Vector Algebra
Magnitude of Difference of Cross Product and Vector
nta_pyq_2024_jan
Grade 12
Question:
Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{b}|=1$ and $|\vec{b}\times\vec{a}|=2$. Then $|(\vec{b}\times\vec{a})-\vec{b}|^2$ is equal to:
Step-by-Step Solution
Key Concept: $|(\vec{b}\times\vec{a})-\vec{b}|^2=|\vec{b}\times\vec{a}|^2+|\vec{b}|^2-2(\vec{b}\times\vec{a})\cdot\vec{b}$. Note $(\vec{b}\times\vec{a})\perp\vec{b}$, so the dot product is 0.
$(\vec{b}\times\vec{a})\cdot\vec{b}=0$. $|(\vec{b}\times\vec{a})-\vec{b}|^2=|\vec{b}\times\vec{a}|^2+|\vec{b}|^2=4+1=5$.
Correct Answer: 2