If \( \vec{a} \) and \( \vec{b} \) are two non-zero vectors such that \( |2\vec{a}+\vec{b}|=|\vec{b}| \) and \( \vec{a}+\vec{b}\ne\vec{0} \), then the angle between \( \vec{a}+\vec{b} \) and \( \vec{a} \), is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 90^{\circ} \)
Squaring both sides of \( |2\vec{a}+\vec{b}|=|\vec{b}| \),
\( \displaystyle 4|\vec{a}|^{2}+4\,\vec{a}\cdot\vec{b}+|\vec{b}|^{2}=|\vec{b}|^{2} \)
\( \displaystyle \vec{a}\cdot\vec{b}=-|\vec{a}|^{2} \)
So the dot product of \( \vec{a}+\vec{b} \) with \( \vec{a} \) is
\( \displaystyle (\vec{a}+\vec{b})\cdot\vec{a}=|\vec{a}|^{2}+\vec{a}\cdot\vec{b} \)
\( \displaystyle =0 \)
Both vectors are non-zero, so they are perpendicular and the angle is \( 90^{\circ} \).