Arbitrary vector

**shilz222** · November 26th 2007, 02:57 PM

Let $\bold{A}$ be an arbitrary vector and let $\bold{\hat{n}}$ be a unit vector in some fixed direction. Show that $\bold{A} = (\bold{A} \cdot \bold{\hat{n}})\bold{\hat{n}} + (\bold{\hat{n}} \times \bold{A}) \times \bold{\hat{n}}$ .

For this type of problem would I just multiply everything out in the RHS and simplify?

So $\bold{A} = A \bold{\hat{n}} + 0$ .

So $(\bold{\hat{n}} \times \bold{A}) \times \bold{\hat{n}} = 0$ ?

**Plato** · November 26th 2007, 03:41 PM

Here is a basic fact: $A \times \left( {B \times C} \right) = \left( {A \cdot C} \right)B - \left( {A \cdot B} \right)C$ .
In other words, $A \times \left( {B \times C} \right)$ is a linear combination of B & C.

Now apply that with this:
$\begin{array}{rcl}<br /> \left( {N \times A} \right) \times N & = & - \left[ {N \times \left( {N \times A} \right)} \right] \\ <br /> & = & - \left[ {\left( {N \cdot A} \right)N - \left( {N \cdot N} \right)A} \right] \\ <br /> & = & A - \left( {N \cdot A} \right)N \\ \end{array}<br />$

Can you see how to finish?

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