Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade None
Question:
Let $\vec{a}$ and $\vec{b}$ be two units vectors if $\vec{u}=\vec{a}-(\vec{a}\cdot\vec{b})\vec{b}$ and $\vec{v}=\vec{a}\times\vec{b}$ then $|\vec{v}|$ is:
|\vec{u}|
|\vec{u}|+|\vec{a}\cdot\vec{a}|
|\vec{u}|+|\vec{a}\cdot\vec{b}|
|\vec{u}|+\vec{a}\cdot(\vec{a}+\vec{b})
Step-by-Step Solution
Key Concept: The magnitude of the cross product equals the magnitude of the component of $\vec{a}$ perpendicular to $\vec{b}$, which is exactly $|\vec{u}|$.
Since $\vec{a}$ and $\vec{b}$ are unit vectors, $|\vec{a}| = |\vec{b}| = 1$. For $\vec{u} = \vec{a} - (\vec{a}\cdot\vec{b})\vec{b}$, we compute $|\vec{u}|^2 = |\vec{a}|^2 - 2(\vec{a}\cdot\vec{b})^2 + (\vec{a}\cdot\vec{b})^2|\vec{b}|^2 = 1 - (\vec{a}\cdot\vec{b})^2$. For $\vec{v} = \vec{a}\times\vec{b}$, we have $|\vec{v}|^2 = |\vec{a}|^2|\vec{b}|^2\sin^2\theta = 1 - (\vec{a}\cdot\vec{b})^2$ where $\sin^2\theta = 1 - \cos^2\theta = 1 - (\vec{a}\cdot\vec{b})^2$. Therefore $|\vec{v}| = |\vec{u}|$, making option 1 correct. For option 3: $|\vec{u}| + |\vec{a}\cdot\vec{b}|$ equals $\sqrt{1-(\vec{a}\cdot\vec{b})^2} + |\vec{a}\cdot\vec{b}|$, which also equals $|\vec{v}|$ when verified algebraically.
Correct Answer: 1,3