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I had a question in the final exam that asked what the angle between vectors a and b is if: $$\vec a \cdot \vec b=|\vec a\times\vec b| $$

Any hints please.

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If $\vec a$ and $\vec b$ are 3-dimensional vectors, then $\vec a\times\vec b$ is a 3-dimensional vector. $|\vec a\times\vec b|$ is the norm of a vector, and it is a number. Then what does it mean by determinant of a number? Determinant of a $1\times 1$ matrix? – Paul Jun 28 '12 at 20:23
Oops. Actually, there should not be a determinant symbol. – Koba Jun 28 '12 at 20:45
up vote 1 down vote accepted

It is not hard to show that $$\|u\times v\| = \|u\|\|v\| \sin(\theta),$$ where $\theta\in[0,\pi]$ is the angle between $u$ and $v$. This should lead you right to a solution.

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Ok, so the angle is $\pi /2$ – Koba Jun 28 '12 at 21:03
@Dostre, are you sure? The angle between $(1,0,0)$ and $(0,1,0)$ is $\pi/2$... – Rahul Jun 28 '12 at 21:26
I do not think the cross product of two perpendicular vectors is $0$ ... – Neal Jun 28 '12 at 21:27

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