Question

Let three vectors \(\vec a,\vec b\) and \(\vec c\) be such that \(|\vec a|=3,|\vec b|=4,|\vec c|=5\) and each of them is perpendicular to the sum of the other two vectors, then find the value of \(|\vec a+\vec b+\vec c|\).

Hint:

Each vector ⊥ sum of the other two gives a·b+b·c+c·a = 0; expand |a+b+c|².

Answer 1

Step 1: Recognising the 3-4-5 clue:
Since \(3^2+4^2=5^2\), the given magnitudes hint the three vectors could form a right-angled configuration — but we solve generally with dot products, not by assuming a specific geometry.

Step 2: Using each perpendicularity condition individually:
\(\vec a\cdot(\vec b+\vec c)=0\), \(\vec b\cdot(\vec c+\vec a)=0\), \(\vec c\cdot(\vec a+\vec b)=0\) — each expands to a pairwise dot-product sum equal to zero.

Step 3: Summing to isolate the cross terms:
Adding all three gives \(2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0\), so all pairwise dot products together vanish.

Step 4: Squaring the resultant vector:
\(|\vec a+\vec b+\vec c|^2=\sum|\vec a|^2+2\sum\vec a\cdot\vec b=(9+16+25)+0=50\).

Final Answer:
\[ \boxed{5\sqrt2} \]