Question

Let \( \bar{a} \) be a vector perpendicular to the plane containing non-zero vectors \( \bar{b} \) and \( \bar{c} \). If \( \bar{a}, \bar{b}, \bar{c} \) are such that

\[ |\bar{a} + \bar{b} + \bar{c}| = \sqrt{|\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2}, \]

then

\[ |(\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{c})| = \]

• A. \( |\bar{a}| + |\bar{b}| + |\bar{c}| \)
• B. \( |\bar{a}| |\bar{b}| |\bar{c}| \)
• C. \( |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 \)
• D. (4) \( |\bar{a}|^2 |\bar{c}|^2 \)

Hint:

When dealing with perpendicular vectors, use the identity \[ (\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{c}) = |\bar{a}|^2 (\bar{b} \cdot \bar{c}) - (\bar{a} \cdot \bar{c}) (\bar{b} \cdot \bar{a}). \] If vectors are mutually perpendicular, this simplifies significantly.

Answer 1

The Correct Option is B

Step 1: Understanding the Given Condition 
We are given: \[ |\bar{a} + \bar{b} + \bar{c}| = \sqrt{|\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2}. \] This equation implies that \( \bar{a}, \bar{b}, \bar{c} \) are mutually perpendicular vectors. That is, \[ \bar{a} \cdot \bar{b} = 0, \quad \bar{b} \cdot \bar{c} = 0, \quad \bar{c} \cdot \bar{a} = 0. \] 

Step 2: Evaluating \( |(\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{c})| \) 
Using the vector identity: \[ (\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{c}) = (\bar{a} \cdot \bar{a}) (\bar{b} \cdot \bar{c}) - (\bar{a} \cdot \bar{c}) (\bar{b} \cdot \bar{a}). \] Since \( \bar{b} \) and \( \bar{c} \) are perpendicular to \( \bar{a} \), we know: \[ \bar{b} \cdot \bar{a} = 0, \quad \bar{b} \cdot \bar{c} = 0, \quad \bar{a} \cdot \bar{c} = 0. \] Thus, the equation simplifies to: \[ |(\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{c})| = |\bar{a}| |\bar{b}| |\bar{c}|. \] 

Step 3: Conclusion 
Since this matches option (2), the correct answer is: \[ \boxed{|\bar{a}| |\bar{b}| |\bar{c}|}. \]