Suppose $a,b$ and $c$ are the components of a vector P along three non-coplanar unit vectors A, B and C respectively. Then, is the below equality true ?
$$\textbf{P}=a\textbf{A}+b\textbf{B}+c\textbf{C}$$
I know that it's true if :
$$\textbf{A}=\hat{i}$$
$$\textbf{B}=\hat{j}$$
$$\textbf{C}=\hat{k}$$
where $\hat{i},\hat{j},\hat{k}$ are unit vectors along $x,y$ and $z$ axis respectively. But I am not sure if this holds for any general case (like what if these unit vectors are not perpendicular etc. etc. ).
Context:
My textbook assumed the above equality true in general to solve a question which was to prove that :
$$\vec{p}\times\vec{q}=\vec{a}[a\space p\space q] +\vec{b}[b\space p\space q] +\vec{c}[c\space p \space q]. $$
where $\vec{a},\vec{b}$ and $\vec{c}$ are any three coplanar unit vectors and $[x\space y\space z]$ denotes the scalar triple product.
Here's the original :
Best Answer
Your intuition is correct: although any three non-coplanar vectors can form a basis for a three-dimensional vector space (meaning you can always write $P = aA + bB + cC$ where $A,B,C$ are the three basis vectors and $P$ is any other vector), in general, projection onto the basis vectors gives the components of a vector (the numbers $a,b,c$) only in the case of an orthonormal basis, not when the basis is any arbitrary set of non-coplanar vectors.
The argument in the book is wrong and it is not hard to find a counterexample to the "fact" that they claim to have proved.
Here is a counterexample: \begin{align} \vec a &= (1, 0, 0), \\ \vec b &= \left(\tfrac1{\sqrt2}, \tfrac1{\sqrt2}, 0\right), \\ \vec c &= \left(\tfrac1{\sqrt2}, 0, \tfrac1{\sqrt2}\right), \\ \vec p &= (2, 3, 4), \\ \vec q &= (-1, 2, 1), \\ [\vec a,\vec p,\vec q] &= -5, \\ [\vec b,\vec p,\vec q] &= -\tfrac{11}{\sqrt2}, \\ [\vec c,\vec p,\vec q] &= \sqrt2, \\ [\vec a,\vec p,\vec q] \vec a + [\vec b,\vec p,\vec q] \vec b + [\vec c,\vec p,\vec q] \vec c &= -5(1, 0, 0) + \left(-\tfrac{11}{\sqrt2}\right) \left(\tfrac1{\sqrt2}, \tfrac1{\sqrt2}, 0\right) + \sqrt2 \left(\tfrac1{\sqrt2}, 0, \tfrac1{\sqrt2}\right) \\ &= (-5, 0, 0) + \left(-\tfrac{11}{2}, -\tfrac{11}{2}, 0\right) + (1, 0, 1) \\ &= \left(-\tfrac{19}{2}, -\tfrac{11}{2}, 1\right), \end{align}
but $$ \vec p \times \vec q = (-5, -6, 7). $$
They claimed the formula always works and here is an example where it clearly doesn't.