Question

Three vectors \(\vec a,\vec b,\vec c\) satisfy the condition \(\vec a+\vec b+\vec c=0\). If \(|\vec a|=3,|\vec b|=4\) and \(|\vec c|=2\), then find the value of \(\mu=\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a\).

Hint:

Square the condition a+b+c=0 and expand using the dot product.

Answer 1

Step 1: Isolating one vector and dotting:
From \(\vec a+\vec b+\vec c=0\), write \(\vec c=-(\vec a+\vec b)\), so \(|\vec c|^2=|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2\), giving \(4=9+2\vec a\cdot\vec b+16\Rightarrow \vec a\cdot\vec b=-\dfrac{21}{2}\).

Step 2: Repeating for b.c using a=-(b+c):
\(|\vec a|^2=|\vec b|^2+2\vec b\cdot\vec c+|\vec c|^2\Rightarrow 9=16+2\vec b\cdot\vec c+4\Rightarrow \vec b\cdot\vec c=-\dfrac{11}{2}\).

Step 3: Repeating for c.a using b=-(c+a):
\(|\vec b|^2=|\vec c|^2+2\vec c\cdot\vec a+|\vec a|^2\Rightarrow16=4+2\vec c\cdot\vec a+9\Rightarrow \vec c\cdot\vec a=\dfrac{3}{2}\).

Step 4: Summing all three:
\(\mu=-\dfrac{21}{2}-\dfrac{11}{2}+\dfrac{3}{2}=-\dfrac{29}{2}\), matching the faster method.

Final Answer:
\[ \boxed{\mu=-\dfrac{29}{2}} \]