Discussion
Let \(\vec{a}, \vec{b}, \vec{c}\) be such that If \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\) .If \(|\vec{a}| = 3\) ,\(|\vec{b}| = 4\) ,\(|\vec{c}| = 5\) then \(|\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}| =\)
Answer: C
The value is 25.
Square the condition \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\) by taking its dot product with itself:
\[ \begin{aligned} |\vec{a}|^{2} + |\vec{b}|^{2} + |\vec{c}|^{2} + 2S &= 0 \\ 9 + 16 + 25 + 2S &= 0 \\ S &= -25 \end{aligned} \]
where \(S = \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}\). Its absolute value is 25.
50 is the sum of squares before halving, which is not offered; 20, 24, 30 and 35 do not come from this identity.
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