For two non-zero vectors \(\vec{a}\) and \(\vec{b}\), the relation \(|\vec{a}+\vec{b}| = |\vec{a}|+|\vec{b}|\) holds if and only if
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\vec{a}\) and \(\vec{b}\) are in the same direction
Let \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\). Using \(|\vec{v}|^2 = \vec{v}\cdot\vec{v}\):
\( |\vec{a}+\vec{b}|^2 = |\vec{a}|^2+2|\vec{a}|\,|\vec{b}|\cos\theta+|\vec{b}|^2 \)
\( \left(|\vec{a}|+|\vec{b}|\right)^2 = |\vec{a}|^2+2|\vec{a}|\,|\vec{b}|+|\vec{b}|^2 \)
The two are equal only when
\( \cos\theta = 1, \quad \text{that is } \theta = 0 \)
So the relation holds exactly when \(\vec{a}\) and \(\vec{b}\) point the same way. In every other case the triangle law gives a real triangle, and one side is shorter than the other two together.
Why the other options are incorrect
Option (1) \(\vec{a}\) and \(\vec{b}\) are perpendicular — Incorrect. Then \(\cos\theta = 0\) and \(|\vec{a}+\vec{b}| = \sqrt{|\vec{a}|^2+|\vec{b}|^2}\), which is less than \(|\vec{a}|+|\vec{b}|\).
Option (3) \(\vec{a}\) and \(\vec{b}\) are in opposite directions — Incorrect. Then \(\cos\theta = -1\) and \(|\vec{a}+\vec{b}|\) is the difference of the two lengths, not their sum.
Option (4) \(|\vec{a}| = |\vec{b}|\) — Incorrect. Equal lengths are not enough: for \(\vec{a} = \hat{\imath}\) and \(\vec{b} = \hat{\jmath}\), \(|\vec{a}+\vec{b}| = \sqrt{2}\) but \(|\vec{a}|+|\vec{b}| = 2\).