If \(\vec a\) and \(\vec b\) are two unit vectors and \(\vec a+\vec b\) is also unit vector, the magnitude of \(\vec a-\vec b\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\sqrt{3}\)
Since \(|\vec a| = |\vec b| = |\vec a + \vec b| = 1\), squaring \(|\vec a + \vec b|\) with the dot product gives \(\vec a \cdot \vec b\):
\(|\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2 + 2\,\vec a \cdot \vec b\)
\(1 = 1 + 1 + 2\,\vec a \cdot \vec b\)
\( \displaystyle \vec a \cdot \vec b = -\frac{1}{2} \)
Then
\(|\vec a - \vec b|^2 = |\vec a|^2 + |\vec b|^2 - 2\,\vec a \cdot \vec b\)
\( \displaystyle |\vec a - \vec b|^2 = 1 + 1 - 2\left(-\frac{1}{2}\right) = 3 \)
\(|\vec a - \vec b| = \sqrt{3}\)