Question

If \( \vec{a}, \vec{b}, \vec{c} \) are three vectors such that \( a \neq 0 \) and \( \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}) \), \( |\vec{a}| = |\vec{c}| = 1 \), \( |\vec{b}| = 4 \) and \( |\vec{b} \times \vec{c}| = \sqrt{15} \), if \( \vec{b} - 2\vec{c} = \lambda \vec{a} \) then \( \lambda^2 \) equals:

• A. \( -4 \)
• B. 16
• C. 1
• D. 4

Hint:

Use vector identities like \( |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 \) to switch between dot and cross products easily.

Answer 1

The Correct Option is B

Step 1: Use given cross product relation.
Given:
\[ \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}). \]
This implies:
\[ \vec{a} \times (\vec{b} - 2\vec{c}) = 0. \]

Step 2: Interpret the result.
If cross product is zero, vectors are parallel. Hence:
\[ \vec{b} - 2\vec{c} = \lambda \vec{a}. \]

Step 3: Take magnitude on both sides.
\[ |\vec{b} - 2\vec{c}| = |\lambda||\vec{a}|. \]
Given \( |\vec{a}| = 1 \), so:
\[ |\vec{b} - 2\vec{c}| = |\lambda|. \]

Step 4: Expand magnitude squared.
\[ |\vec{b} - 2\vec{c}|^2 = |\vec{b}|^2 + 4|\vec{c}|^2 - 4(\vec{b} \cdot \vec{c}). \]

Step 5: Use given values.
\[ |\vec{b}| = 4,\quad |\vec{c}| = 1. \]
So:
\[ |\vec{b} - 2\vec{c}|^2 = 16 + 4 - 4(\vec{b} \cdot \vec{c}). \]

Step 6: Find \( \vec{b} \cdot \vec{c} \) using cross product.
We use identity:
\[ |\vec{b} \times \vec{c}|^2 = |\vec{b}|^2 |\vec{c}|^2 - (\vec{b} \cdot \vec{c})^2. \]
\[ 15 = 16(1) - (\vec{b} \cdot \vec{c})^2. \]
\[ (\vec{b} \cdot \vec{c})^2 = 1 \Rightarrow \vec{b} \cdot \vec{c} = \pm 1. \]

Step 7: Compute \( \lambda^2 \).
Take \( \vec{b} \cdot \vec{c} = 1 \):
\[ |\vec{b} - 2\vec{c}|^2 = 16 + 4 - 4(1) = 16. \]
Thus:
\[ \lambda^2 = 16. \]
Final Answer:
\[ \boxed{16}. \]