Question

If \( |\vec{a}| = 8 \), \( |\vec{b}| = 3 \) and \( |\vec{a} \times \vec{b}| = 12 \), then the value of \( |\vec{a} \cdot \vec{b}| \) is

• A. \( 6\sqrt{3} \)
• B. \( 8\sqrt{3} \)
• C. \( 12\sqrt{3} \)
• D. \( 3\sqrt{12} \)

Hint:

Tip 1: Lagrange's Identity is the most efficient way to solve problems involving both dot and cross products.
Tip 2: Always look for perfect square factors like 144 when simplifying square roots in competitive exams.

Answer 1

The Correct Option is C

Concept:
• Lagrange's Identity: For any two vectors \( \vec{a} \) and \( \vec{b} \), the relationship between their cross product and dot product is given by: \[ |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \]
• Alternatively, using the definitions: \[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \quad \text{and} \quad \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]

Step 1: List the given magnitudes
We are provided with the following values: \[ |\vec{a}| = 8 \] \[ |\vec{b}| = 3 \] \[ |\vec{a} \times \vec{b}| = 12 \]

Step 2: Apply Lagrange's Identity to relate the products
Using the identity: \[ (12)^2 + (\vec{a} \cdot \vec{b})^2 = (8)^2 \times (3)^2 \] \[ 144 + (\vec{a} \cdot \vec{b})^2 = 64 \times 9 \]

Step 3: Calculate the square of the dot product
\[ 144 + (\vec{a} \cdot \vec{b})^2 = 576 \] Subtract 144 from both sides: \[ (\vec{a} \cdot \vec{b})^2 = 576 - 144 \] \[ (\vec{a} \cdot \vec{b})^2 = 432 \]

Step 4: Find the magnitude of the dot product
To find \( |\vec{a} \cdot \vec{b}| \), take the square root of 432: \[ |\vec{a} \cdot \vec{b}| = \sqrt{432} \] Factorize 432 to simplify the radical: \[ |\vec{a} \cdot \vec{b}| = \sqrt{144 \times 3} \] \[ |\vec{a} \cdot \vec{b}| = 12\sqrt{3} \]