Question

Let \( \vec{a} \) and \( \vec{b} \) be two vectors such that \( |\vec{b}| = 1 \) and \( |\vec{b} \times \vec{a}| = 2 \). Then \( \left| (\vec{b} \times \vec{a}) - \vec{b} \right|^2 \) is equal to

• A. 3
• B. 5
• C. 1
• D. 4

Answer 1

The Correct Option is B

To solve the problem, we need to find the square of the magnitude of the vector \( (\vec{b} \times \vec{a}) - \vec{b} \). Given that \( |\vec{b}| = 1 \) and \( |\vec{b} \times \vec{a}| = 2 \), let's proceed step by step:

1. Start with the expression whose magnitude we need to find: \((\vec{b} \times \vec{a}) - \vec{b}\)
2. Calculate the magnitude squared, which is given by the dot product of the vector with itself: \(\left| (\vec{b} \times \vec{a}) - \vec{b} \right|^2 = \left[ (\vec{b} \times \vec{a}) - \vec{b} \right] \cdot \left[ (\vec{b} \times \vec{a}) - \vec{b} \right]\)
3. Expand the dot product: \(= (\vec{b} \times \vec{a}) \cdot (\vec{b} \times \vec{a}) - 2(\vec{b} \times \vec{a}) \cdot \vec{b} + \vec{b} \cdot \vec{b}\)
4. We know:
   • \(| \vec{b} \times \vec{a} | = 2\)
   • \(\vec{b} \cdot (\vec{b} \times \vec{a}) = 0\) (This is because the dot product of a vector and a cross product of the same vector with another vector is zero.)
   • \(\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1^2 = 1\)
5. Substitute these values into the expression: \(\left| (\vec{b} \times \vec{a}) - \vec{b} \right|^2 = 2^2 - 2 \times 0 + 1 = 4 + 1 = 5\)

Therefore, the magnitude squared of the vector \( (\vec{b} \times \vec{a}) - \vec{b} \) is 5.

Answer 2

Given \( |\vec{b}| = 1 \) and \( |\vec{b} \times \vec{a}| = 2 \), we need to find \( |(\vec{b} \times \vec{a}) - \vec{b}|^2 \).

Expanding \( |(\vec{b} \times \vec{a}) - \vec{b}|^2 \) using \( |\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2\vec{u} \times \vec{v} \):

\[ |(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2 - 2(\vec{b} \times \vec{a}) \times \vec{b}. \]

Since \( |\vec{b} \times \vec{a}| = 2 \), we get \( |\vec{b} \times \vec{a}|^2 = 4 \), and \( |\vec{b}|^2 = 1 \).

The cross product \( (\vec{b} \times \vec{a}) \times \vec{b} = 0 \) because \( \vec{b} \times \vec{a} \) is perpendicular to \( \vec{b} \).

Substituting these values:

\[ |(\vec{b} \times \vec{a}) - \vec{b}|^2 = 4 + 1 = 5. \]