The cross product $\vec{a} \times \vec{b}$ of two vectors has magnitude:
A$|\vec{a}||\vec{b}|\cos\theta$
B$|\vec{a}||\vec{b}|\sin\theta$
C$|\vec{a}| + |\vec{b}|$
D$|\vec{a}|^2 - |\vec{b}|^2$
Answer & Solution
Correct answer: B. $|\vec{a}||\vec{b}|\sin\theta$
$|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$. The cross product is a vector perpendicular to both $\vec{a}$ and $\vec{b}$, with direction given by the right-hand rule. Its magnitude equals the area of the parallelogram spanned by $\vec{a}$ and $\vec{b}$.
Geometric reading: when $\theta = 0$ or $\pi$ (parallel or anti-parallel), $\sin\theta = 0$ and the cross product vanishes, which is why parallel vectors have zero cross product.
Related questions
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