Question

Let \( \bar{a}, \bar{b}, \bar{c} \) be three vectors such that \( \bar{a} + \bar{b} + \bar{c} = \bar{0}, |\bar{a}| = 3, |\bar{b}| = 4, |\bar{c}| = 5 \), then \( \bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a} = \)}

• A. 25
• B. -25
• C. 50
• D. -50

Hint:

If the sum of vectors is zero, the sum of their dot products is $-\frac{1}{2}\sum |\bar{v}|^2$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We use the square of the sum of three vectors identity.
Step 2: Key Formula or Approach:
\( |\bar{a} + \bar{b} + \bar{c}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) \).
Step 3: Detailed Explanation:
Given \( \bar{a} + \bar{b} + \bar{c} = \bar{0} \), so \( |\bar{a} + \bar{b} + \bar{c}|^2 = 0 \).
\[ 0 = 3^2 + 4^2 + 5^2 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) \] \[ 0 = 9 + 16 + 25 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) \] \[ 0 = 50 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) \] \[ 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = -50 \] \[ \bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a} = -25 \] Step 4: Final Answer:
The value is -25.