If \( \vec{a},\vec{b} \) and \( \vec{c} \) are three vectors such that \( |\vec{a}|=\displaystyle\frac{1}{\sqrt{2}} \), \( |\vec{b}|=\displaystyle\frac{1}{\sqrt{3}} \), \( |\vec{c}|=\displaystyle\frac{1}{\sqrt{6}} \), \( |\vec{a}+\vec{b}+\vec{c}|=1 \) and \( \vec{c}=\lambda(\vec{a}\times\vec{b}) \). The angle between \( \vec{a} \) and \( \vec{b} \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{\pi}{2} \)
Since \( \vec{c}=\lambda(\vec{a}\times\vec{b}) \), the vector \( \vec{c} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \), so \( \vec{a}\cdot\vec{c}=\vec{b}\cdot\vec{c}=0 \).
Squaring the given modulus,
\( \displaystyle |\vec{a}+\vec{b}+\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) \)
\( \displaystyle 1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+2\,\vec{a}\cdot\vec{b}+0 \)
\( \displaystyle 1=1+2\,\vec{a}\cdot\vec{b} \)
\( \displaystyle \vec{a}\cdot\vec{b}=0\ \Rightarrow\ \theta=\frac{\pi}{2} \)