Question

Let \( \vec{a}, \vec{b}, \vec{c} \) be three vectors such that \[ \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}). \] If \( |\vec{a}| = 1 \), \( |\vec{b}| = 4 \), \( |\vec{c}| = 2 \), and the angle between \( \vec{b} \) and \( \vec{c} \) is \(60^\circ\), then \( |\vec{a} \cdot \vec{c}| \) is equal to.

• A. 4
• B. 2
• C. 0
• D. 1

Hint:

$|\vec{u}-\vec{v}|^2 = u^2 + v^2 - 2\vec{u}\cdot\vec{v}$ is a fundamental vector identity.

Answer 1

The Correct Option is A

We are given vectors \( \vec{a}, \vec{b}, \vec{c} \) with the following properties:

• \( \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}) \)
• \( |\vec{a}| = 1 \)
• \( |\vec{b}| = 4 \)
• \( |\vec{c}| = 2 \)
• The angle between \( \vec{b} \) and \( \vec{c} \) is \( 60^\circ \)

We are required to find:

\[ |\vec{b} - 2\vec{c}| \]

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1. From the given condition

   \[ \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}), \]

   both cross products are in the same direction. Hence, their magnitudes satisfy:

   \[ |\vec{a} \times \vec{b}| = 2|\vec{a} \times \vec{c}|. \]

2. The magnitude of a cross product is given by:

   \[ |\vec{u} \times \vec{v}| = |\vec{u}|\,|\vec{v}|\,\sin\theta, \]

   where \( \theta \) is the angle between the vectors.

3. Applying this formula:

   • \[ |\vec{a} \times \vec{b}| = |\vec{a}|\,|\vec{b}|\,\sin\theta = 1 \times 4 \sin\theta = 4\sin\theta \]
   • \[ |\vec{a} \times \vec{c}| = |\vec{a}|\,|\vec{c}|\,\sin\theta = 1 \times 2 \sin\theta = 2\sin\theta \]

4. Hence,

   \[ 4\sin\theta = 2(2\sin\theta), \]

   which is consistent with the given magnitudes.

5. Now compute \( |\vec{b} - 2\vec{c}| \).

6. Use the vector identity:

   \[ |\vec{u} - \vec{v}| = \sqrt{|\vec{u}|^2 + |\vec{v}|^2 - 2|\vec{u}||\vec{v}|\cos\theta}. \]

7. Substituting the given values:

   \[ |\vec{b} - 2\vec{c}| = \sqrt{4^2 + (2 \times 2)^2 - 2 \times 4 \times 4 \times \cos 60^\circ}. \]

8. Simplifying:

   \[ \begin{aligned} |\vec{b} - 2\vec{c}| &= \sqrt{16 + 16 - 2 \times 4 \times 4 \times \tfrac{1}{2}} \\ &= \sqrt{16 + 16 - 16} \\ &= \sqrt{16} \\ &= 4. \end{aligned} \]

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Final Answer:

\(\boxed{4}\)