Vector Algebra
Vector Algebra
nta_abhyas_2025
Grade None
Question:
Let $\vec{u}$, $\vec{v}$ and $\vec{w}$ are vectors such that $\vec{u} + \vec{v} + \vec{w} = \vec{0}$. If $|\vec{u}| = 3$, $|\vec{v}| = 4$ and $|\vec{w}| = 5$, then the value of $\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}$ is
Step-by-Step Solution
Key Concept: When three vectors sum to zero, squaring the vector equation yields a relationship between the magnitudes and pairwise dot products.
Given $\vec{u} + \vec{v} + \vec{w} = \vec{0}$ with $|\vec{u}| = 3$, $|\vec{v}| = 4$, and $|\vec{w}| = 5$. Squaring the constraint equation: $(\vec{u} + \vec{v} + \vec{w})^2 = 0 \Rightarrow |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) = 0$. This gives $9 + 16 + 25 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) = 0$, so $\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u} = -25$.
Correct Answer: A