Find the vector products of the unit vectors of a…
Find the vector products of the unit vectors of a right-handed coordinate system. Compute $\big[(\vec i\times\vec j)\times\vec i\big]\times\vec i$.
Explanation of Method
Vector products of the coordinate unit vectors and an iterated product.
Step 1: Table of products
$\vec i,\vec j,\vec k$ are mutually perpendicular unit vectors. The same vector gives $\sin0^\circ=0$, different ones give $\sin90^\circ=1$, and the direction follows the right-hand rule.
| $\times$ | $\vec i$ | $\vec j$ | $\vec k$ |
|---|---|---|---|
| $\vec i$ | $\vec 0$ | $\vec k$ | $-\vec j$ |
| $\vec j$ | $-\vec k$ | $\vec 0$ | $\vec i$ |
| $\vec k$ | $\vec j$ | $-\vec i$ | $\vec 0$ |
(Row $\times$ column, e.g. $\vec i\times\vec j=\vec k$.)
Step 2: The iterated product
\[ \vec i\times\vec j=\vec k,\qquad \vec k\times\vec i=\vec j,\qquad \vec j\times\vec i=-\vec k. \]
\[\big[(\vec i\times\vec j)\times\vec i\big]\times\vec i=-\vec k\]
Step 3: Non-associativity
The vector product is not associative, for example
\[ (\vec i\times\vec i)\times\vec j=\vec 0\times\vec j=\vec 0,\qquad \vec i\times(\vec i\times\vec j)=\vec i\times\vec k=-\vec j. \]
Result
\[\boxed{\,\big[(\vec i\times\vec j)\times\vec i\big]\times\vec i=-\vec k\,}\]
Remark
Wording. The source says "clockwise = positive". Whether the cycle looks clockwise depends on how the circle is drawn. The rule is the cyclic order $\vec i\to\vec j\to\vec k\to\vec i$: products taken in that order are $+$, products taken in the reverse order are $-$.
Correction. The source compares $\vec i\times(\vec j\times\vec i)=\vec j$ with the final result $-\vec k$. These are different expressions, so the comparison proves nothing. In fact $\vec i\times(\vec j\times\vec i)=\vec j=(\vec i\times\vec j)\times\vec i$, which is a coincidence of this particular case. The example above shows the non-associativity correctly.