 (direct vector formula):
\[ |\vec a \times \vec b|^2 = |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2 = (10^2)(2^2) - 12^2 = 400 - 144 = 256. \] \[ |\vec a \times \vec b| = \sqrt{256} = 16. \]

Area of the parallelogram: \( \boxed{16} \) (square units).