If \( \vec{a} \) and \( \vec{b} \) are two unit vectors and \( |\vec{a}-\vec{b}|=\sqrt{3} \), then the value of \( |\vec{a}+\vec{b}| \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1
Since \( |\vec{a}|=|\vec{b}|=1 \), squaring the given modulus gives \( \vec{a}\cdot\vec{b} \):
\( \displaystyle |\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2\,\vec{a}\cdot\vec{b} \)
\( \displaystyle 3=1+1-2\,\vec{a}\cdot\vec{b} \)
\( \displaystyle \vec{a}\cdot\vec{b}=-\frac{1}{2} \)
\( \displaystyle |\vec{a}+\vec{b}|^2=1+1+2\left(-\frac{1}{2}\right)=1 \)
\( \displaystyle |\vec{a}+\vec{b}|=1 \)
Shortcut: Parallelogram law: \( |\vec{a}+\vec{b}|^2+|\vec{a}-\vec{b}|^2=2\left(|\vec{a}|^2+|\vec{b}|^2\right)=4 \), so \( |\vec{a}+\vec{b}|^2=4-3=1 \).