Question

If the lengths of three vectors $\bar{a}, \bar{b}$ and $\bar{c}$ are 5, 12, 13 units respectively, and each one is perpendicular to the sum of the other two, then $|\bar{a} + \bar{b} + \bar{c}| = ..............$

• A. $\sqrt{338}$
• B. 169
• C. 338
• D. 676

Hint:

If $\vec{a} \perp (\vec{b}+\vec{c})$, then $|\vec{a}+\vec{b}+\vec{c}|^2 = \sum |\vec{a}|^2$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The perpendicularity condition translates to dot products being zero. We use the expansion of the square of a vector sum to find the magnitude.
Step 2: Key Formula or Approach:
The magnitude of the sum is given by:
\[ |\bar{a} + \bar{b} + \bar{c}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 + 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) \] Given $|\bar{a}| = 5, |\bar{b}| = 12, |\bar{c}| = 13$.
Step 3: Detailed Explanation:
The perpendicularity conditions are:
1. $\bar{a} \cdot (\bar{b} + \bar{c}) = 0 \implies \bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c} = 0$
2. $\bar{b} \cdot (\bar{a} + \bar{c}) = 0 \implies \bar{b} \cdot \bar{a} + \bar{b} \cdot \bar{c} = 0$
3. $\bar{c} \cdot (\bar{a} + \bar{b}) = 0 \implies \bar{c} \cdot \bar{a} + \bar{c} \cdot \bar{b} = 0$
Summing these three equations:
\[ 2(\bar{a} \cdot \bar{b} + \bar{b} \cdot \bar{c} + \bar{c} \cdot \bar{a}) = 0 \] Now, substitute the magnitudes:
\[ |\bar{a} + \bar{b} + \bar{c}|^2 = 5^2 + 12^2 + 13^2 + 0 = 25 + 144 + 169 = 338 \] \[ |\bar{a} + \bar{b} + \bar{c}| = \sqrt{338} \] Step 4: Final Answer:
The magnitude of the sum is $\sqrt{338}$.