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: Given Conditions
$\vec{a} \cdot (\vec{b}+\vec{c}) = 0 \implies \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$.
Similarly, $\vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{c} = 0$ and $\vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0$.
Step 2: Sum of dot products
Adding all three: $2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$.
Step 3: Expansion
$|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\sum \vec{a}\cdot\vec{b})$
$= 5^2 + 12^2 + 13^2 + 0 = 25 + 144 + 169 = 338$.
$|\vec{a}+\vec{b}+\vec{c}| = \sqrt{338}$.
Final Answer: (A)