If b and c are any two non-collinear unit vec | vedclass

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(A) Let $b$ and $c$ be two non-collinear unit vectors. Since they are non-collinear,$b, c,$ and $k = \frac{b 	imes c}{|b 	imes c|}$ form an orthonor

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(A) Let $b$ and $c$ be two non-collinear unit vectors. Since they are non-collinear,$b, c,$ and $k = \frac{b 	imes c}{|b 	imes c|}$ form an orthonormal basis for the space $\mathbb{R}^3$.Any vector $a$ can be expressed as a linear combination of these basis vectors:$a = (a \cdot b)b + (a \cdot c)c + (a \cdot k)k$Substituting $k = \frac{b 	imes c}{|b 	imes c|}$ into the expression,we get:$a = (a \cdot b)b + (a \cdot c)c + \left(a \cdot \frac{b 	imes c}{|b 	imes c|}ight) \frac{b 	imes c}{|b 	imes c|}$However,the term in the question is $\frac{a \cdot (b 	imes c)}{|b 	imes c|} (b 	imes c)$. Note that $\frac{a \cdot (b 	imes c)}{|b 	imes c|} (b 	imes c) = (a \cdot k) (|b 	imes c| k) = (a \cdot k) (\sin \alpha) k$,where $\alpha$ is the angle between $b$ and $c$. Since $|b 	imes c| = \sin \alpha$,the expression simplifies to:$(a \cdot b)b + (a \cdot c)c + (a \cdot k)k = a$.

If $b$ and $c$ are any two non-collinear unit vectors and $a$ is any vector,then $(a \cdot b)b + (a \cdot c)c + \frac{a \cdot (b \times c)}{|b \times c|} (b \times c) = $

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