Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
Let $\vec{u}, \vec{v}, \vec{w}$ be such that $|\vec{u}| = 1$, $|\vec{v}| = 2$, $|\vec{w}| = 3$. If the projection $\vec{v}$ along $\vec{u}$ is equal to that of $\vec{w}$ along $\vec{u}$ and $\vec{v}, \vec{w}$ are perpendicular to each other, then $\frac{|\vec{u} - \vec{v}|^2}{2}$ equals _______.
Step-by-Step Solution
Key Concept: Orthogonal vectors have zero dot products; use this to simplify expansion of squared magnitudes.
Given $\vec{v} \cdot \vec{u} = 0$, $\vec{u} \cdot \vec{v} \cdot \vec{w} = 0$, we establish orthogonality conditions. From $|\vec{u} - \vec{v} + \vec{w}|^2 = 14$, expanding the squared magnitude using orthogonality yields $|\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 = 14$ after simplification.
Correct Answer: I need to find $\frac{|\vec{u} - \vec{v}|^2}{2}$ given the conditions.
**Given Information:**
- $|\vec{u}| = 1$, $|\vec{v}| = 2$, $|\vec{w}| = 3$
- Projection of $\