Let \(\vec{a}, \vec{b}\) be two vectors such that \(|\vec{a}| = 2, |\vec{b}| = 3, \vec{a} \cdot \vec{b} = 4\), then \(|\vec{a} - \vec{b}|\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\sqrt{5}\)
Since \(|\vec{v}|^2 = \vec{v} \cdot \vec{v}\) for any vector, the dot product of \(\vec{a} - \vec{b}\) with itself gives
\(|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})\)
\(= |\vec{a}|^2 - 2\,\vec{a} \cdot \vec{b} + |\vec{b}|^2\)
\(= 4 - 8 + 9 = 5\)
Taking the positive square root:
\(|\vec{a} - \vec{b}| = \sqrt{5}\)
Why the other options are incorrect
Option (1) 0 — Incorrect. \(|\vec{a} - \vec{b}| = 0\) would need \(\vec{a} = \vec{b}\), which is impossible since \(|\vec{a}| \ne |\vec{b}|\).
Option (3) 5 — Incorrect. This is \(|\vec{a} - \vec{b}|^2\). The square root has not been taken.
Option (4) 1 — Incorrect. This is \(|\vec{b}| - |\vec{a}|\). The magnitude of a difference is not the difference of the magnitudes.