Vector Algebra
Relationship between dot product and cross product magnitudes
nta_pyq_2023_jan
Grade 12
Question:
Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=\sqrt{14}$, $|\vec{b}|=\sqrt{6}$ and $|\vec{a}\times\vec{b}|=\sqrt{48}$. Then $(\vec{a}\cdot\vec{b})^2$ is equal to _____.
Step-by-Step Solution
Key Concept: Use the identity $|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$.
$(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 - |\vec{a}\times\vec{b}|^2 = 84-48=36$. Answer: 36
Correct Answer: 36