alleryThe angle $\theta$ between two planes with normal vectors $\vec{n}_1$ and $\vec{n}_2$ satisfies
A$\cos\theta=\dfrac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$
B$\cos\theta=\dfrac{\vec{n}_1\times\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$
C$\sin\theta=\dfrac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$
D$\tan\theta=\dfrac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$
Answer & Solution
Correct answer: A. $\cos\theta=\dfrac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$
The angle between two planes is equal to the angle between their normal vectors.
**Therefore $\cos\theta=\dfrac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$.**
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