If \(\vec a,\vec b,\vec c\) are vectors such that \(\vec a+\vec b+\vec c=0\) and \(|\vec a| = 1,|\vec b|=2,|\vec c|= 5\), then the expression \(\vec a.\vec b+\vec b.\vec c+\vec c.\vec a\) equals
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → −15
Here 0 is the zero vector. Taking the dot product of the sum with itself:
\(|\vec a + \vec b + \vec c|^2 = 0\)
\(|\vec a|^2 + |\vec b|^2 + |\vec c|^2 + 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a) = 0\)
\(1 + 4 + 25 + 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a) = 0\)
\(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a = -15\)
Note: It seems that there is a mistake in the original NTA paper, which prints \(|\vec c| = 5\) together with \(|\vec a| = 1\) and \(|\vec b| = 2\). Since \(\vec c = -(\vec a + \vec b)\), the triangle inequality gives \(|\vec c| \le |\vec a| + |\vec b| = 3\), so no such vectors exist. The question can still be solved, because the method above uses only the three given magnitudes, and it gives −15, Option (2).