Question

Three vectors \(\vec a, \vec b\) and \(\vec c\) satisfy the condition \(\vec a+\vec b+\vec c=0\). If \(|\vec a|=3, |\vec b|=4\) and \(|\vec c|=5\), then find the value of \(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a\).

Hint:

Square both sides of a+b+c=0 using the dot product with itself.

Answer 1

Step 1: Key Formula or Approach:
Since \(\vec a+\vec b+\vec c=0\), taking the dot product of both sides with itself gives \((\vec a+\vec b+\vec c)\cdot(\vec a+\vec b+\vec c) = 0\). Expanding this will bring out exactly the terms we need.

Step 2: Expanding the dot product:
\[ |\vec a|^2+|\vec b|^2+|\vec c|^2 + 2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a) = 0 \]

Step 3: Substituting the magnitudes:
\(|\vec a|^2=9,\ |\vec b|^2=16,\ |\vec c|^2=25\), so:
\[ 9+16+25+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0 \]
\[ 50 + 2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a) = 0 \]

Step 4: Solving for the required sum:
\[ \vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a = \frac{-50}{2} = -25 \]

Final Answer:
The value is \(-25\). \[ \boxed{-25} \]