If \(\vec{a}\) and \(\vec{b}\) are non-zero vectors such that \(|\vec{a}+\vec{b}| = |\vec{a}|\), then which of the following vectors is perpendicular to \(\vec{b}\)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2\vec{a}+\vec{b}\)
Squaring the given condition and using \(|\vec{v}|^2 = \vec{v}\cdot\vec{v}\):
\(|\vec{a}+\vec{b}|^2 = |\vec{a}|^2\)
\(|\vec{a}|^2+2\,\vec{a}\cdot\vec{b}+|\vec{b}|^2 = |\vec{a}|^2\)
\(2\,\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{b} = 0\)
By the distributive law of the scalar product, the left side is one product:
\((2\vec{a}+\vec{b})\cdot\vec{b} = 0\)
A zero scalar product of non-zero vectors means they are perpendicular, so \(2\vec{a}+\vec{b}\) is perpendicular to \(\vec{b}\).
Why the other options are incorrect
Here \(\displaystyle\vec{a}\cdot\vec{b} = -\frac{1}{2}|\vec{b}|^2\).
Option (1) \(\vec{a}+2\vec{b}\) — Incorrect. \((\vec{a}+2\vec{b})\cdot\vec{b} = -\frac{1}{2}|\vec{b}|^2+2|\vec{b}|^2 = \frac{3}{2}|\vec{b}|^2 \ne 0\).
Option (3) \(\vec{a}+\vec{b}\) — Incorrect. \((\vec{a}+\vec{b})\cdot\vec{b} = -\frac{1}{2}|\vec{b}|^2+|\vec{b}|^2 = \frac{1}{2}|\vec{b}|^2 \ne 0\).
Option (4) \(\vec{a}-\vec{b}\) — Incorrect. \((\vec{a}-\vec{b})\cdot\vec{b} = -\frac{1}{2}|\vec{b}|^2-|\vec{b}|^2 = -\frac{3}{2}|\vec{b}|^2 \ne 0\).