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I have to prove the following properties

  • $(a \times b) \cdot a = (a \times b) \cdot b = 0$
  • $|a \times b|^2 = |a|^2 |b|^2 - (a \cdot b)^2$
  • $(a \times b) \times c = (a \cdot c)b - (b \cdot c)a$

The way I am proving them is by literally saying:

$$a = (A, B, C)$$ $$b = (D, E, F)$$ $$c = (G, H, I)$$

and then just showing both sides of the equals sign are the same. Is this the best way or is there a quicker way?

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Sometimes you just have to sit down and do some unpleasant computations! But, if you can show that cross-product is invariant under rotations, then you can try rotating one of the vectors to be of the form $(A,0,0)$ and that will simplify things a bit. – Zach L. Jan 30 '13 at 16:41

It depends on what you mean is the best way. At times with vectors, there isn't a need to resort to horrendous amounts of computations, if you understand what is happening. For example, to show that 2 vectors $a, b$ are the same, one way of doing that is to show that $a \cdot v = b \cdot v$ for all vectors $v$, in particular the unit vectors $i, j, k$.

Here are some suggestions

  1. Prove by the geometric interpretation of cross and dot product. The cross product produces a vector that is orthogonal to both $a$ and $b$, hence the dot product is 0.

  2. Prove by the trigonometric identity, use Pythagorean Theorem.

  3. Prove by considering the $x, y, z$ coordinates separately.

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In the first one you can argue that $(a*b)$ is perpendicular to both $a$ and $b$ so its dot product with either of them must be zero.

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In the 2nd one u can use the fact that $|a*b|^2 = |a|^2|b|^2(sin \theta)^2$ and similarly for dot product i.e. $|a.b|^2 = |a|^2|b|^2(cos \theta)^2$,$\theta$ being the angle between a,b.

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