Vector Algebra
Magnitude of Cross Product of Linear Combinations
nta_pyq_2023_apr
Grade 12

Question:

$|\vec{a}|=2$, $|\vec{b}|=3$, angle between $\vec{a},\vec{b}=\frac{\pi}{4}$. Then $|(\vec{a}+2\vec{b})\times(2\vec{a}-3\vec{b})|^2$ is equal to
441
482
841
882

Step-by-Step Solution

Key Concept: $(\vec{a}+2\vec{b})\times(2\vec{a}-3\vec{b})=-3(\vec{a}\times\vec{b})+4(\vec{b}\times\vec{a})=-7(\vec{a}\times\vec{b})$. Wait: expand: $2(\vec{a}\times\vec{a})-3(\vec{a}\times\vec{b})+4(\vec{b}\times\vec{a})-6(\vec{b}\times\vec{b})=-7(\vec{a}\times\vec{b})$.
$49\times18=882$.
Correct Answer: 4

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