Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
Let $\vec{\lambda} = \vec{a} \times (\vec{b} + \vec{c})$, $\vec{\mu} = \vec{b} \times (\vec{c} + \vec{a})$ and $\vec{v} = \vec{c} \times (\vec{a} + \vec{b})$. Then:
\vec{\lambda} + \vec{\mu} = \vec{v}
\vec{\lambda}, \vec{\mu} and \vec{v} are coplanar
\vec{\lambda} + \vec{\mu} + \vec{v} = \vec{0}
\vec{\lambda} + \vec{v} = \vec{\mu}
Step-by-Step Solution
Key Concept: The sum $\vec{\lambda} + \vec{\mu} + \vec{v}$ simplifies to zero due to the antisymmetric property of cross products, implying coplanarity.
Expand each vector product using the distributive property: $\vec{\lambda} = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$, $\vec{\mu} = \vec{b} \times \vec{c} + \vec{b} \times \vec{a}$, and $\vec{v} = \vec{c} \times \vec{a} + \vec{c} \times \vec{b}$. Adding these three expressions: $\vec{\lambda} + \vec{\mu} + \vec{v} = (\vec{a} \times \vec{b} + \vec{b} \times \vec{a}) + (\vec{b} \times \vec{c} + \vec{c} \times \vec{b}) + (\vec{c} \times \vec{a} + \vec{a} \times \vec{c}) = \vec{0}$ since $\vec{p} \times \vec{q} = -\vec{q} \times \vec{p}$. This shows option 3 is correct. Since $\vec{\lambda} + \vec{\mu} + \vec{v} = \vec{0}$, we have $\vec{v} = -(\vec{\lambda} + \vec{\mu})$, meaning $\vec{v}$ is a linear combination of $\vec{\lambda}$ and $\vec{\mu}$, so all three vectors are coplanar (option 2).
Correct Answer: 2,3