Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
Given that the vectors $\vec{a}, \vec{b}$ and $\vec{c}$ (no two of them are collinear). Further if $(\vec{a} + \vec{b})$ is collinear with $\vec{c}, (\vec{b} + \vec{c})$ is collinear with $\vec{a}$ and $|\vec{a}| = |\vec{b}| = |\vec{c}| = \sqrt{2}$. Then the value of $|\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}|$ is ______.
Step-by-Step Solution
Key Concept: Linear dependence of vectors combined with dot product conditions yields constraints on their mutual dot products.
Given $\vec{a} + \vec{b} = \lambda\vec{c}$ and $\vec{b} + \vec{c} = \mu\vec{a}$, subtracting gives $(\vec{a} - \vec{c}) = \lambda\vec{c} - \mu\vec{a}$, so $\vec{a}(1 + \mu) = \vec{c}(\lambda + 1)$. This implies $\mu = -1, \lambda = -1$, giving $\vec{a} + \vec{b} + \vec{c} = 0$. From $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$ with $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -3$.
Correct Answer: 3